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Contents
Physical World, Units and Measurements
P-1 – P-13
1.
Topic 1 : Unit of Physical Quantities
Topic 2 : Dimensions of Physical Quantities
Topic 3 : Errors in Measurements
Motion in a Straight Line
P-14 – P-25
2.
Topic 1 : Distance, Displacement & Uniform Motion
Topic 2 : Non-uniform Motion
Topic 3 : Relative Velocity
Topic 4 : Motion Under Gravity
Motion in a Plane
P-26 – P-35
3.
Topic 1 : Vectors
Topic 2 : Motion in a Plane with Constant Acceleration
Topic 3 : Projectile Motion
Topic 4 : Relative Velocity in Two Dimensions & Uniform Circular Motion
Laws of Motion
P-36 – P-53
4.
Topic 1 : Ist, IInd & IIIrd Laws of Motion
Topic 2 : Motion of Connected Bodies, Pulley & Equilibrium of Forces
Topic 3 : Friction
Topic 4 : Circular Motion, Banking of Road
Work, Energy and Power
P-54 – P-75
5.
Topic 1 : Work
Topic 2 : Energy
Topic 3 : Power
6.
System of Particles and Rotational Motion
Topic 4 : Collisions
Topic 1 : Centre of Mass, Centre of Gravity & Principle of Moments
Topic 2 : Angular Displacement, Velocity and Acceleration
Topic 3 : Torque, Couple and Angular Momentum
Topic 4 : Moment of Inertia and Rotational K.E.
Topic 5 : Rolling Motion
P-76 – P-112
Gravitation
P-113 – P-130
7.
Topic 1 : Kepler’s Laws of Planetary Motion
Topic 2 : Newton’s Universal Law of Gravitation
Topic 3 : Acceleration due to Gravity
Topic 4 : Gravitational Field and Potential Energy
Topic 5 : Motion of Satellites, Escape Speed and Orbital Velocity
Mechanical Properties of Solids
8.
P-131 – P-138
Topic 1 : Hooke’s Law & Young’s Modulus
Topic 2 : Bulk and Rigidity Modulus and Work Done in Stretching a Wire
Mechanical Properties of Fluids
9.
P-139 – P-154
Topic 1 : Pressure, Density, Pascal’s Law and Archimedes’ Principle
Topic 2 : Fluid Flow, Reynold’s Number and Bernoulli’s Principle
Topic 3 : Viscosity and Terminal Velocity
Topic 4 : Surface Tension, Surface Energy and Capillarity
10. Termal Properties of Matter
P-155 – P-168
Topic 1 : Termometer & Termal Expansion
Topic 2 : Calorimetry and Heat Transfer
Topic 3 : Newton’s Law of Cooling
11. Termodynamics
P-169 – P-185
Topic 1 : First Law of Termodynamics
Topic 2 : Specifc Heat Capacity and Termodynamical Processes
Topic 3 : Carnot Engine, Refrigerators and Second Law of Termodynamics
12. Kinetic Teory
P-186 – P-198
Topic 1 : Kinetic Teory of an Ideal Gas and Gas Laws
Topic 2 : Speed of Gas, Pressure and Kinetic Energy
Topic 3 : Degree of Freedom, Specifc Heat Capacity, and Mean Free Path
13. Oscillations
P-199 – P-218
Topic 1 : Displacement, Phase, Velocity and Acceleration in S.H.M.
Topic 2 : Energy in Simple Harmonic Motion
Topic 3 : Time Period, Frequency, Simple Pendulum and Spring Pendulum
14. Waves
Topic 4 : Damped, Forced Oscillations and Resonance
Topic 1 : Basic of Mechanical Waves, Progressive and Stationary Waves
Topic 2 : Vibration of String and Organ Pipe
P-219 – P-234
Topic 3 : Beats, Interference and Superposition of Waves
Topic 4 : Musical Sound and Doppler’s Efect
P-235 – P-253
15. Electric Charges and Fields
Topic 1 : Electric Charges and Coulomb’s Law
Topic 2 : Electric Field and Electric Field Lines
16. Electrostatic Potential and Capacitance
Topic 3 : Electric Dipole, Electric Flux and Gauss’s Law
P-254 – P-280
Topic 1 : Electrostatic Potential and Equipotential Surfaces
Topic 2 : Electric Potential Energy and Work Done in Carrying a Charge
Topic 3 : Capacitors, Grouping of Capacitor and Energy Stored in a Capacitor
P-281 – P-311
17. Current Electricity
Topic 1 : Electric Current, Drif of Electrons, Ohm’s Law, Resistance and Resistivity
Topic 2 : Combination of Resistances
Topic 3 : Kirchhof ’s Laws, Cells, Termo e.m.f. Electrolysis
Topic 4 : Heating Efect of Current
18. Moving Charges and Magnetism
Topic 5 : Wheatstone Bridge and Diferent Measuring Instruments
P-312 – P-339
Topic 1 : Motion of Charged Particle in Magnetic Field
Topic 2 : Magnetic Field Lines, Biot-Savart’s Law and Ampere’s Circuital Law
Topic 3 : Force and Torque on Current Carrying Conductor
Topic 4 : Galvanometer and its Conversion into Ammeter and Voltmeter
P-340 – P-347
19. Magnetism and Matter
Topic 1 : Magnetism, Gauss’s Law, Magnetic Moment, Properties of Magnet
Topic 2 : Te Earth Magnetism, Magnetic Materials and their Properties
20. Electromagnetic Induction
Topic 3 : Magnetic Equipment
P-348 – P-360
Topic 1 : Magnetic Flux, Faraday’s and Lenz’s Law
21. Alternating Current
Topic 2 : Motional and Static EMI and Application of EMI
Topic 1 : Alternating Current, Voltage and Power
Topic 2 : AC Circuit, LCR Circuit, Quality and Power Factor
Topic 3 : Transformers and LC Oscillations
P-361 – P-376
P-377 – P-388
22. Electromagnetic Waves
Topic 1 : Electromagnetic Waves, Conduction and Displacement Current
Topic 2 : Electromagnetic Spectrum
P-389 – P-414
23. Ray Optics and Optical Instruments
Topic 1 : Plane Mirror, Spherical Mirror and Refection of Light
Topic 2 : Refraction of Light at Plane Surface and Total Internal Refection
Topic 3 : Refraction at Curved Surface Lenses and Power of Lens
Topic 4 : Prism and Dispersion of Light
Topic 5 : Optical Instruments
P-415 – P-432
24. Wave Optics
Topic 1 : Wavefront, Interference of Light, Coherent and Incoherent Sources
Topic 2 : Young’s Double Slit Experiment
Topic 3 : Difraction, Polarisation of Light and Resolving Power
P-433 – P-448
25. Dual Nature of Radiation and Matter
Topic 1 : Matter Waves, Cathode and Positive Rays
Topic 2 : Photon, Photoelectric Efect X-rays and Davisson-Germer Experiment
P-449 – P-460
26. Atoms
Topic 1 : Atomic Structure and Rutherford’s Nuclear Model
Topic 2 : Bohr’s Model and the Spectra of the Hydrogen Atom
P-461 – P-472
27. Nuclei
Topic 1 : Composition and Size of the Nuclei
Topic 2 : Mass-Energy Equivalence and Nuclear Reactions
Topic 3 : Radioactivity
28. Semiconductor Electronics : Materials, Devices and Simple Circuits
P-473 – P-493
Topic 1 : Solids, Semiconductors and P-N Junction Diode
Topic 2 : Junction Transistor
29. Communication Systems
Topic 3 : Digital Electronics and Logic Gates
P-494 – P-500
Topic 1 : Communication Systems
MT-1 – MT-8
Mock Test 2 with Solutions
MT-9 – MT-16
Mock Test 1 with Solutions
Opening/ Closing Rank for TOP NITs & List of NITs in India
Opening/ Closing Rank for Top NITS
College Name
NIT Trichy
NIT Rourkela
NIT Surathkal
NIT Warangal
NIT Calicut
NIT Kurukshetra
NIT Durgapur
MNIT Allahabad
NIT Silchar
MNIT Jaipur
OR/CR
ORF
CR
OR
CR
OR
CR
OR
CR
OR
CR
OR
CSE
2060
5317
2253
9420
960
3181
978
2341
2201
10222
2268
ECE
5325
8011
8571
12009
3378
5608
2919
2919
8023
14769
8320
ME
4154
12970
11662
20304
6315
11788
4340
10209
10629
20480
11195
EE/EEE
5708
10353
4084
19168
3456
6801
5270
8152
9703
18966
9454
CR
6170
12067
18115
16273
OR
CR
OR
CR
OR
CR
OR
5611
12095
1449
4051
8699
23882
1148
12509
16098
3600
7128
17899
40841
3881
14511
22753
5884
11145
21851
49215
9277
13595
19325
5879
8790
32579
56958
4119
CR
3831
7868
11426
9179
List of NITs in India
After IITs, NITs form the second layer of topmost engineering Institutes in India
Rank Of (Amongst NITs)
Name
State
NIRF Score
NIRF Ranking
1
NIT Trichy
Tamil Nadu
61.62
10
2
NIT Rourkela
Odisha
57.75
16
3
NIT Karnataka
Karnataka
55.25
21
4
NIT Warangal
Telangana
53.21
26
5
NIT Calicut
Kerala
52.69
28
6
V-NIT
Maharashtra
51.27
31
7
NIT Kurukshetra
Haryana
47.58
41
7
MN-NIT
Uttar Pradesh
47.49
42
8
NIT Durgapur
West Bengal
46.47
46
9
NIT Silchar
Assam
45.61
51
10
M-NIT
Rajasthan
45.20
53
11
SV-NIT
Gujarat
41.88
58
12
NIT Hamirpur
Himachal Pradesh
41.48
60
13
MA-NIT
Madhya Pradesh
40.98
62
14
NITIE
Maharashtra
40.48
66
15
NIT Meghalaya
Meghalaya
40.32
67
16
NIT Agartala
Tripura
39.53
70
17
NIT Raipur
Chattisgarh
39.09
74
18
NIT Goa
Goa
37.06
87
FAQs - Frequently Asked Questions
QUESTION: Which Colleges other than IITs accept JEE Advanced scores?
at the end of each paper of the examination, given to them at the start
of the paper.
ANSWER:
QUESTION: During examination can I change my answers?
1.
Institute of Science (IISc), Bangalore
2.
Indian Institute of Petroleum and Energy (IIPE), Visakhapatnam
ANSWER: Candidate will have the option to change previously
saved answer of any question, anytime during the entire duration of
the test.
3. Indian Institutes of Science Education and Research (IISER),
Bhopal
4. Indian Institutes of Science Education and Research (IISER),
Mohali
5. Indian Institutes of Science Education and Research (IISER),
Kolkata
6. Indian Institutes of Science Education and Research (IISER),
Pune
7. Indian Institutes of Science Education and Research (IISER),
Thiruvananthapuram
8. Indian Institute of Space Science and Technology (IIST), Thiruvananthapuram
QUESTION: How can I change a previously saved answer during the
CBT of JEE (Advanced)-2018?
ANSWER: To change the answer of a question that has already been
answered and saved, first select the corresponding question from
the Question Palette, then click on “Clear Response” to clear the
previously entered answer and subsequently follow the procedure for
answering that type of question.
QUESTION: Will I be given a printout/hard copy of the questions
papers along with my responses to questions in Paper-I and Paper-II
after the completion of the respective papers?
ANSWER: No.
9. Rajiv Gandhi Institute of Petroleum Technology (RGIPT), Rae
Bareli
QUESTION: How will I be getting a copy of the questions papers
and my responses to questions in Paper-I and Paper-II?
QUESTION: If I am absent in one of the papers (Paper 1, Paper 2),
will my result be declared?
ANSWER: The responses of all the candidates who have appeared
for both Paper 1 and Paper 2, recorded during the exam, along with
the questions of each paper, will be electronically mailed to their
registered email ids, by Friday, May 25, 2018, 10:00 IST.
ANSWER: NO. You will be considered absent in JEE
(Advanced)-2018 and the result will not be prepared/declared. It is
compulsory to appear in both the papers for result preparation.
QUESTION: Do I have to choose my question paper language at the
time of JEE (Advanced)-2018 registration?
ANSWER: NO. There is no need to indicate question paper language
at the time of JEE (Advanced)-2018 registration. Candidates will
have the option to choose their preferred language (English or Hindi),
as the default language for viewing the questions, at the start of the
Computer Based Test (CBT) examination of JEE (Advanced)-2018.
QUESTION: Can I change the language (from English to Hindi
and vice versa) of viewing the questions during the CBT of JEE
(Advanced)-2018?
ANSWER: Questions will be displayed on the screen of the Candidate
in the chosen default language (English or Hindi). Further, the
candidate can also switch/toggle between English or Hindi languages,
as the viewing language of any question, anytime during the entire
period of the examination. The candidate will also be having the
option of changing default question viewing language anytime during
the examination.
QUESTION: Will I be given rough sheets for my calculations during
the CBT of JEE (Advanced)-2018?
ANSWER: Yes, you will be given “Scribble Pad” (containing
blank sheets, for rough work) at the start of every paper of JEE
(Advanced)-2018. You can do all your calculations inside this
“Scribble Pad”. Candidates MUST submit their signed Scribble Pads
QUESTION: Suppose two candidates have same JEE
(Advanced)-2018 aggregate marks. Will the two candidates be given
the same rank?
ANSWER: If the aggregate marks scored by two or more candidates
are the same, then the following tie-break policy will be used for
awarding ranks: Step 1: Candidates having higher positive marks
will be awarded higher rank. If the tie breaking criterion at Step 1
fails to break the tie, then the following criterion at Step 2 will be
followed. Step 2: Higher rank will be assigned to the candidate who
has obtained higher marks in Mathematics. If this does not break the
tie, higher rank will be assigned to the candidate who has obtained
higher marks in Physics. If there is a tie even after this, candidates will
be assigned the same rank.
QUESTION: I have read in newspapers that for the academic year
2018-2019, supernumerary seats for female candidates would be there
in IITs. Does this mean that the non-females will get reduced number
of seats in IITs in 2018?
ANSWER: A decision has been taken at the level of the IIT Council to,
inter alia, improve the gender balance in the undergraduate programs
at the IITs from the current (approximately) 8% to 14% in 2018-19
by creating supernumerary seats specifically for female candidates,
without any reduction in the number of seats that was made available
to non-female candidates in the previous academic year (i.e. academic
year 2017-2018).
1
P-1
Physical World, Units and Measurements
Physical World, Units
and Measurements
6.
TOPIC 1 Unit of Physical Quantities
1.
2.
The density of a material in SI unit is 128 kg m–3. In
certain units in which the unit of length is 25 cm and
the unit of mass is 50 g, the numerical value of density
of the material is:
[10 Jan. 2019 I]
(a) 40
(b) 16
(c) 640
(d) 410
A metal sample carrying a current along X-axis with density Jx is
subjected to a magnetic field Bz (along z-axis). The electric field
Ey developed along Y-axis is directly proportional to Jx as well
as Bz. The constant of proportionality has SI unit
[Online April 25, 2013]
3
2
As
m
m2
m
(a)
(b)
(c)
(d)
As
As
m3
A
Dimensions of Physical
TOPIC 2
Quantities
3.
4.
5.
1
E
1
, y=
and z =
are
B
CR
m0 e0
defined where C-capacitance, R-Resistance, l-length,
E-Electric field, B-magnetic field and e 0 , m 0 , - free
space permittivity and permeability respectively. Then :
[Sep. 05, 2020 (II)]
(a) x, y and z have the same dimension.
(b) Only x and z have the same dimension.
(c) Only x and y have the same dimension.
(d) Only y and z have the same dimension.
Dimensional formula for thermal conductivity is (here K
denotes the temperature :
[Sep. 04, 2020 (I)]
–2
–2
(a) MLT K
(b) MLT K–2
–3
(c) MLT K
(d) MLT–3 K–1
A quantity x is given by (IFv2/WL4) in terms of moment of
inertia I, force F, velocity v, work W and Length L. The
dimensional formula for x is same as that of :
[Sep. 04, 2020 (II)]
(a) planck’s constant
(b) force constant
(c) energy density
(d) coefficient of viscosity
The quantities x =
7.
8.
Amount of solar energy received on the earth's surface
per unit area per unit time is defined a solar constant.
Dimension of solar constant is : [Sep. 03, 2020 (II)]
(a) ML2T–2
(b) ML0T–3
(c) M2L0T–1
(d) MLT–2
If speed V, area A and force F are chosen as fundamental
units, then the dimension of Young's modulus will be :
[Sep. 02, 2020 (I)]
2
–1
2
–3
(a) FA V
(b) FA V
(c) FA2V–2
(d) FA–1V0
If momentum (P), area (A) and time (T) are taken to be the
fundamental quantities then the dimensional formula for
energy is :
[Sep. 02, 2020 (II)]
(a) [P2AT –2]
(b) [PA–1T–2]
(c) [PA1/ 2T-1 ]
9.
(d) [P1/ 2AT-1]
Which of the following combinations has the dimension
of electrical resistance (Î0 is the permittivity of vacuum
and m0 is the permeability of vacuum)?
[12 April 2019 I]
(a)
m0
e0
(b)
m0
e0
(c)
e0
m0
e0
(d) m
0
10. In the formula X = 5YZ2, X and Z have dimensions of
capacitance and magnetic field, respectively. What are
the dimensions of Y in SI units ?
[10 April 2019 II]
–3 –2 8 4
–1 –2 4 2
(a) [M L T A ]
(b) [M L T A ]
(c) [M–2 L0 T–4 A–2]
(d) [M–2 L–2 T6 A3]
11. In SI units, the dimensions of
Î0
is: [8 April 2019 I]
m0
(a) A–1TML3
(b) AT2 M–1L–1
(c) AT–3ML3/2
(d) A2T3 M–1L–2
12. Let l, r, c and v represent inductance, resistance,
capacitance and voltage, respectively. The dimension of
l
in SI units will be :
[12 Jan. 2019 II]
rcv
–
2
–1
(a) [LA ]
(b) [A ]
(c) [LTA]
(d) [LT2]
P-2
Physics
13. The force of interaction between two atoms is given by
æ x2 ö
F = ab exp ç ç akT ÷÷ ; where x is the distance, k is the
è
ø
Boltzmann constant and T is temperature and a and b are
two constants. The dimensions of b is: [11 Jan. 2019 I]
(a) M0L2T–4
(b) M2LT–4
(c) MLT–2
(d) M2 L2 T–2
14. If speed (V), acceleration (A) and force (F) are considered
as fundamental units, the dimension of Young’s modulus
will be :
[11 Jan. 2019 II]
(a)
V–2A2F–2
V–2A2F2
(b)
(c)
V–4 A–2 F
(d) V–4A2F
hc5
where c is speed of
G
light, G universal gravitational constant and h is the Planck’s
constant. Dimension of f is that of :
[9 Jan. 2019 I]
(a) area
(b) energy
(c) momentum
(d) volume
16. Expression for time in terms of G (universal gravitational
constant), h (Planck's constant) and c (speed of light) is
proportional to:
[9 Jan. 2019 II]
15. A quantity f is given by f =
(a)
(c)
hc5
G
(b)
Gh
(d)
5
c3
Gh
Gh
c
c3
17. The dimensions of stopping potential V0 in photoelectric
effect in units of Planck’s constant ‘h’, speed of light ‘c’
and Gravitational constant ‘G’ and ampere A is:
[8 Jan. 2019 I]
(a) hl/3G2/3cl/3 A –1
(b) h2/3c5/3G1/3A –1
(c) h–2/3 e–1/3 G4/3 A–1
(d) h2G3/2C1/3 A–1
2
B
, where B is magnetic field and m0
18. The dimensions of
2m 0
is the magnetic permeability of vacuum, is:
[8 Jan. 2019 II]
(a) MLT–2
(b) ML2T–1
(c) ML2T–2
(d) ML–1T–2
19. The characteristic distance at which quantum gravitational
effects are significant, the Planck length, can be determined
from a suitable combination of the fundamental physical
constants G, h and c. Which of the following correctly
gives the Planck length?
[Online April 15, 2018]
1
ö2
1
æ Gh
2 3
(a) G2hc (b) ç 3 ÷
(c)
2 h 2 c (d) Gh c
G
èc ø
20. Time (T), velocity (C) and angular momentum (h) are
chosen as fundamental quantities instead of mass, length
and time. In terms of these, the dimensions of mass would
be :
[Online April 8, 2017]
(a) [ M ]=[ T–1 C–2 h ]
(b) [ M ]=[ T–1 C2 h ]
–1
–2
–1
(c) [ M ]=[ T C h ] (d) [ M ]=[ T C–2 h ]
21. A, B, C and D are four different physical quantities having
different dimensions. None of them is dimensionless. But
we know that the equation AD = C ln (BD) holds true.
Then which of the combination is not a meaningful quantity ?
[Online April 10, 2016]
C AD 2
(b) A2 –B2C2
BD
C
A
(A - C)
-C
(c)
(d)
B
D
22. In the following 'I' refers to current and other symbols
have their usual meaning, Choose the option that
corresponds to the dimensions of electrical conductivity :
[Online April 9, 2016]
(a) M–1 L–3T 3 I
(b) M–1 L–3 T3 I2
(c) M–1L3T3 I
(d) ML–3 T–3 I2
23. If electronic charge e, electron mass m, speed of light in
vacuum c and Planck’s constant h are taken as fundamental
quantities, the permeability of vacuum m0 can be expressed
in units of :
[Online April 11, 2015]
(a)
æ hc ö
(b) ç 2 ÷
è me ø
æ h ö
(a) ç 2 ÷
è me ø
æ mc 2 ö
(d) çç 2 ÷÷
è he ø
24. If the capacitance of a nanocapacitor is measured in terms
of a unit ‘u’ made by combining the electric charge ‘e’,
Bohr radius ‘a0’, Planck’s constant ‘h’ and speed of light
‘c’ then:
[Online April 10, 2015]
æ h ö
(c) ç 2 ÷
è ce ø
(a) u =
e2 h
a0
(b) u =
(c) u =
e2c
ha 0
(d) u =
hc
2
e a0
e2 a 0
hc
25. From the following combinations of physical constants
(expressed through their usual symbols) the only
combination, that would have the same value in different
systems of units, is:
[Online April 12, 2014]
(a)
(b)
(c)
(d)
ch
2peo2
e2
2pe o Gme2
(me = mass of electron)
m o eo G
c2 he 2
2p m o eo h
G
ce2
P-3
Physical World, Units and Measurements
26. In terms of resistance R and time T, the dimensions of ratio
m
of the permeability m and permittivity e is:
e
[Online April 11, 2014]
(a) [RT–2] (b) [R2T–1] (c) [R2]
(d) [R2T2]
27. Let [ Î0 ] denote the dimensional formula of the permittivity
of vacuum. If M = mass, L = length, T = time and A =
electric current, then:
[2013]
–1
–3
2
–1
–3
4
(a) Î0 = [M L T A] (b) Î0 = [M L T A2]
(c) Î0 = [M1 L2 T1 A2] (d) Î0 = [M1 L2 T1 A]
28. If the time period t of the oscillation of a drop of liquid of
density d, radius r, vibrating under surface tension s is given
by the formula t = r 2b s c d a / 2 . It is observed that the
d
. The value of b
s
[Online April 23, 2013]
time period is directly proportional to
should therefore be :
(a)
3
4
(b)
3
2
3
(d)
3
2
29. The dimensions of angular momentum, latent heat and
capacitance are, respectively. [Online April 22, 2013]
(c)
(a) ML2 T1A 2 , L2 T -2 , M -1L-2 T 2
(b) ML2 T -2 , L2 T 2 , M -1L-2 T 4 A 2
(c) ML2 T -1, L2 T -2 , ML2 TA 2
(d) ML2 T -1 , L2 T -2 , M -1L-2 T 4 A 2
30. Given that K = energy, V = velocity, T = time. If they are
chosen as the fundamental units, then what is dimensional
formula for surface tension?
[Online May 7, 2012]
(a) [KV–2T –2 ]
(b) [K2 V2T–2 ]
(c) [K2V–2 T–2 ]
(d) [KV2 T2 ]
31. The dimensions of magnetic field in M, L, T and C
(coulomb) is given as
[2008]
(a) [MLT–1 C–1]
(b) [MT2 C–2]
(c) [MT–1 C–1]
(d) [MT–2 C–1]
32. Which of the following units denotes the dimension
ML2
, where Q denotes the electric charge?
[2006]
Q2
2
(a) Wb/m
(b) Henry (H)
(c) H/m2
(d) Weber (Wb)
33. Out of the following pair, which one does NOT have
identical dimensions ?
[2005]
(a) Impulse and momentum
(b) Angular momentum and planck’s constant
(c) Work and torque
(d) Moment of inertia and moment of a force
34. Which one of the following represents the correct
dimensions of the coefficient of viscosity?
[2004]
-1 -1
(a) éë ML T ùû
-1
(b) éë MLT ùû
-1 -2
(c) éë ML T ùû
-2 -2
(d) éë ML T ùû
35. Dimensions of
meaning, are
1 , where symbols have their usual
mo eo
[2003]
(b) [L-2 T 2 ]
(a) [L-1T]
(c) [L2 T -2 ]
(d) [LT -1 ]
36. The physical quantities not having same dimensions are
(a) torque and work
[2003]
(b) momentum and planck’s constant
(c) stress and young’s modulus
(d) speed and (m o e o ) -1 / 2
37. Identify the pair whose dimensions are equal
[2002]
(a) torque and work
(b) stress and energy
(c) force and stress
(d) force and work
TOPIC 3 Errors in Measurements
38. A screw gauge has 50 divisions on its circular scale. The
circular scale is 4 units ahead of the pitch scale marking,
prior to use. Upon one complete rotation of the circular
scale, a displacement of 0.5 mm is noticed on the pitch
scale. The nature of zero error involved, and the least
count of the screw gauge, are respectively :
[Sep. 06, 2020 (I)]
(a) Negative, 2 mm
(b) Positive, 10 mm
(c) Positive, 0.1 mm
(d) Positive, 0.1 mm
39. The density of a solid metal sphere is determined by
measuring its mass and its diameter. The maximum error in
æ x ö
the density of the sphere is çè
÷ %. If the relative errors
100 ø
in measuring the mass and the diameter are 6.0% and 1.5%
respectively, the value of x is ______.
[NA Sep. 06, 2020 (I)]
40. A student measuring the diameter of a pencil of circular
cross-section with the help of a vernier scale records the
following four readings 5.50 mm, 5.55 mm, 5.45 mm,
5.65 mm, The average of these four reading is 5.5375 mm
and the standard deviation of the data is 0.07395 mm. The
average diameter of the pencil should therefore be recorded as :
[Sep. 06, 2020 (II)]
(a) (5.5375 ± 0.0739) mm (b) (5.5375 ± 0.0740) mm
(c) (5.538 ± 0.074) mm
(d) (5.54 ± 0.07) mm
41. A physical quantity z depends on four observables a, b, c
and d, as z =
2
2 3
a b
. The percentages of error in the meacd 3
surement of a, b, c and d are 2%, 1.5%, 4% and 2.5% respectively. The percentage of error in z is :
[Sep. 05, 2020 (I)]
P-4
42.
43.
44.
45.
46.
47.
48.
Physics
(a) 12.25%
(b) 16.5%
(c) 13.5%
(d) 14.5%
Using screw gauge of pitch 0.1 cm and 50 divisions on its
circular scale, the thickness of an object is measured. It
should correctly be recorded as : [Sep. 03, 2020 (I)]
(a) 2.121 cm
(b) 2.124 cm
(c) 2.125 cm
(d) 2.123 cm
The least count of the main scale of a vernier callipers is
1 mm. Its vernier scale is divided into 10 divisions and
coincide with 9 divisions of the main scale. When jaws are
touching each other, the 7th division of vernier scale
coincides with a division of main scale and the zero of
vernier scale is lying right side of the zero of main scale.
When this vernier is used to measure length of a cylinder
the zero of the vernier scale betwen 3.1 cm and 3.2 cm and
4th VSD coincides with a main scale division. The length
of the cylinder is : (VSD is vernier scale division)
[Sep. 02, 2020 (I)]
(a) 3.2 cm
(b) 3.21 cm
(c) 3.07 cm
(d) 2.99 cm
If the screw on a screw-gauge is given six rotations, it moves
by 3 mm on the main scale. If there are 50 divisions on the
circular scale the least count of the screw gauge is:
[9 Jan. 2020 I]
(a) 0.001 cm
(b) 0.02 mm
(c) 0.01 cm
(d) 0.001 mm
For the four sets of three measured physical quantities as
given below. Which of the following options is correct?
[9 Jan. 2020 II]
(A) A1 = 24.36, B1 = 0.0724, C1 = 256.2
(B) A2 = 24.44, B2 = 16.082, C2 = 240.2
(C) A3 = 25.2, B3 = 19.2812, C3 = 236.183
(D) A4 = 25, B4 = 236.191, C4 = 19.5
(a) A4 + B4 + C4 < A1 + B1 + C1 < A3 + B3 + C3 < A2 + B2 + C2
(b) A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3 = A4 + B4 + C4
(c) A4 + B4 + C4 < A1 + B1 + C1 = A2 + B2 + C2 = A3 + B3 + C3
(d) A1 + B1 + C1 < A3 + B3 + C3 < A2 + B2 + C2 < A4 + B4 + C4
A simple pendulum is being used to determine the value
of gravitational acceleration g at a certain place. The length
of the pendulum is 25.0 cm and a stop watch with 1 s
resolution measures the time taken for 40 oscillations to
be 50 s. The accuracy in g is:
[8 Jan. 2020 II]
(a) 5.40%
(b) 3.40%
(c) 4.40%
(d) 2.40%
In the density measurement of a cube, the mass and edge
length are measured as (10.00 ± 0.10) kg and (0.10 ± 0.01)
m, respectively. The error in the measurement of density
is:
[9 April 2019 I]
3
(a) 0.01 kg/m
(b) 0.10 kg/m3
3
(c) 0.013 kg/m
(d) 0.07 kg/m3
The area of a square is 5.29 cm 2. The area of 7 such
squares taking into account the significant figures is:
[9 April 2019 II]
(a) 37cm2
49.
50.
51.
52.
53.
54.
(b) 37.030 cm2
(c) 37.03 cm2
(d) 37.0 cm2
In a simple pendulum experiment for determination of
acceleration due to gravity (g), time taken for 20
oscillations is measured by using a watch of 1 second
least count. The mean value of time taken comes out to be
30 s. The length of pendulum is measured by using a
meter scale of least count 1 mm and the value obtained is
55.0 cm. The percentage error in the determination of g is
close to :
[8 April 2019 II]
(a) 0.7% (b) 0.2%
(c) 3.5%
(d) 6.8%
The least count of the main scale of a screw gauge is 1 mm.
The minimum number of divisions on its circular scale
required to measure 5 µm diameter of a wire is:
[12 Jan. 2019 I]
(a) 50
(b) 200
(c) 100
(d) 500
The diameter and height of a cylinder are measured by
a meter scale to be 12.6 ± 0.1 cm and 34.2 ± 0.1 cm,
respectively. What will be the value of its volume in
appropriate significant figures?
[10 Jan. 2019 II]
3
(a) 4264 ± 81 cm
(b) 4264.4 ± 81.0 cm3
3
(c) 4260 ± 80 cm
(d) 4300 ± 80 cm3
The pitch and the number of divisions, on the circular
scale for a given screw gauge are 0.5 mm and 100
respectively. When the screw gauge is fully tightened
without any object, the zero of its circular scale lies 3
division below the mean line.
The readings of the main scale and the circular scale, for
a thin sheet, are 5.5 mm and 48 respectively, the
thickness of the sheet is:
[9 Jan. 2019 II]
(a) 5.755 mm
(b) 5.950 mm
(c) 5.725 mm
(d) 5.740 mm
The density of a material in the shape of a cube is
determined by measuring three sides of the cube and its
mass. If the relative errors in measuring the mass and
length are respectively 1.5% and 1%, the maximum error
in determining the density is:
[2018]
(a) 2.5% (b) 3.5%
(c) 4.5%
(d) 6%
The percentage errors in quantities P, Q, R and S are 0.5%,
1%, 3% and 1.5% respectively in the measurement of a
physical quantity A =
P 3Q 2
.
RS
The maximum percentage error in the value of A will be
[Online April 16, 2018]
(a) 8.5%
(b) 6.0%
(c) 7.5%
(d) 6.5%
55. The relative uncertainty in the period of a satellite orbiting
around the earth is 10–2. If the relative uncertainty in the
radius of the orbit is negligible, the relative uncertainty in
the mass of the earth is
[Online April 16, 2018]
(a) 3×10–2
(b) 10–2
(c) 2 × 10–2
(d) 6 × 10–2
P-5
Physical World, Units and Measurements
56. The relative error in the determination of the surface area
of a sphere is a. Then the relative error in the determination
of its volume is
[Online April 15, 2018]
2
2
3
a (b)
a
a
(c)
(d) a
3
3
2
In a screw gauge, 5 complete rotations of the screw cause
it to move a linear distance of 0.25 cm. There are 100 circular
scale divisions. The thickness of a wire measured by this
screw gauge gives a reading of 4 main scale divisions and
30 circular scale divisions. Assuming negligible zero error,
the thickness of the wire is:
[Online April 15, 2018]
(a) 0.0430 cm
(b) 0.3150 cm
(c) 0.4300 cm
(d) 0.2150 cm
The following observations were taken for determining
surface tensiton T of water by capillary method :
Diameter of capilary, D = 1.25 × 10–2 m
rise of water, h = 1.45 × 10–2 m
Using g = 9.80 m/s2 and the simplified relation
rhg
T=
´ 10 3 N/m, the possible error in surface tension
2
is closest to :
[2017]
(a) 2. 4 % (b) 10 %
(c) 0.15%
(d) 1.5%
A physical quantity P is described by the relation
P = a1/2 b2 c3 d –4
If the relative errors in the measurement of a, b, c and d
respectively, are 2%, 1%, 3% and 5%, then the relative
error in P will be :
[Online April 9, 2017]
(a) 8%
(b) 12%
(c) 32%
(d) 25%
A screw gauge with a pitch of 0.5 mm and a circular scale
with 50 divisions is used to measure the thickness of a
thin sheet of Aluminium. Before starting the measurement,
it is found that wen the two jaws of the screw gauge are
brought in contact, the 45th division coincides with the
main scale line and the zero of the main scale is barely
visible. What is the thickness of the sheet if the main scale
reading is 0.5 mm and the 25th division coincides with the
main scale line?
[2016]
(a) 0.70 mm
(b) 0.50 mm
(c) 0.75 mm
(d) 0.80 mm
A student measures the time period of 100 oscillations of
a simple pendulum four times. The data set is 90 s, 91 s, 95
s, and 92 s. If the minimum division in the measuring clock
is 1 s, then the reported mean time should be:
[2016]
(a) 92 ± 1.8 s
(b) 92 ± 3s
(c) 92 ± 1.5 s
(d) 92 ± 5.0 s
(a)
57.
58.
59.
60.
61.
62. The period of oscillation of a simple pendulum is
L
. Measured value of L is 20.0 cm known to 1 mm
g
accuracy and time for 100 oscillations of the pendulum is
found to be 90 s using a wrist watch of 1s resolution. The
accuracy in the determination of g is :
[2015]
(a) 1%
(b) 5%
(c) 2%
(d) 3%
T = 2p
63. Diameter of a steel ball is measured using a Vernier callipers
which has divisions of 0.1 cm on its main scale (MS) and
10 divisions of its vernier scale (VS) match 9 divisions on
the main scale. Three such measurements for a ball are
given as:
[Online April 10, 2015]
S.No. MS(cm) VS divisions
1.
0.5
8
2.
3.
0.5
0.5
4
6
If the zero error is – 0.03 cm, then mean corrected diameter
is:
(a) 0.52 cm
(b) 0.59 cm
(c) 0.56 cm
(d) 0.53 cm
64. The current voltage relation of a diode is given by
I = ( e1000V T - 1) mA, where the applied voltage V is in
volts and the temperature T is in degree kelvin. If a student
makes an error measuring ±0.01 V while measuring the
current of 5 mA at 300 K, what will be the error in the
value of current in mA?
[2014]
(a) 0.2 mA (b) 0.02 mA (c) 0.5 mA (d) 0.05 mA
65. A student measured the length of a rod and wrote it as 3.50
cm. Which instrument did he use to measure it?
[2014]
(a) A meter scale.
(b) A vernier calliper where the 10 divisions in vernier
scale matches with 9 division in main scale and main
scale has 10 divisions in 1 cm.
(c) A screw gauge having 100 divisions in the circular
scale and pitch as 1 mm.
(d) A screw gauge having 50 divisions in the circular scale
and pitch as 1 mm.
66. Match List - I (Event) with List-II (Order of the time interval
for happening of the event) and select the correct option
from the options given below the lists:
[Online April 19, 2014]
List - I
(1) Rotation
period of earth
List - II
(i) 10 5 s
(2) Revolution
(ii) 10 7 s
period of earth
(3) Period of light (iii) 10 –15 s
wave
(4) Period of
(iv) 10 –3 s
sound wave
(a) (1)-(i), (2)-(ii), (3)-(iii), (4)-(iv)
(b) (1)-(ii), (2)-(i), (3)-(iv), (4)-(iii)
(c) (1)-(i), (2)-(ii), (3)-(iv), (4)-(iii)
(d) (1)-(ii), (2)-(i), (3)-(iii), (4)-(iv)
P-6
Physics
67. In the experiment of calibration of voltmeter, a standard cell
of e.m.f. 1.1 volt is balanced against 440 cm of potential wire.
The potential difference across the ends of resistance is
found to balance against 220 cm of the wire. The
corresponding reading of voltmeter is 0.5 volt. The error in
the reading of volmeter will be: [Online April 12, 2014]
(a) – 0. 15 volt
(b) 0.15 volt
(c) 0.5 volt
(d) – 0.05 volt
68. An experiment is performed to obtain the value of
acceleration due to gravity g by using a simple pendulum of
length L. In this experiment time for 100 oscillations is
measured by using a watch of 1 second least count and the
value is 90.0 seconds. The length L is measured by using a
meter scale of least count 1 mm and the value is 20.0 cm. The
error in the determination of g would be:
[Online April 9, 2014]
(a) 1.7% (b) 2.7%
(c) 4.4%
(d) 2.27%
69. Resistance of a given wire is obtained by measuring the
current flowing in it and the voltage difference applied across
it. If the percentage errors in the measurement of the current
and the voltage difference are 3% each, then error in the
value of resistance of the wire is
[2012]
(a) 6%
(b) zero
(c) 1%
(d) 3%
70. A spectrometer gives the following reading when used to
measure the angle of a prism.
Main scale reading : 58.5 degree
Vernier scale reading : 09 divisions
Given that 1 division on main scale corresponds to 0.5
degree. Total divisions on the Vernier scale is 30 and match
with 29 divisions of the main scale. The angle of the prism
from the above data is
[2012]
(a) 58.59 degree
(b) 58.77 degree
(c) 58.65 degree
(d) 59 degree
71. N divisions on the main scale of a vernier calliper coincide
with (N + 1) divisions of the vernier scale. If each division of
main scale is ‘a’ units, then the least count of the instrument
is
[Online May 19, 2012]
(a) a
(c)
N
´a
N +1
(b)
a
N
(d)
a
N +1
72. A student measured the diameter of a wire using a screw
gauge with the least count 0.001 cm and listed the
measurements. The measured value should be recorded
as
[Online May 12, 2012]
(a) 5.3200 cm
(b) 5.3 cm
(c) 5.32 cm
(d) 5.320 cm
73. A screw gauge gives the following reading when used to
measure the diameter of a wire.
Main scale reading : 0 mm
Circular scale reading : 52 divisions
Given that 1mm on main scale corresponds to 100 divisions
of the circular scale. The diameter of wire from the above
data is
[2011]
(a) 0.052 cm
(b) 0.026 cm
(c) 0.005 cm
(d) 0.52 cm
74. The respective number of significant figures for the
numbers 23.023, 0.0003 and 2.1 × 10–3 are
[2010]
(a) 5, 1, 2
(b) 5, 1, 5
(c) 5, 5, 2
(d) 4, 4, 2
75. In an experiment the angles are required to be measured
using an instrument, 29 divisions of the main scale exactly
coincide with the 30 divisions of the vernier scale. If the
smallest division of the main scale is half- a degree
(= 0.5°), then the least count of the instrument is: [2009]
(a) half minute
(b) one degree
(c) half degree
(d) one minute
76. A body of mass m = 3.513 kg is moving along the x-axis
with a speed of 5.00 ms–1. The magnitude of its momentum
is recorded as
[2008]
(a) 17.6 kg ms–1
(b) 17.565 kg ms–1
(c) 17.56 kg ms–1
(d) 17.57 kg ms–1
77. Two full turns of the circular scale of a screw gauge cover a
distance of 1mm on its main scale. The total number of
divisions on the circular scale is 50. Further, it is found that
the screw gauge has a zero error of – 0.03 mm. While
measuring the diameter of a thin wire, a student notes the
main scale reading of 3 mm and the number of circular scale
divisions in line with the main scale as 35. The diameter of
the wire is
[2008]
(a) 3.32 mm
(b) 3.73 mm
(c) 3.67 mm
(d) 3.38 mm
P-7
Physical World, Units and Measurements
1.
(a) Density of material in SI unit,
=
=
2.
128 ( 50 g )( 20)
( 25cm ) ( 4)
3
3
128
( 20) = 40 units
64
=
=
6.
(b) According to question
E y µ J x BZ
\ Constant of proportionality
K=
Ey
=
BZ J x
C m3
=
J x As
I
E
]
= C (speed of light) and J =
Area
B
(a) We know that
1
Speed of light, c =
m 0 e0
=x
7.
8.
dQ
dT
= kA
dt
dx
\x =
[ML2 T -3 ]
IFv 2
WL4
(d) Young's modulus, Y =
F
A
stress
strain
Dl
= FA –1V0
l0
(c) Energy, E µ AaT b Pc
E = kAaT b P c
...(i)
M 1L2T -2 = M c L2a + cT b - c
by comparison
c=1
2a + c = 2
b – c = –2
c = 1, a = 1/2, b = –1
æ dQ ö
çè
÷
dt ø
Þk=
æ dT ö
Aç ÷
è dx ø
5.
= M1L0 T -3 .
TL2
Dimension of momentum, P = M 1L1T -1
Dimension of area, A = L2
Dimension of time, T = T 1
Putting these value in equation (i), we get
l
l
= = Speed
Rc t
Thus, x, y, z will have the same dimension of speed.
= [MLT -3K -1 ]
[L2 ][KL-1 ]
(c) Dimension of Force F = M1L1T–2
Dimension of velocity V = L1T–1
Dimension of work = M1L2T–2
Dimension of length = L
Moment of inertia = ML2
= M1L-1T -2 = Energy density
where k is a dimensionless constant and a, b and c are the
exponents.
\z =
[k ] =
M1L2 T -2
or,
E
=y
B
Time constant, t = Rc = t
(d) From formula,
L3
Energy
Time Area
Dimension of Energy, E = ML2T–2
Dimension of Time = T
Dimension of Area = L2
\ Dimension of Solar constant
ÞY =
Also, c =
4.
M1L-2 T -2
(b) Solar constant =
=
[As
3.
(M1L2 T -2 )(L4 )
128kg
m3
Density of material in new system
=
(M1L2 )(M1L1T -2 )(L1T -2 )2
\ E = A1/ 2T -1 P1
9.
(a)
m0
=
e0
m 20
= m0 c
e0 m 0
m0c ® MLT–2A–2 × LT–1
ML2T–3A–2
Dimensions of resistance
10. (a) X = 5YZ2
X
ÞY µ 2
Z
æ 1
ö
= c÷
ç
è m0e0
ø
...(i)
P-8
Physics
X = Capacitance =
Q Q2
[ A2T 2 ]
=
=
V W
[ ML2T -2 ]
X = [M–1L–2T4A2]
F
IL
Z = [MT–2A–1]
[Q F = ILB]
15. (b) Dimension of [h] = [ML2T–1]
[ M -1 L-2T 4 A2 ]
[G] = [M–1L3T–2]
Hence dimension of
(Using (i))
é e0 ù
é e0 ù
e 02
ú = e C[LT
11. (d) ê m ú = m e = ê
T–1]×[e0]
0
êë m 0 e 0 úû
0 0
ëê 0 úû
é
ù
1
= Cú
êQ
m0 e0
êë
úû
2
q
Q F=
4pe 0 r 2
Þ [e 0 ] =
[ AT ]2
-2
2
= [ A2 M -1 L-3T 4 ]
[ MLT ] ´ [ L ]
é e0 ù
-1
2
-1 -3 4
\ ê
ú = [ LT ] ´ [ A M L T ]
m
ëê 0 úû
= [ M -1L-2T 3 A2 ]
12. (b) As we know,
élù
êë r úû = [ T ] and [ cv] = [ AT ]
é l ù é T ù
\ê ú=ê
= é A –1 ù
ë rcv û ë AT úû ë û
13. (b) Force of interaction between two atoms,
æ -x2 ö
ç
÷
ç akT ÷
ø
abeè
Since exponential terms are dimensionless
é x2 ù
\ ê akT ú = M0L0T0
êë
úû
Þ
so [Y] = [V–4FA2] = [V–4A2F]
[C] = [LT–1]
[ MT -2 A-1 ]2
Y = [M–3L–2T8A4]
F=
solving above equations we get:
a = – 4, b = 1, c = 2
Z =B=
Y=
b = 1, a + b + c = – 1, –a –2b –2c = – 2
L2
= M0 L0T 0
[a ]ML2T-2
Þ [a] = M–1T2
[F] = [a] [b]
MLT–2 = M–1T2[b]
Þ [b] = M2LT–4
14. (d) Let [Y] = [V]a [F]b [A]c
[ML–1T–2] = [LT–1]a [MLT–2]b [LT –2]c
[ML–1T–2] = [MbLa+b+c T–a–2b–2c]
Comparing power both side of similar terms we get,
é hC 5 ù é ML2T -1 ù × é L5T -5 ù
û ë
û
ê
ú=ë
é M -1 L3T -2 ù
ê G ú
ë
û
ë
û
= [ML2T–2] = energy
16. (c) Let t µ Gx hy Cz
Dimensions of G = [M–1L3T–2],
h = [ML2T–1] and C = [LT–1]
[T] = [M–1L3T–2]x[ML2T–1]y[LT–1]z
[M0L0T1] = [M–x+y L3x+2y+z T–2x–y–z]
By comparing the powers of M, L, T both the sides
–x+ y=0 Þx=y
3x + 2y + z = 0 Þ 5x + z = 0
..... (i)
–2x – y –z = 1
Þ 3x + z = –1
..... (ii)
Solving eqns. (i) and (ii),
Gh
1
5
x=y= ,z=- \ tµ
2
2
C5
17. (None)
Stopping potential (V0 ) µ h x I yG Z C r
Here,h = Planck’s constant = éë ML2T -1 ùû
I = current = [A]
G = Gravitational constant = [M–1L3T–2]
and c = speed of light = [LT–1]
V0 = potential= [ML2T–3A–1]
\ [ML2T–3A–1]=[ML2T–1]x [A]y[M–1L3T–2]z[LT–1]r
Mx – z; L2x+3z+r; T–x–2z–r; Ay
Comparing dimension of M, L, T, A, we get
y = –1, x = 0, z = – 1 , r = 5
\ V0 µ h0 I –1G –1C 5
18. (d) The quantity
field.
Þ
B2
is the energy density of magnetic
2m0
é B 2 ù Energy Force ´ displacement
=
ê
ú=
(displacement)3
êë 2m0 úû Volume
é ML2T –2 ù
–1 –2
=ê
ú = ML T
3
êë L
úû
P-9
Physical World, Units and Measurements
19. (b) Plank length is a unit of length, lp = 1.616229 × 10–35 m
lp =
hG
c3
20. (a) Let mass, related as M µ Tx Cy hz
M1L0 T0 = (T')x (L1T–1)y (M1L2T–1)z
M1L0 T0 = Mz Ly + 2z + Tx– y–z
z= 1
y + 2z = 0
x – y– z = 0
y = –2
x+ 2–1=0
x = –1
–1 –2 1
M = [T C h ]
21. (d) Dimension of A ¹ dimension of (C)
Hence A – C is not possible.
22. (b) We know that resistivity
RA
θ<
l
1
l
Conductivity = resistivity < RA
lI
<
( Q V = RI)
VA
[L][I]
W W
<
QV<
<
é [ML2 T,2 ù
q
it
ê
ú ´[L2 ]
ê [I][T] ú
êë
úû
< [M,1L,3T 3 ][I2 ] < [M,1L,3T3I2 ]
23. (c) Let µ0 related with e, m, c and h as follows.
m0 = keambcchd
[MLT–2A–2] = [AT]a [M]b [LT–1]c [ML2T–1]d
= [Mb + d Lc + 2d Ta – c – d Aa]
On comparing both sides we get
a = – 2
...(i)
b+ d = 1
...(ii)
c + 2d = 1
...(iii)
a – c – d = –2
...(iv)
By equation (i), (ii), (iii) & (iv) we get,
a = – 2, b = 0, c = – 1, d = 1
é h ù
\ [m 0 ] = ê 2 ú
ë ce û
24. (d) Let unit ‘u’ related with e, a0, h and c as follows.
[u] = [e]a [a0]b [h]c [C]d
Using dimensional method,
[M–1L–2T+4A+2] = [A1T1]a[L]b[ML2T–1]c[LT–1]d
[M–1L–2T+4A+2] = [Mc Lb+2c+d Ta–c–d Aa]
a = 2, b = 1, c = – 1, d = – 1
e 2 a0
hc
25. (b) The dimensional formulae of
\
u=
0 0 1 1ù
e = éM L T A
ë
û
e0 = é M -1L3T 4 A 2 ù
ë
û
G = é M -1L3T -2 ù and me = é M1L0T 0 ù
ë
û
ë
û
Now,
=
e2
2 pe0 Gm e2
é M0 L0 T1A1 ù
ë
û
2
2 p é M -1L-3T 4 A 2 ù é M -1L3T -2 ù é M1L0 T 0 ù
ë
ûë
ûë
û
2
éT 2A 2 ù
ë
û
=
2p éM -1-1+ 2 L-3+ 3 T 4 - 2 A 2 ù
ë
û
=
éT 2 A 2 ù
1
ë
û
=
0 0 2 2ù
2p
é
2p M L T A
ë
û
1
e2
is dimensionless thus the combination
2p
2 pe 0 Gm e2
would have the same value in different systems of units.
26. (c) Dimensions of m = [MLT–2A–2]
Dimensions of Î = [M–1L–3T4A2]
Dimensions of R = [ML2T–3A–2]
Q
[MLT -2 A -2 ]
Dimensions of m
=
Dimensions of Î [M -1L-3T 4 A 2 ]
= [M2L4T–6A–4 ] = [R2]
1 q1q 2
27. (b) As we know, F =
4 pe0 R 2
\
Þ e0 =
q1q2
4pFR 2
Hence, e0 =
C2
[AT]2
=
N.m2 [MLT -2 ][L2 ]
= [M–1 L–3 T4 A2]
28. (c)
29. (d) Angular momentum = m × v × r = ML2 T–1
Q ML2 T -2
=
= L2T–2
m
M
Charge
= M -1L-2 T 4 A 2
Capacitance C =
P.d.
Latent heat L =
30. (a) Surface tension, T =
T2
F F l T2
= . .
l
l l T2
-2
=V )
l2
Therefore, surface tension = [KV–2T–2]
31. (c) Magnitude of Lorentz formula F = qvB sin q
(As, F.l = K (energy);
B=
F
MLT -2
=
= [ MT -1C -1]
qv C ´ LT -1
P-10
Physics
32. (b) Mutual inductance =
f BA
=
I
I
[ MT -1Q -1L2 ]
= ML2Q -2
[QT -1 ]
33. (d) Moment of Inertia, I = MR2
[I] = [ML2]
r
r r
Moment of force, t = r ´ F
r
t = [ L][ MLT -2 ] = [ML2T -2 ]
34. (a) According to, Stokes law,
F
F = 6phrv Þ h =
6pr v
[Henry] =
h=
[ MLT –2 ]
–1
Þ h = [ ML-1T -1 ]
[ L][ LT ]
35. (c) As we know, the velocity of light in free space is
given by
c=
1
1
\
= e 2 = Z12T 2
m 0 e0
mo eo
1
2
2
mo eo = C [m/s]
= [LT–1]2
= [M0L2T–2]
36. (b) Momentum, = mv = [MLT–1]
Planck’s constant,
2 –2
E = [ ML T ]
= [ML2T–1]
[T –1 ]
v
r r
37. (a) Work W = F × s = Fs cos q
r r
Q A × B = AB cos q
h=
= [ MLT -2 ][ L] = [ML2T -2 ] ;
r r r
Torque, t = r ´ F Þ t = rF sin q
r r
Q A ´ B = AB sin q
= [ L ] [MLT -2 ] = [ ML2T -2 ]
38. (b) Given : No. of division on circular scale of screw gauge = 50
Pitch = 0.5 mm
Least count of screw gauge
=
Pitch
No. of division on circular scale
0.5
mm = 1 ´ 105 m = 10 mm
50
And nature of zero error is positive.
39. (1050)
=
Density, r =
6
M
M
Þ r = MD -3
=
3
p
V
4 æ Dö
pç ÷
3 è 2ø
æ Dr ö Dm
æ DD ö
\%ç ÷ =
+ 3ç
= 6 + 3 ´ 1.5 = 10.5%
è D ÷ø
è rø
m
æ Dr ö 1050
æ x ö
%ç ÷ =
%=ç
%
è 100 ÷ø
è r ø 100
\ x = 1050.00
40. (d) Average diameter, dav = 5.5375 mm
Deviation of data, Dd = 0.07395 mm
As the measured data are upto two digits after decimal,
therefore answer should be in two digits after decimal.
\ d = (5.54 ± 0.07) mm
41. (d) Given : Z =
a 2b 2/ 3
cd 3
Percentage error in Z,
=
DZ 2 Da 2 Db 1 Dc 3Dd
=
+
+
+
Z
a
3 b 2 c
d
2
1
´ 1.5 + ´ 4 + 3 ´ 2.5 = 14.5%.
3
2
42. (a) Thickness = M.S. Reading + Circular Scale Reading
(L.C.)
= 2´ 2+
Here LC =
Pitch
0.1
=
= 0.002 cm per
Circular scale division 50
division
So, correct measurement is measurement of integral
multiple of L.C.
43. (c) L.C. of vernier callipers = 1 MSD – 1 VSD
9ö
æ
= ç1 - ÷ ´ 1 = 0.1 mm = 0.01 cm
10
è
ø
Here 7th division of vernier scale coincides with a division
of main scale and the zero of vernier scale is lying right
side of the zero of main scale.
Zero error = 7 × 0.1 = 0.7 mm = 0.07 cm.
Length of the cylinder = measured value – zero error
= (3.1 + 4 × 0.01) – 0.07 = 3.07 cm.
44. (d) When screw on a screw-gauge is given six rotations,
it moves by 3mm on the main scale
3
= 0.5mm
6
Pitch 0.5 mm
=
\ Least count L.C. =
CSD
50
1
mm = 0.01 mm = 0.001cm
=
100
45. (None)
D1 = A1 + B1 + C1 = 24.36 + 0.0724 + 256.2 = 280.6
D2 = A2 + B2 + C2 = 24.44 + 16.082 + 240.2 = 280.7
D3 = A3 + B3 + C3 = 25.2 + 19.2812 + 236.183= 280.7
\
Pitch =
P-11
Physical World, Units and Measurements
D4 = A4 + B4 + C4 = 25 + 236.191 + 19.5 = 281
None of the option matches.
46. (c) Given, Length of simple pendulum, l = 25.0 cm
Time of 40 oscillation, T = 50s
Time period of pendulum
T = 2p
l
g
4p2 l
4p 2l
Þg= 2
g
T
Dg Dl 2DT
=
+
Þ Fractional error in g =
g
l
T
Þ T2 =
Dg æ 0.1 ö
æ 1ö
=ç
+ 2 ç ÷ = 0.044
÷
è 50 ø
g è 25.0 ø
Dg
\ Percentage error in g =
´ 100 = 4.4%
g
47. (Bonus) d = M = M = Ml -3
V
l3
Þ
0.10
æ 0.01ö
Dd DM
Dl
3
=
+ 3ç
=
+3
è 0.10 ÷ø = 0.31 kg/m ,
10.00
d
M
l
48. (d) A = 7 × 5.29 = 37.03 cm2
The result should have three significant figures, so
A = 37.0 cm2
49. (d) We have
T = 2p
l
2 l
or g = 4p 2
g
T
Dg
DR
DT
´ 100 =
´ 100 + 2
´ 100
g
Q
T
0.1
æ 1ö
´ 100 + 2 ç ÷ ´ 100
è 30 ø
55
= 0.18 + 6.67 = 6.8%
50. (b) Least count of main scale of screw gauge = 1 mm
Least count of screw gauge
=
=
Pitch
Number of division on circular scale
10-3
N
Þ N = 200
51. (c)
52. (c) Least count of screw gauge,
5 ´ 10 -6 =
Pitch
LC = No. of division
= 0.5 × 10–3 = 0.5 × 10–2 mm + ve error = 3 × 0.5 × 10–2 mm
= 1.5 × 10–2 mm = 0.015 mm
Reading = MSR + CSR – (+ve error)
= 5.5 mm + (48 × 0.5 × 10–2) – 0.015
= 5.5 + 0.24 – 0.015 = 5.725 mm
53. (c) = 1.5 % + 3 (1%) = 4.5%
54. (d) Maximum percentage error in A
= 3(% error in P) + 2(% error in Q)
1
+ (% error in R) + 1(% error in S)
2
1
= 3 ´ 0.5 + 2 ´ 1 + ´ 3 + 1 ´ 1.5
2
= 1.5 + 2 + 1.5 + 1.5 = 6.5%
55. (c) From Kepler's law, time period of a satellite,
r3
4p 2 3
T2 =
r
Gm
GM
Relative uncertainty in the mass of the earth
T = 2p
DM
DT
=2
= 2 ´ 10-2
M
T
(Q 4p & G constant and
Dr
negligible)
r
Ds
Dr
56. (c) Relative error in Surface area,
= 2 ´ = a and
s
r
Dv
Dr
relative error in volume,
= 3´
v
r
\ Relative error in volume w.r.t. relative error in area,
Dv 3
= a
v
2
Value of 1 part on main scale
57. (d) Least count =
Number of parts on vernier scale
0.25
cm = 5 × 10–4 cm
=
5×100
Reading = 4 × 0.05 cm + 30 × 5 × 10–4 cm
= (0.2 + 0.0150) cm = 0.2150 cm (Thickness of wire)
relative uncertainty in radius
rhg
´ 103
2
Relative error in surface tension,
58. (d) Surface tension, T =
DT Dr Dh
=
+
+ 0 (Q g, 2 & 103 are constant)
T
r
h
Percentage error
DT æ 10 –2 ´ 0.01 10 –2 ´ 0.01ö
= ç
+
÷ 100
T
è 1.25 ´ 10 –2 1.45 ´ 10 –2 ø
= (0.8 + 0.689)
= (1.489) = 1.489% @ 1.5%
59. (c) Given, P = a1/2 b2 c2 d–4,
Maximum relative error,
DP 1 Da
Db
Dc
Dd
=
+2
+3 +4
P
2 a
b
c
d
1
= ´ 2 + 2 ´ 1 + 3 ´ 3 + 4 ´ 5 = 32%
2
0.5
60. (d) L.C. =
= 0.01 mm
50
Zero error = 5 × 0.01 = 0.05 mm (Negative)
Reading = (0.5 + 25 × 0.01) + 0.05 = 0.80 mm
100 ´
P-12
Physics
| DT1 | + | DT2 | + | DT3 | + | DT4 |
4
2 +1+ 3 + 0
=
= 1.5
4
As the resolution of measuring clock is 1.5 therefore
the mean time should be 92 ± 1.5
L
62. (d) As, g = 4p2
T2
Dg
DL
DT
So,
´100 =
´100 + 2
´100
g
L
T
61. (c) DT =
0.1
1
´100 + 2 ´ ´100 = 2.72 ; 3%
=
20
90
0.1
63. (b) Least count =
= 0.01 cm
10
d1 = 0.5 + 8 × 0.01 + 0.03 = 0.61 cm
d2 = 0.5 + 4 × 0.01 + 0.03 = 0.57 cm
d3 = 0.5 + 6 × 0.01 + 0.03 = 0.59 cm
0.61 + 0.57 + 0.59
Mean diameter =
3
68. (b) According to the question.
t = (90 ± 1) or,
l = (20 ± 0.1) or,
t = 2p
69.
When, I = 5mA, e1000 V /T = 6mA
1000
T
Error = ± 0.01 (By exponential function)
)´
1000
´ (0.01) = 0.2 mA
300
65. (b) Measured length of rod = 3.50 cm
For Vernier Scale with 1 Main Scale Division = 1 mm
9 Main Scale Division = 10 Vernier Scale Division,
Least count = 1 MSD –1 VSD = 0.1 mm
66. (a) Rotation period of earth is about 24 hrs ; 105 s
Revolution period of earth is about 365 days ; 107 s
Speed of light wave C = 3 × 108 m/s
Wavelength of visible light of spectrum
l = 4000 – 7800 Å
70.
= (6 mA) ´
1
C = f l æç and T = ö÷
è
fø
Therefore period of light wave is 10–15 s (approx)
67. (d) In a voltmeter
V µl
V = kl
Now, it is given E = 1.1 volt for l1 = 440 cm
and V = 0.5 volt for l2 = 220 cm
Let the error in reading of voltmeter be DV then,
1.1 = 400 K and (0.5 – DV) = 220 K.
Þ
1.1 0.5 - DV
=
440
220
\
DV = -0.05 volt
l
4p 2l
Þ g= 2
g
t
1 ö
æ 0.1
Dg
Dt ö
æ Dl
+ 2 ´ ÷ = 0.027
= ± ç +2 ÷ = ç
90 ø
è l
è 20
g
t ø
Dg
% = 2.7%
\
g
(a) According to ohm’s law, V = IR
V
R=
I
Absolute error
´102
\ Percentage error =
Measurement
DV
DI
´100 =
´ 100 = 3%
where,
V
I
DR
DV
DI
´ 100 =
´ 102 +
´ 102
then,
R
V
I
= 3% + 3% = 6%
(c) Q Reading of Vernier = Main scale reading
+ Vernier scale reading × least count.
Main scale reading = 58.5
Vernier scale reading = 09 division
least count of Vernier = 0.5°/30
0.5°
Thus, R = 58.5° + 9 ×
30
R = 58.65°
(d) No. of divisions on main scale = N
No. of divisions on vernier scale = N + 1
size of main scale division = a
Let size of vernier scale division be b
then we have
aN
aN = b (N + 1) Þ b =
N +1
aN
Least count is a – b = a –
N +1
a
é N +1 - N ù
= aê
ú = N +1
ë N +1 û
(d) The least count (L.C.) of a screw guage is the smallest
length which can be measured accurately with it.
or,
I = (e1000 V /T - 1) mA (given)
V /T
Dl 0.1
=
l
20
Dg
%=?
g
As we know,
= 0.59 cm
64. (a) The current voltage relation of diode is
Also, dI = (e1000
Dt 1
=
t
90
71.
72.
1
cm
1000
Hence measured value should be recorded upto 3 decimal
places i.e., 5.320 cm
As least count is 0.001 cm =
P-13
Physical World, Units and Measurements
73. (a) Least count, L.C. =
1
mm
100
Diameter of wire = MSR + CSR × L.C.
Q 1 mm = 0.1 cm
= 0+
1
× 52 = 0.52 mm = 0.052 cm
100
74. (a) Number of significant figures in 23.023 = 5
Number of significant figures in 0.0003 = 1
Number of significant figures in 2.1 × 10–3 = 2
So, the radiation belongs to X-rays part of the spectrum.
75. (d) 30 Divisions of V.S. coincide with 29 divisions of M.S.
29
\ 1 V.S.D =
MSD
30
L.C. = 1 MSD – 1VSD
29
= 1 MSD MSD
30
1
MSD
30
1
´ 0.5° = 1 minute.
=
30
76. (a) Momentum, p = m × v
=
Given, mass of a body = 3.513 kg speed of body
= (3.513) × (5.00) = 17.565 kg m/s
= 17.6 (Rounding off to get three significant figures)
77. (d) Least count of screw gauge = 0.01 mm
Q
0.5
mm
50
Reading = [M.S.R. + C.S.R. × L.C.] – (zero error)
= [3 + 35 × 0.01] – (–0.03) = 3.38 mm
14
2
Physics
Motion in a Straight
Line
Distance, Displacement &
TOPIC 1
Uniform Motion
1.
TOPIC 2 Non-uniform Motion
A particle is moving with speed v = b x along positive
x-axis. Calculate the speed of the particle at time t = t (assume
that the particle is at origin at t = 0).
[12 Apr. 2019 II]
6.
The velocity (v) and time (t) graph of a body in a straight
line motion is shown in the figure. The point S is at 4.333
seconds. The total distance covered by the body in 6 s is:
[05 Sep. 2020 (II)]
b2 t
b2 t
b2 t
(b)
(c) b2 t
(d)
2
4
2
All the graphs below are intended to represent the same
motion. One of them does it incorrectly. Pick it up.
[2018]
v (m/s) 4
2
0
–2
(a)
2.
distance
velocity
(a)
3.
4.
5.
B
S
D
t (in s)
1 2 3 4 5 6
C
37
49
m (b) 12 m
(c) 11 m
(d)
m
3
4
The speed verses time graph for a particle is shown in the
figure. The distance travelled (in m) by the particle during
the time interval t = 0 to t = 5 s will be __________.
[NA 4 Sep. 2020 (II)]
(a)
position
(b)
time
7.
velocity
position
(c)
A
time
(d)
10
8
u
–1 6
(ms )
4
2
time
A car covers the first half of the distance between two
places at 40 km/h and other half at 60 km/h. The average
speed of the car is
[Online May 7, 2012]
(a) 40 km/h
(b) 45 km/h
(c) 48 km/h
(d) 60 km/h
The velocity of a particle is v = v0 + gt + ft2. If its position
is x = 0 at t = 0, then its displacement after unit time (t =
1) is
[2007]
(a) v0 + g /2 + f
(b) v0 + 2g + 3f
(c) v0 + g /2 + f/3
(d) v0 + g + f
A particle located at x = 0 at time t = 0, starts moving
along with the positive x-direction with a velocity 'v' that
varies as v = a x . The displacement of the particle
varies with time as
[2006]
2
1/2
3
(a) t
(b) t
(c) t
(d) t
1 2 3 4 5
time
(s)
8.
9.
The distance x covered by a particle in one dimensional
motion varies with time t as x2 = at2 + 2bt + c. If the
acceleration of the particle depends on x as x–n, where n
is an integer, the value of n is ______. [NA 9 Jan 2020 I]
A bullet of mass 20g has an initial speed of 1 ms–1, just
before it starts penetrating a mud wall of thickness 20 cm.
If the wall offers a mean resistance of 2.5×10–2 N, the speed
of the bullet after emerging from the other side of the wall
is close to :
[10 Apr. 2019 II]
(a) 0.1 ms–1
(b) 0.7 ms–1
(c) 0.3 ms–1
(d) 0.4 ms–1
P-15
Motion in a Straight Line
(a) a +
b2
4c
(b) a +
b2
3c
b2
b2
(d) a +
c
2c
11. A particle starts from origin O from rest and moves with a
uniform acceleration along the positive x-axis. Identify all
figures that correctly represents the motion qualitatively
(a = acceleration, v = velocity, x = displacement, t = time)
[8 Apr. 2019 II]
(c) a +
14. An automobile, travelling at 40 km/h, can be stopped at a
distance of 40 m by applying brakes. If the same automobile
is travelling at 80 km/h, the minimum stopping distance, in
metres, is (assume no skidding) [Online April 15, 2018]
(a) 75 m
(b) 160 m (c) 100 m (d) 150 m
15. The velocity-time graphs of a car and a scooter are shown
in the figure. (i) the difference between the distance
travelled by the car and the scooter in 15 s and (ii) the time
at which the car will catch up with the scooter are,
respectively
[Online April 15, 2018]
(a) 337.5m and 25s
(b) 225.5m and 10s
(c) 112.5m and 22.5s
A Car B
45
Velocity (ms –1) ®
10. The position of a particle as a function of time t, is given
by
x(t) = at + bt2 – ct3
where, a, b and c are constants. When the particle attains
zero acceleration, then its velocity will be:
[9 Apr. 2019 II]
30
(B)
(C)
(D)
G
15
O
0
(A)
F
Scooter
E
5
C
D
10 15 20 25
Time in (s) ®
(d) 11.2.5m and 15s
16. A man in a car at location Q on a straight highway is moving
with speed v. He decides to reach a point P in a field at a
distance d from highway (point M) as shown in the figure.
Speed of the car in the field is half to that on the highway.
What should be the distance RM, so that the time taken to
reach P is minimum?
[Online April 15, 2018]
P
d
(a) (B), (C)
(b) (A)
(c) (A), (B), (C)
(d) (A), (B), (D)
12. A particle starts from the origin at time t = 0 and moves
along the positive x-axis. The graph of velocity with
respect to time is shown in figure. What is the position
of the particle at time t = 5s?
[10 Jan. 2019 II]
v
(m/s)
3
Q
M
d
d
(c)
(d) d
2
2
17. Which graph corresponds to an object moving with a
constant negative acceleration and a positive velocity ?
[Online April 8, 2017]
(a)
d
3
R
(b)
(a)
2
(b)
Velocity
Velocity
1
0
13.
1 2 3 4 5 6 7 8 9 10
(a) 10 m
(b) 6 m
(c) 3 m
(d) 9 m
In a car race on straight road, car A takes a time t less
than car B at the finish and passes finishing point with
a speed 'v' more than of car B. Both the cars start from
rest and travel with constant acceleration a1 and a2
respectively. Then 'v' is equal to:
[9 Jan. 2019 II]
2a1 a 2
(a)
(b)
t
2a1 a 2 t
a1 + a 2
a1 + a 2
(c)
a1 a 2 t
(d)
t
2
Time
(c)
Time
(d)
Velocity
Velocity
Distance
Distance
18. The distance travelled by a body moving along a line in
time t is proportional to t3.
The acceleration-time (a, t) graph for the motion of the
body will be
[Online May 12, 2012]
P-16
Physics
(x1 – x2)
(x1 – x2)
a
a
(a)
(b)
(c)
t
t
a
(d)
t
The graph of an object’s motion (along the x-axis) is shown
in the figure. The instantaneous velocity of the object at
points A and B are vA and vB respectively. Then
[Online May 7, 2012]
23.
x(m)
15
10
B
5
Dt = 8
0
20.
24.
A
Dx = 4 m
10
25.
20 t (s)
(a) vA = vB = 0.5 m/s
(b) vA = 0.5 m/s < vB
(c) vA = 0.5 m/s > vB
(d) vA = vB = 2 m/s
An object, moving with a speed of 6.25 m/s, is decelerated
at a rate given by
26.
dv
= -2.5 v where v is the instantaneous speed. The time
dt
21.
taken by the object, to come to rest, would be: [2011]
(a) 2 s
(b) 4 s
(c) 8 s
(d) 1 s
A body is at rest at x = 0. At t = 0, it starts moving in the
positive x-direction with a constant acceleration. At the
same instant another body passes through x = 0 moving
in the positive x-direction with a constant speed. The
position of the first body is given by x1(t) after time ‘t’;
and that of the second body by x2(t) after the same time
interval. Which of the following graphs correctly
describes (x1 – x2) as a function of time ‘t’?
[2008]
(x1 – x2)
(x1 – x2)
(a)
O
t
(b)
(d)
O
t
f
to come to rest. If the
2
total distance traversed is 15 S , then
[2005]
1
(a) S = ft 2
(b) S = f t
6
1 2
1 2
ft
(c) S = ft
(d) S =
4
72
A particle is moving eastwards with a velocity of 5 ms–1.
In 10 seconds the velocity changes to 5 ms–1 northwards.
The average acceleration in this time is
[2005]
1 -2
(a)
ms towards north
2
1
(b)
ms - 2 towards north - east
2
1
(c)
ms - 2 towards north - west
2
(d) zero
The relation between time t and distance x is t = ax2 + bx
where a and b are constants. The acceleration is [2005]
(a) 2bv3
(b) –2abv 2 (c) 2av2
(d) –2av 3
An automobile travelling with a speed of 60 km/h, can
brake to stop within a distance of 20m. If the car is going
twice as fast i.e., 120 km/h, the stopping distance will be
[2004]
(a) 60 m
(b) 40 m
(c) 20 m
(d) 80 m
A car, moving with a speed of 50 km/hr, can be stopped by
brakes after at least 6 m. If the same car is moving at a
speed of 100 km/hr, the minimum stopping distance is
[2003]
(a) 12 m
(b) 18 m
(c) 24 m
(d) 6 m
If a body looses half of its velocity on penetrating 3 cm in
a wooden block, then how much will it penetrate more
before coming to rest?
[2002]
(a) 1 cm
(b) 2 cm
(c) 3 cm
(d) 4 cm.
Speeds of two identical cars are u and 4u at the specific
instant. The ratio of the respective distances in which the
two cars are stopped from that instant is
[2002]
(a) 1 : 1
(b) 1 : 4
(c) 1 : 8
(d) 1 : 16
and then decelerates at the rate
t
19.
t
22. A car, starting from rest, accelerates at the rate f through a
distance S, then continues at constant speed for time t
a
(c)
O
O
27.
28.
TOPIC 3 Relative Velocity
t
29. Train A and train B are running on parallel tracks in the
opposite directions with speeds of 36 km/hour and 72
km/hour, respectively. A person is walking in train A in
the direction opposite to its motion with a speed of 1.8
P-17
Motion in a Straight Line
km/hour. Speed (in ms–1) of this person as observed from
train B will be close to : (take the distance between the
tracks as negligible)
[2 Sep. 2020 (I)]
(a) 29.5 ms–1
(b) 28.5 ms–1
(c) 31.5 ms–1q
(d) 30.5 ms–1
30. A passenger train of length 60 m travels at a speed of 80
km/hr. Another freight train of length 120 m travels at a
speed of 30 km/h. The ratio of times taken by the
passenger train to completely cross the freight train when:
(i) they are moving in same direction, and (ii) in the
opposite directions is:
[12 Jan. 2019 II]
11
5
3
25
(a)
(b)
(c)
(d)
5
2
2
11
31. A person standing on an open ground hears the sound of
a jet aeroplane, coming from north at an angle 60º with
ground level. But he finds the aeroplane right vertically
above his position. If v is the speed of sound, speed of the
plane is:
[12 Jan. 2019 II]
32.
2v
v
3
v
(a)
(b)
(c) v
(d)
3
2
2
A car is standing 200 m behind a bus, which is also at rest.
The two start moving at the same instant but with different forward accelerations. The bus has acceleration 2 m/s2
and the car has acceleration 4 m/s2. The car will catch up
with the bus after a time of :
[Online April 9, 2017]
(a)
33.
34.
110 s
(b)
(c)
u+v
2
uv
(b)
1 2
u + v2
2
(d)
æ u 2 + v2 ö
ç
÷
2 ø
è
TOPIC 4 Motion Under Gravity
35.
2 æhö
ç ÷
3 ègø
h
g
(b) t = 1.8
æhö
(c) t = 3.4 ç ÷
ègø
(d) t =
2h
3g
36. A Tennis ball is released from a height h and after freely
falling on a wooden floor it rebounds and reaches height
h
. The velocity versus height of the ball during its motion
2
may be represented graphically by :
(graph are drawn schematically and on not to scale)
[4 Sep. 2020 (I)]
v
v
h/2
(a)
h/2
h
h(v) (b)
v
v
(c)
h(v)
h
h
h/2
h(v) (d)
h
h/2
h(v)
120 s
(c) 10 2 s
(d) 15 s
A person climbs up a stalled escalator in 60 s. If standing
on the same but escalator running with constant velocity
he takes 40 s. How much time is taken by the person to
walk up the moving escalator? [Online April 12, 2014]
(a) 37 s
(b) 27 s
(c) 24 s
(d) 45 s
A goods train accelerating uniformly on a straight railway
track, approaches an electric pole standing on the side of
track. Its engine passes the pole with velocity u and the
guard’s room passes with velocity v. The middle wagon of
the train passes the pole with a velocity.
[Online May 19, 2012]
(a)
(a) t =
A helicopter rises from rest on the ground vertically upwards with a constant acceleration g. A food packet is
dropped from the helicopter when it is at a height h. The
time taken by the packet to reach the ground is close to
[g is the accelertion due to gravity] : [5 Sep. 2020 (I)]
37. A ball is dropped from the top of a 100 m high tower on a
1
planet. In the last s before hitting the ground, it covers a
2
distance of 19 m. Acceleration due to gravity (in ms–2) near
the surface on that planet is _______.
[NA 8 Jan. 2020 II]
38. A body is thrown vertically upwards. Which one of the
following graphs correctly represent the velocity vs time?
[2017]
(a)
(c)
(b)
(d)
39. Two stones are thrown up simultaneously from the edge
of a cliff 240 m high with initial speed of 10 m/s and 40
m/s respectively. Which of the following graph best
represents the time variation of relative position of the
second stone with respect to the first ?
P-18
Physics
(Assume stones do not rebound after hitting the ground
and neglect air resistance, take g = 10 m/ s2)
[2015]
y
h
(The figures are schematic and not drawn to scale)
(a)
240
(b)
(y2 – y1) m
240
(c)
(c)
240
41.
240
12
t(s)
t(s)
From a tower of height H, a particle is thrown vertically
upwards with a speed u. The time taken by the particle, to
hit the ground, is n times that taken by it to reach the
highest point of its path. The relation between H, u and n
is:
[2014]
2
2
2
2
(a) 2gH = n u
(b) gH = (n – 2) u d
(c) 2gH = nu2 (n – 2)
(d) gH = (n – 2)u2
Consider a rubber ball freely falling from a height h = 4.9 m
onto a horizontal elastic plate. Assume that the duration
of collision is negligible and the collision with the plate is
totally elastic.
Then the velocity as a function of time and the height as
a function of time will be :
[2009]
v
(a)
y
h
t
O
–v1
t
v
+v1
(b)
O
–v1
y
t1
2t1
4t1
t
t
O
t
t
42.
(y2 – y1) m
12
+v1
y
h
t(s)
(d)
12
t
2t1
v
v1
8
(y2 – y1 ) m
t® 8
40.
t(s)
12
t1
t
(y2 – y1) m
(d)
8
O
h
t
A parachutist after bailing out falls 50 m without friction. When
parachute opens, it decelerates at 2 m/s2 . He reaches the ground
with a speed of 3 m/s. At what height, did he bail out ? [2005]
(a) 182 m
(b) 91 m
(c) 111m
(d) 293m
43. A ball is released from the top of a tower of height h meters.
It takes T seconds to reach the ground. What is the position
T
of the ball at
second
[2004]
3
8h
(a)
meters from the ground
9
7h
meters from the ground
(b)
9
(c)
h
meters from the ground
9
17 h
meters from the ground
18
44. From a building two balls A and B are thrown such that A is
thrown upwards and B downwards (both vertically). If vA
and vB are their respective velocities on reaching the
ground, then
[2002]
(a) vB > vA
(b) vA = vB
(c) vA > vB
(d) their velocities depend on their masses.
(d)
P-19
Motion in a Straight Line
1.
x
(b) Given, v = b x
dx
= b x1/2
or
dt
x
or
òx
-1/2
0
t
x
Þ 2 x = at Þ x =
dx = ò bdt
6.
0
(a)
1/2
x
b2t 2
or
= 6t
or x =
1/ 2
4
Differentiating w. r. t. time, we get
dx b2 ´ 2t
=
dt
4
(t = t)
a2 2
t
4
4 A B
v(m/s) 2
O
S D
0 1 2 3 4 5 6
–2
C
t
(in s)
1 13
=
3 3
1 5
SD = 2 - =
3 3
Distance covered by the body = area of v-t graph
= ar (OABS) + ar (SCD)
OS = 4 +
2
b t
2
(b) Graphs in option (c) position-time and option (a)
velocity-position are corresponding to velocity-time graph
option (d) and its distance-time graph is as given below.
Hence distance-time graph option (b) is incorrect.
distance
or v =
2.
t
é2 x ù
ò x = a ò dt ; ê 1 ú = a[t ]t0
0
0
ë
û0
dx
1 æ 13 ö
1 5
32 5 37
+ =
m
ç + 1÷ ´ 4 + ´ ´ 2 =
2è 3 ø
2 3
3 3 3
u
(20)
A
8
=
7.
time
3.
(c)
Average speed =
Total distance travelled x
=
Total time taken
T
x
= 48 km/h
x
x
+
2 ´ 40 2 ´ 60
dx
(c) We know that, v =
dt
Þ dx = v dt
B t
5
Distance travelled = Area of speed-time graph
O
=
4.
x
t
0
0
8.
5.
(3) Distance X varies with time t as x2 = at2 + 2bt + c
dx
= 2at + 2b
dt
dx
dx (at + b)
Þx
= at + b Þ
=
dt
dt
x
Þ 2x
Integrating, ò dx = ò v dt
t
1
´ 5 ´ 8 = 20 m
2
=
t
é
gt 2 ft 3 ù
or x = ò (v0 + gt + ft ) dt = êv0 t +
+
ú
2
3 úû
ëê
0
0
gt 2 ft 3
or, x = v0 t +
+
2
3
g f
At t = 1, x = v0 + + .
2 3
(a) v = a x ,
dx
dx
=a x Þ
Þ
= a dt
dt
x
Integrating both sides,
2
2
Þx
d 2 x æ dx ö
+ç ÷ =a
dt 2 è dt ø
2
æ dx ö
æ at + b ö
a -ç ÷
a -ç
÷
2
d x
è dt ø =
è x ø
Þ
=
x
x
dt 2
=
Þ
ax 2 - ( at + b )
x3
a µ x–3
2
=
ac - b 2
x3
Hence, n = 3
2
P-20
9.
Physics
(b) From the third equation of motion
But, a =
a1 – a 2
Putting this value of t0 in equation (i)
F
m
2
=
é 2.5 ´ 10-2 ù 20
Þ v 2 = (1)2 - (2) ê
ú
-3
ëê 20 ´10 ûú 100
Þ v2 = 1 –
Þv=
1
m/s = 0.7m/s
2
10. (b) x = at + bt2 – ct3
dx d
= (at + bt 2 + ct 3 )
dt dt
= a + 2bt – 3ct2
11.
dv
d
= (a + 2bt - 3ct 2 )
dt
dt
æb ö
\ t =ç ÷
è 3c ø
2ö
2 æ
æb ö
æ b ö = a+b
a
+
2
b
3
c
ç
÷
and v =
ç ÷
ç ÷
3c ø
è
è 3c ø
è 3c ø
(d) For constant acceleration, there is straight line
®
parallel to t-axis on a - t .
®
®
Inclined straight line on v - t , and parabola on x - t .
12. (d) Position of the particle,
S = area under graph (time t = 0 to 5s)
1
= ´ 2 ´ 2 + 2 ´ 2 + 3´1= 9m
2
13. (c) Let time taken by A to reach finishing point is t0
\ Time taken by B to reach finishing point = t0 + t
x
vA = a1t0
vB = a2(t0 + t)
vA – vB = v
Þ v = a1 t0 – a2 (t0 + t) = (a1 – a2)t0–a2t ...(i)
1
1
x B = x A = a1 t 02 = a 2 (t 0 + t) 2
2
2
Þ a1 t 0 = a 2 ( t 0 + t )
(
)
a2 t – a 2t =
a1a 2 t + a 2 t – a 2 t
v –u
1 2
and S = ut + at
t
2
1 (45)
Distance travelled by car in 15 sec =
(15)2
2 15
675
=
m
2
Distance travelled by scooter in 15 seconds = 30 × 15 = 450
(Q distance = speed × time)
Difference between distance travelled by car and scooter
in 15 sec, 450 – 337.5 = 112.5 m
Let car catches scooter in time t;
675
+ 45(t –15) = 30t
2
337.5 + 45t – 675 = 30t
Þ 15t = 337.5
Þ t = 22.5 sec
16. (a) Let the car turn of the highway at a distance 'x' from
the point M. So, RM = x
And if speed of car in field is v, then time taken by the car
to cover the distance QR = QM – x on the highway,
QM - x
.....(i)
2v
Time taken to travel the distance 'RP ' in the field
t1 =
d 2 + x2
..... (ii)
v
Total time elapsed to move the car from Q to P
t2 =
u=0
Þ
a1 + a 2
– a 2t
15. (c) Using equation, a =
Velocity, v =
or 0 = 2b – 3c × 2t
(
a1 – a 2
or, v = a1a 2 t
14. (b) According to question, u1 = 40 km/h, v1 = 0 and s1 = 40 m
using v2 – u2 = 2as; 02 – 402 = 2a × 40 ...(i)
Again, 02 – 802 = 2as
...(ii)
From eqn. (i) and (ii)
Stopping distance, s = 160 m
1
2
Acceleration,
a2t
v =( al – a2 )
æFö
\v = u - 2ç ÷ S
èmø
2
a 2t
Þ to =
v2 – u2 = 2aS
)
a1 – a 2 t 0 = a 2 t
QM - x
d 2 + x2
+
v
2v
dt
=0
For 't ' to be minimum
dx
ù
1é 1
x
ê- +
ú =0
v ëê 2
d 2 + x 2 ûú
t = t1 + t2 =
d
d
=
or x =
3
22 - 1
Q
P
d
R
M
P-21
Motion in a Straight Line
17. (c) According to question, object is moving with
constant negative acceleration i.e., a = – constant (C)
vdv
= -C
dx
vdv = – Cdx
v2
v2 k
= - Cx + k
x=+
2
2C C
Hence, graph (3) represents correctly.
18. (b) Distance along a line i.e., displacement (s)
= t3 (Q s µ t 3 given)
By double differentiation of displacement, we get
acceleration.
3
2
ds dt
dv d 3t
=
= 3t 2 and a =
=
= 6t
dt
dt
dt
dt
a = 6t or a µ t
Hence graph (b) is correct.
V=
19. (a)
Instantaneous velocity v =
Dx
Dt
Dx A 4 m
=
= 0.5 m/s
Dt A
8s
Dx
8m
and vB = B =
= 0.5 m/s
Dt B 16s
i.e., vA = vB = 0.5 m/s
From graph, vA =
dv
20. (a) Given,
= -2.5 v
dt
dv
Þ
= – 2.5 dt
v
0
-½
This equation is of parabola.
For t <
For t =
v
; the slope is negative
a
v
; the slope is zero
a
v
; the slope is positive
a
These characteristics are represented by graph (b).
22. (d) Let car starts from A from rest and moves up to point
B with acceleration f.
For t >
1 2
ft1
2
Distance, BC = (ft1)t
Distance, AB = S =
t
dv = -2.5ò dt
0
A f B
t1
Þ
0
é v +½ ù
t
= -2.5 [ t ]0
Þ ê (½) ú
êë
úû 6.25
C f /2 D
2t 1
t
............. (i)
f t1t = 12 S
1 2
f t1 = S
2
Þ – 2(6.25)½ = – 2.5t
Þ – 2 × 2.5 = –2.5t
............ (ii)
Dividing (i) by (ii), we get t1 =
Þ t = 25
21. (b) For the body starting from rest, distance travelled
(x1) is given by
1
x1 = 0 + at2
2
1 2
Þ x1 = at
2
Þ S=
2
23. (c)
v2
N
D v = v 2 + (-v 1 )
W
t
t
6
1 ætö
f t2
fç ÷ =
2 è 6ø
72
x1 – x2
v/a
( ft1 )2
u2
=
= ft12 = 2S
2a 2( f / 2)
Distance, CD =
15 S
Total distance, AD = AB + BC + CD = 15S
AD = S + BC + 2S
Þ S + f t1t + 2 S = 15 S
Integrating,
ò6.25 v
For the body moving with constant speed
x2 = vt
1
\ x1 - x2 = at 2 - vt
2
at t = 0, x1 – x2 = 0
90°
- v1
uur
Initial velocity, v1 = 5iˆ,
v1
S
E
P-22
Physics
uur
Final velocity, v2 = 5 ˆj,
uur ur
ur
Change in velocity D v = (v 2 - v 1 )
=
5 2 + 52 + 0 = 5 2m/s
[As | v1 | = | v2 | = 5 m/s]
uur
Dv
Avg. acceleration =
t
250 ´ 250
» = –16 ms–2.
324 ´ 2 ´ 6
Case-2 : Initial velocity, u = 100 km/hr
a=–
5 2
1
=
m / s2
10
2
5
tan q =
= -1
-5
which means q is in the second quadrant.
(towards north-west)
24. (d) Given, t = ax2 + bx;
Diff. with respect to time (t)
d
d
dx
dx
(t ) = a ( x 2 ) + b
= a.2 x + b.v.
dt
dt
dt
dt
Þ 1 = 2axv + bv = v(2ax + b)(v = velocity)
1
2ax + b = .
v
Again differentiating, we get
500 ´ 500
= 24m
324 ´ 32
27. (a) In first case
s=
u1 = u ; v1 =
u
, s = 3 cm, a1 = ?
2 1
...(i)
2
æ dx
ö
= v÷
çèQ
ø
dt
–u2
8
In second case: Assuming the same retardation
Þ
a=
u2 = u /2 ; v2 = 0 ; s2 = ?; a2 =
v22 - u 22 = 2a2 ´ s2
…(i)
5
18
100
m/s
3
and (0)2 – u¢2 = –2ad¢
or u¢2 = 2ad¢
…(ii)
(ii) divided by (i) gives,
d'
4 = Þ d ' = 4 ´ 20 = 80m
d
26. (c) Fir first case : Initial velocity,
5
u = 50 ´ m / s,
18
v = 0,s = 6m, a = a
Using, v 2 - u 2 = 2as
2
5ö
æ
Þ - ç 100 ´ ÷ = 2 × (–16) × 5
18 ø
è
æ uö
2
çè ÷ø - u = 2 × a × 3
2
In second case speed, u¢ = 120 ´
=
2
5ö
æ
Þ 02 - ç100 ´ ÷ = 2as
18 ø
è
Using, v12 - u12 = 2a1s1
dx
1 dv
+0= - 2
dt
v dt
dv
= –2av3
dt
25. (d) In first case speed,
5
50
u = 60 ´ m/s =
m/s
18
3
d = 20m,
Let retardation be a then
(0)2 – u2 = –2ad
or u2 = 2ad
5
m/sec
18
v = 0, s = s, a = a
As v2 – u2 = 2as
= 100 ´
=
Þ a=
2
5ö
æ
Þ - ç 50 ´ ÷ = 2 ´ a ´ 6
è
18 ø
v12 + v22 + 2v1v2 cos 90
=
2a
2
5ö
æ
Þ 02 - ç 50 ´ ÷ = 2 ´ a ´ 6
è
18 ø
-u 2
8
...(ii)
æ –u2 ö
u2
= 2ç
÷ ´ s2
4
è 8 ø
Þ s2 = 1 cm
28. (d) For first car
u1 = u, v1 = 0, a1 = – a, s1 = s1
\ 0-
As v12 - u12 = 2a1s1
Þ –u2 = –2as1
Þ u2 = 2as1
u2
2a
For second car
u2 = 4u, v1 = 0, a2 = – a, s2 = s2
Þ s1 =
\
v22 - u22 = 2a2 s2
Þ
–(4u)2 = 2(–a)s2
...(i)
P-23
Motion in a Straight Line
Þ 16 u2 = 2as2
8u 2
Þ s2 =
a
Dividing (i) and (ii),
29.
...(ii)
s1 u 2 a
1
=
× 2 =
s2 2a 8u
16
(a) According to question, train A and B are running on
parallel tracks in the opposite direction.
1.8 km/h
36 km/h
A
B
v
R (Observer)
Distance, PQ = vp × t (Distance = speed × time)
Distance, QR = V.t
PQ
QR
120
= 24 second.
5
34 (d) Let 'S' be the distance between two ends 'a' be the
constant acceleration
As we know v2 – u2 = 2aS
vc2 = u 2 + aS
vc2 = u 2 +
v2 - u 2
2
u 2 + v2
2
35. (c) For upward motion of helicopter,
vc =
v2 = u 2 + 2 gh Þ v 2 = 0 + 2 gh Þ v = 2gh
Now, packet will start moving under gravity.
Let 't' be the time taken by the food packet to reach the
ground.
1
s = ut + at 2
2
1
1
Þ -h = 2 gh t - gt 2 Þ gt 2 - 2 gh t - h = 0
2
2
or, t =
Bus
2´
2 m/sec2
200 m
Given, uC = uB = 0, aC = 4 m/s2, aB = 2 m/s2
hence relative acceleration, aCB = 2 m/sec2
1
Now, we know, s = ut + at 2
2
1
200 = ´ 2t 2 Q u = 0
2
Hence, the car will catch up with the bus after time
t = 10 2 second
1
1
15 "escalator"
+
=
60 40 120
second
2 gh ± 2 gh + 4 ´
1 vp ´ t
v
=
Þ vp =
2
V.t
2
4 m/sec2
Car
So, the person’s speed is
v2 - u2
2
Let v be velocity at mid point.
S
2
2
Therefore, vc - u = 2a
2
VMA = –1.8 km/h = –0.5 m/s
Vman, B = Vman, A + VA, B
= Vman, A + VA – VB = –0.5 + 10 – (–20)
= – 0.5 + 30 = 29.5 m/s.
30. (a)
vP
P
31. (d) Q
o
60
32. (c)
1 "escalator"
40 second
Walking with the escalator going, the speed add.
or, aS =
VB = -72 km/h = –20 m/s
cos 60° =
Person’s speed walking only is
So, the time to go up the escalator t =
VA = 36 km/h = 10 m/s
72 km/h
1 "escalator"
60 second
Standing the escalator without walking the speed is
33. (c)
or, t =
g
´h
2
g
2
2 gh
(1 + 2) Þ t =
g
or, t = 3.4
2h
(1 + 2)
g
h
g
36. (c) For uniformly accelerated/ deaccelerated motion :
v 2 = u 2 ± 2 gh
As equation is quadratic, so, v-h graph will be a parabola
P-24
Physics
v
at t= 0, h = d
2
1 ® 2 : V increases downwards
h 2 ® velocity changes its direction
2 ® 3 : V decreases upwards
d
1
3
collision
takes 2
place
v=
Using, S = ut +
Þ S = 0´t +
1 2
gt
2
1 2
gt
2
Þ 200 = gt2
Þt=
In last
200
g
[Q 2S = 100m]
…(i)
1
s, body travels a distance of 19 m, so in
2
æ 1ö
çt – ÷
è 2ø
distance travelled = 81
2
1 æ 1ö
g ç t – ÷ = 81
2 è 2ø
2
æ 1ö
\ g ç t – ÷ = 81´ 2
è 2ø
81´ 2
æ 1ö
Þ çt – ÷ =
2
g
è
ø
Þ
u + 2 gh
H
37. (08.00) Let the ball takes time t to reach the ground
\
u
2
Now, v = u + at
Initially velocity is downwards (–ve) and then after
collision it reverses its direction with lesser magnitude, i.e.
velocity is upwards (+ve).
Note that time t = 0 corresponds to the point on the graph
where h = d.
Next time collision takes place at 3.
Now,
y1 = 10t – 5t2 ; y2 = 40t – 5t2
y1 = – 240m, t = 8s
y2 – y1 = 30t for t < 8s.
t > 8s,
1
y2 – y1 = 240 – 40t – gt2
2
40. (c) Speed on reaching ground
39. (b)
for
\
for
1
1
=
( 200 – 81 ´ 2)
2
g
using (i)
g = 2(10 2 – 9 2)
Þ g =2 2
\ g = 8 m/s2
38. (a) For a body thrown vertically upwards acceleration
remains constant (a = – g) and velocity at anytime t is
given by V = u – gt
During rise velocity decreases linearly and during fall
velocity increases linearly and direction is opposite to
each other.
Hence graph (a) correctly depicts velocity versus time.
Þ
u 2 + 2 gh = -u + gt
Time taken to reach highest point is t =
Þt =
u + u 2 + 2 gH
g
(from question)
=
u
,
g
nu
g
Þ 2gH = n(n –2)u2
41. (b) For downward motion v = –gt
The velocity of the rubber ball increases in downward
direction and we get a straight line between v and t with a
negative slope.
1 2
Also applying y - y0 = ut + at
2
1 2
1 2
We get y - h = - gt Þ y = h - gt
2
2
The graph between y and t is a parabola with y = h at t = 0.
As time increases y decreases.
For upward motion.
The ball suffer elastic collision with the horizontal elastic
plate therefore the direction of velocity is reversed and the
magnitude remains the same.
Here v = u – gt where u is the velocity just after collision.
As t increases, v decreases. We get a straight line between v
and t with negative slope.
1 2
Also y = ut - gt
2
All these characteristics are represented by graph (b).
42. (d) Initial velocity of parachute
after bailing out,
u=
2gh
u = 2 ´ 9.8 ´ 50 = 14 5
The velocity at ground,
v = 3m/s
v2 - u2
32 - 980
» 243 m
S=
=
2´2
4
Initially he has fallen 50 m.
\ Total height from where
he bailed out = 243 + 50 = 293 m
50 m
v
a = - 2 m / s2
3m / s
P-25
Motion in a Straight Line
43. (a)
We have s = ut +
44. (b)
1 2
gt ,
2
velocity u
1
Þ h = 0 × T + gT2
2
1 2
Þ h = gT
2
Vertical distance moved in time
Ball A is thrown upwards with
from the building. During its
downward journey when it comes
back to the point of throw, its
speed is equal to the speed of
throw
(u). So, for the journey of both
the balls from point A to B.
T
is
3
2
1 æTö
1 gT 2 h
h' = g ç ÷ Þ h' = ´
=
2 è 3ø
2
9
9
\ Position of ball from ground = h -
h
8h
=
9
9
We can apply v2 – u2 = 2gh.
As u, g, h are same for both the
balls, vA = vB
A
h
B
u
u
P-26
3
Physics
Motion in a Plane
5.
TOPIC 1 Vectors
1.
2.
3.
®
A force F = (i$ + 2 $j + 3k$ ) N acts at a point (4$i + 3 $j - k$ ) m.
Then the magnitude of torque about the point ($i + 2 $j + k$ ) m
will be x N-m. The value of x is ______.
(
[NA 7 Jan. 2020 II]
uuur
uur uuur
A 2 = 5 and A1 + A 2 = 5. The value of
uur uuur
3A1 - 2A 2 is :
[8 April 2020 II]
)(
(a) – 106.5
4.
(c) 90°
6.
[NA Sep. 05, 2020 (I)]
r
r
r
r
The sum of two forces P and Q is R such that | R | =
r
r
| P | . The angle q (in degrees) that the resultant of 2 P
r
r
and Q will make with Q is _______.
uur
Let A1 = 3,
uur uuur
2A1 + 3A 2 ·
)
Two forces P and Q, of magnitude 2F and 3F, respectively,
are at an angle q with each other. If the force Q is
doubled, then their resultant also gets doubled. Then, the
angle q is:
[10 Jan. 2019 II]
(a) 120°
(b) 60°
(d) 30°
ur
ur
Two vectors A and B have equal magnitudes. The
ur ur
ur ur
magnitude of A + B is ‘n’ times the magnitude of A - B .
ur
ur
The angle between A and B is:
[10 Jan. 2019 II]
(
é n 2 - 1ù
(a) cos -1 ê 2 ú
ë n + 1û
7.
(b) – 99.5
(c) – 112.5
(d)
– 118.5
In the cube of side ‘a’ shown in the figure, the vector
from the central point of the face ABOD to the central
point of the face BEFO will be:
[10 Jan. 2019 I]
8.
(
)
1 ˆ ˆ
a i -k
(b)
2
(c)
(
)
(d)
1
a ˆj - iˆ
2
(
)
(
)
1
a ˆj - kˆ
2
9.
(
)
é n - 1ù
(b) cos -1 ê
ë n + 1úû
é n2 -1ù
é n -1ù
(c) sin -1 ê 2 ú
(d) sin -1 ê
n
+
1
ë n + 1úû
ë
û
r
r
Let A = (iˆ + ˆj) and B = (iˆ - ˆj) . The magnitude of a
r
r r r r r r
coplanar vector C such that A.C = B.C = A.B is given
by
[Online April 16, 2018]
(a)
5
9
(b)
10
9
(c)
20
9
(d)
9
12
ur
A vector A is rotated by a small angle Dq radian (Dq << 1)
ur
ur ur
to get a new vector B . In that case B - A is :
[Online April 11, 2015]
ur
ur
ur
(a) A Dq
(b) B Dq - A
ur æ Dq2 ö
A çç 1 (d) 0
÷
2 ÷ø
è
r r r r
If A ´ B = B ´ A, then the angle between A and B is [2004]
p
p
(a)
(b)
2
3
p
(c) p
(d)
4
(c)
1
a kˆ - iˆ
(a)
2
)
P-27
Motion in a Plane
15. A particle starts from the origin at t = 0 with an initial
velocity of 3.0 iˆ m/s and moves in the x-y plane with a
Motion in a Plane with
TOPIC 2
Constant Acceleration
10.
A balloon is moving up in air vertically above a point A on the
ground. When it is at a height h1, a girl standing at a distance
d (point B) from A (see figure) sees it at an angle 45º with
respect to the vertical. When the balloon climbs up a further
height h2, it is seen at an angle 60º with respect to the vertical
if the girl moves further by a distance 2.464 d (point C). Then the
height h2 is (given tan 30º = 0.5774):
[Sep. 05, 2020 (I)]
is given as v . Then v (in m/s) is____ [NA 8 Jan. 2020 I]
r
17. A particle moves such that its position vector r (t) = cos
h2
h1
A
60°
45°
d
(a) 1.464 d
11.
B 2.464d C
(b) 0.732 d
(c) 0.464 d
(d) d
Starting from the origin at time t = 0, with initial velocity
5 ˆj ms–1, a particle moves in the x–y plane with a constant
acceleration of (10iˆ + 4 ˆj) ms–2. At time t, its coordiantes
are (20 m, y0 m). The values of t and y0 are, respectively :
[Sep. 04, 2020 (I)]
(a) 2 s and 18 m
(b) 4 s and 52 m
(c) 2 s and 24 m
(d) 5 s and 25 m
12.
The position vector of a particle changes with time
r
according to the relation r (t) = 15 t 2 $i + (4 - 20 t 2 ) $j.
What is the magnitude of the acceleration at t = 1?
(a) 40
13.
(b) 25
[9 April 2019 II]
(d) 50
(c) 100
(
)
A particle moves from the point 2.0iˆ + 4.0 ˆj m , at t = 0,
(
)
with an initial velocity 5.0iˆ + 4.0 ˆj ms -1 . It is acted upon
by a constant force which produces a constant acceleration
4.0iˆ + 4.0 ˆj ms -2 . What is the distance of the particle
(
14.
constant acceleration (6.0 iˆ + 4.0 ˆj) m/ s 2. The xcoordinate of the particle at the instant when its ycoordinate is 32 m is D meters. The value of D is:
[9 Jan. 2020 II]
(a) 32
(b) 50
(c) 60
(d) 40
16. A particle is moving along the x-axis with its coordinate
with time ‘t’ given by x(t) = 10 + 8t – 3t2. Another particle is
moving along the y-axis with its coordinate as a function of
time given by y(t) = 5 – 8t3. At t = 1 s, the speed of the
second particle as measured in the frame of the first particle
)
from the origin at time 2s?
[11 Jan. 2019 II]
(a) 15 m
(b) 20 2m
(c) 5 m
(d) 10 2m
A particle is moving with a velocity vr = K (y iˆ + x ĵ ),
where K is a constant. The general equation for its path
is:
[9 Jan. 2019 I]
2
2
(a) y = x + constant
(b) y = x + constant
(c) y2 = x2 + constant
(d) xy = constant
wt iˆ + sin wt ĵ where w is a constant and t is time. Then
r
which of the following statements is true for the velocity v
r
(t) and acceleration a (t) of the particle: [8 Jan. 2020 II]
r
r
r
(a) v is perpendicular to r and a is directed away from
the origin
r
r
r
(b) v and a both are perpendicular to r
r
r
r
(c) v and a both are parallel to r
r
r
r
(d) v is perpendicular to r and a is directed towards
the origin
r
18. A particle is moving with velocity n = k ( yiˆ + xjˆ) , where k
is a constant. The general equation for its path is [2010]
(a) y = x2 + constant
(b) y2 = x + constant
(c) xy = constant
(d) y2 = x2 + constant
19. A particle has an initial velocity of 3iˆ + 4 ˆj and an
acceleration of 0.4iˆ + 0.3 ˆj . Its speed after 10 s is : [2009]
(b) 7 units
(a) 7 2 units
(c) 8.5 units
(d) 10 units
20. The co-ordinates of a moving particle at any time ‘t’are
given by x = a t 3 and y = b t 3 . The speed of the particle
at time ‘t’ is given by
[2003]
(a) 3t a 2 + b2
(b) 3t 2 a 2 + b2
(c) t 2 a 2 + b2
(d)
a 2 + b2
TOPIC 3 Projectile Motion
21. A particle of mass m is projected with a speed u from the
p
w.r.t. horizontal (x-axis). When
3
it has reached its maximum height, it collides completely
ground at an angle q =
inelastically with another particle of the same mass and
velocity uiˆ. The horizontal distance covered by the combined
mass before reaching the ground is:
[9 Jan. 2020 II]
P-28
Physics
(a)
3 3 u2
8 g
(b)
3 2 u2
4 g
5 u2
u2
(d) 2 2
g
8 g
The trajectory of a projectile near the surface of the earth
is given as y = 2x – 9x2. If it were launched at an angle q0
with speed v0 then (g = 10 ms–2):
[12 April 2019 I]
(c)
22.
(a) q0 = sin –1
1
5
and v0 =
5
ms–1
3
24.
25.
26.
(a) 1.0 m (b) 4.2 m
(c) 6.1 m
(d) 9.8 m
æ 2 ö
3
(b) q0 = cos–1 çè 5 ÷ø and v0 = ms–1
5
29. The position of a projectile launched from the origin at t =
r
0 is given by r = 40iˆ + 50 ˆj m at t = 2s. If the projectile
æ 1 ö
9
(c) q0 = cos–1 çè ÷ø and v0 = ms–1
5
3
was launched at an angle q from the horizontal, then q is
(take g = 10 ms–2)
[Online April 9, 2014]
2
-1
-1 3
(a) tan
(b) tan
3
2
4
7
-1
-1
(c) tan
(d) tan
5
4
30. A projectile is given an initial velocity of (iˆ + 2 ˆj ) m/s,
æ 2 ö
3
(d) q0 = sin –1 çè ÷ø and v0 = ms–1
5
5
23.
27. Two guns A and B can fire bullets at speeds 1 km/s and
2 km/s respectively. From a point on a horizontal
ground, they are fired in all possible directions. The
ratio of maximum areas covered by the bullets fired by
the two guns, on the ground is:
[10 Jan. 2019 I]
(a) 1 : 16 (b) 1 : 2
(c) 1 : 4
(d) 1 : 8
28. The initial speed of a bullet fired from a rifle is 630 m/s. The
rifle is fired at the centre of a target 700 m away at the same
level as the target. How far above the centre of the target ?
[Online April 11, 2014]
A shell is fired from a fixed artillery gun with an initial
speed u such that it hits the target on the ground at a
distance R from it. If t1 and t2 are the values of the time
taken by it to hit the target in two possible ways, the
product t1t2 is :
[12 April 2019 I]
(a) R/4g (b) R/g
(c) R/2g
(d) 2R/g
Two particles are projected from the same point with the
same speed u such that they have the same range R, but
different maximum heights, h1 and h2. Which of the
following is correct ?
[12 April 2019 II]
2
2
(a) R = 4 h1h2
(b) R =16 h1h2
2
(c) R = 2 h1h2
(d) R2 = h1h2
A plane is inclined at an angle a = 30o with respect to the
horizontal. A particle is projected with a speed u = 2 ms–1,
from the base of the plane, as shown in figure. The distance
from the base, at which the particle hits the plane is close to
: (Take g=10 ms–2)
[10 April 2019 II]
(a) 20 cm (b) 18 cm
(c) 26 cm
(d) 14 cm
A body is projected at t = 0 with a velocity 10 ms–1 at an
angle of 60° with the horizontal. The radius of curvature
of its trajectory at t = 1s is R. Neglecting air resistance
and taking acceleration due to gravity g = 10 ms–2, the
value of R is:
[11 Jan. 2019 I]
(a) 10.3 m
(b) 2.8 m
(c) 2.5 m
(d) 5.1 m
(
)
where iˆ is along the ground and ĵ is along the vertical.
If g = 10 m/s2 , the equation of its trajectory is : [2013]
(a) y = x - 5 x 2
(b) y = 2 x - 5 x 2
(c) 4 y = 2 x - 5 x 2
(d) 4 y = 2 x - 25 x 2
31. The maximum range of a bullet fired from a toy pistol
mounted on a car at rest is R0= 40 m. What will be the acute
angle of inclination of the pistol for maximum range when
the car is moving in the direction of firing with uniform
velocity v = 20 m/s, on a horizontal surface ? (g = 10 m/s 2)
[Online April 25, 2013]
(a) 30°
(b) 60°
(c) 75°
(d) 45°
32. A ball projected from ground at an angle of 45° just clears
a wall in front. If point of projection is 4 m from the foot of
wall and ball strikes the ground at a distance of 6 m on the
other side of the wall, the height of the wall is :
[Online April 22, 2013]
(a) 4.4 m (b) 2.4 m
(c) 3.6 m
(d) 1.6 m
33. A boy can throw a stone up to a maximum height of 10 m.
The maximum horizontal distance that the boy can throw
the same stone up to will be
[2012]
(a) 20 2 m
(b) 10 m
(c) 10 2 m
(d) 20 m
34. A water fountain on the ground sprinkles water all around
it. If the speed of water coming out of the fountain is v, the
total area around the fountain that gets wet is:
[2011]
(a) p
v4
g
2
p v4
(b) 2 2
g
(c) p
v2
g
2
(d) p
v2
g
P-29
Motion in a Plane
35. A projectile can have the same range ‘R’ for two angles
of projection. If ‘T1’ and ‘T2’ to be time of flights in the
two cases, then the product of the two time of flights is
directly proportional to.
[2004]
1
1
(a) R
(b)
(c) 2
(d) R2
R
R
36. A ball is thrown from a point with a speed ' v0 ' at an
elevation angle of q. From the same point and at the same
' v0 '
2
to catch the ball. Will the person be able to catch the ball? If
yes, what should be the angle of projection q?
[2004]
(a) No
(b) Yes, 30°
(c) Yes, 60°
(d) Yes, 45°
A boy playing on the roof of a 10 m high building throws
a ball with a speed of 10m/s at an angle of 30º with the
horizontal. How far from the throwing point will the ball be
at the height of 10 m from the ground ?
[2003]
instant, a person starts running with a constant speed
37.
1
3
[ g = 10m/s , sin 30 = , cos 30o =
]
2
2
(a) 5.20m (b) 4.33m
(c) 2.60m (d) 8.66m
2
o
Relative Velocity in Two
TOPIC 4 Dimensions & Uniform
Circular Motion
38. A clock has a continuously moving second's hand of 0.1
m length. The average acceleration of the tip of the hand
(in units of ms–2) is of the order of: [Sep. 06, 2020 (I)]
(a) 10 –3
(b) 10 –4
(c) 10 –2
(d) 10 –1
39. When a carsit at rest, its driver sees raindrops falling on
it vertically. When driving the car with speed v, he sees
that raindrops are coming at an angle 60º from the horizontal. On furter increasing the speed of the car to (1 +
b)v, this angle changes to 45º. The value of b is close to:
[Sep. 06, 2020 (II)]
(a) 0.50
(b) 0.41
(c) 0.37
(d) 0.73
40. The stream of a river is flowing with a speed of 2 km/h.
A swimmer can swim at a speed of 4 km/h. What should
be the direction of the swimmer with respect to the flow
of the river to cross the river straight? [9 April 2019 I]
(a) 90°
(b) 150°
(c) 120°
(d) 60°
41. Ship A is sailing towards north-east with velocity km/hr
where points east and , north. Ship B is at a distance of 80
km east and 150 km north of Ship A and is sailing towards
west at 10 km/hr. A will be at minimum distance from B in:
[8 April 2019 I]
(a) 4.2 hrs.
(b) 2.6 hrs.
(c) 3.2 hrs.
(d) 2.2 hrs.
42. Two particles A, B are moving on two concentric circles
of radii R1 and R2 with equal angular speed w. At t = 0,
their positions and direction of motion are shown in the
figure :
[12 Jan. 2019 II]
Y
A
X
R1
B
R2
(a) w(R1 + R2) iˆ
p
is given by:
2w
(b) –w(R1 + R2) iˆ
(c) w(R2 – R1) iˆ
(d) w(R1 – R2) iˆ
®
® and t =
The relative velocity vA
- vB
43. A particle is moving along a circular path with a constant
speed of 10 ms–1. What is the magnitude of the change in
velocity of the particle, when it moves through an angle of
60° around the centre of the circle?
[Online April 10, 2015]
(a)
(b) zero
10 3m/s
(c) 10 2m/s
(d) 10 m/s
44. If a body moving in circular path maintains constant speed
of 10 ms–1, then which of the following correctly describes
relation between acceleration and radius?
[Online April 10, 2015]
a
a
(a)
(b)
r
r
a
a
(c)
(d)
r
r
P-30
1.
Physics
(195)
4 F2 + 36F2 + 24F2 cos q
= 4 (13F2 +12F2cosq)= 52 F2 + 48 F2 cosq
r
Given : F = (iˆ + 2 ˆj + 3kˆ) N
r
And, r = [(4iˆ + 3 ˆj - kˆ) - (iˆ + 2 ˆj + kˆ)] = 3iˆ + ˆj - 2kˆ
r r
Torque, t = r ´ F = (3iˆ + ˆj - 2kˆ) ´ (iˆ + 2 ˆj + 3kˆ)
iˆ
ˆj
6.
kˆ
t = 3 1 -2 = 7iˆ - 11 ˆj + 5kˆ
1 2
2.
3
r
Magnitude of torque, | t | = 195.
(90)
Given,
r
r
r r
r
R = P Þ P+Q = P
P2 + Q2 + 2PQ. cosq = P2
Þ Q + 2P cosq = 0
= a 2 + a 2 – 2a 2 cos q
and accroding to question,
r r
r r
| A + B| =
n | A–B|
2P + Q
2P
or,
q
a
Þ
Þ a = 90°
(d) Using,
R2 = A12 + A22 + 2A1A2cos q
52 = 32 + 52 + 2 × 3 × 5 cos q
or cos q = – 0.3
®ö
æ ®
2
A
+
3
A
1
2÷
çè
ø
4.
7.
a 2 (1+ 1 + 2cos q)
2
r aˆ a ˆ
rG = i + k
2 2
r aˆ a ˆ
rH = j + k
2 2
r r æ a ˆ a ˆö æ a ˆ a ˆö a ˆ ˆ
\ rH – rG = ç j + k÷ – ç i + k ÷ = j – i
è2 2 ø è2 2 ø 2
(a) Using, R2 = P2 + Q2 + 2PQcosq
4 F2 + 9F2 + 12F2 cos q = R2
When forces Q is doubled,
4 F2 + 36F2 + 24F2 cos q = 4R2
(1+ cos q ) = n 2
(1– cos q)
..... (i)
2a – b = 1
..... (ii)
Solving equation (i) and (ii) we get
1
2
a= ,b=
3
3
®ö
æ ®
. ç 3 A1 - 2 A2 ÷ = 2A × 3A
1
1
è
ø
+ (3A2) (3A1) cos q – (2A1)(2A2) cos q – 3A2 × 2A2
= 6A12 + 9A1A2 cos q – 4A1A2cos q – 6A22
= 6A12 6A22 + 5A-1A2 cos q
= 6 × 32 – 6 × 52 + 5 × 3 × 5 (– 0.3)
= – 118.5
(c) From figure,
n2 Þ
æ n 2 –1 ö
q = cos –1 ç 2 ÷
è n +1 ø
r r r r
r
(a) If C = aiˆ + bjˆ then A.C = A.B
a+b=1
rr r r
B.C = A.B
( )
5.
a 2 + a 2 + 2a 2 cos q
= n2
a 2 + a 2 – 2a 2 cos q
a (1+ 1 – 2cos q)
using componendo and dividendo theorem, we get
Q
Q
..(i)
2P
2 P sin q
tan a =
= ¥ (Q 2 P cos q + Q = 0)
Q + 2 P cos q
Þ cos q = –
3.
12F2
1
Þ q = 120o
=–
24F2
2
r
r
(a) Let magnitude of two vectors A and B = a
r r
| A + B | = a 2 + a 2 + 2a 2 cos q and
r r
| A – B | = a 2 + a 2 – 2a 2 éëcos (180° – q ) ùû
\ cos q = –
8.
r
1 4
5
+ =
\ Magnitude of coplanar vector, C =
9 9
9
(a) Arc length = radius × angle
ur ur ur
So, | B – A |=| A | D q
B
A–B
q
9.
A
r r r r
r r r r
(c) A ´ B - B ´ A = 0 Þ A ´ B + A ´ B = 0
r r
\ A´ B = 0
Angle between them is 0, p, or 2 p
from the given options, q = p
P-31
Motion in a Plane
10.
(d) From figure/ trigonometry,
h1
= tan 45°
d
14. (c) From given equation,
r
V = K yiˆ + xjˆ
h2
h1
45°
A
And,
d
30°
B 2.464d C
h1 + h2
= tan 30°
d + 2.464 d
Þ (h1 + h2 ) ´ 3 = 3.46d
Þ (h1 + h2 ) =
3.46d
3
Þ d + h2 =
3.46d
3
\ h2 = d
11.
r
(a) Given : u = 5 ˆj m/s
r
Acceleration, a = 10iˆ + 4 ˆj and
final coordinate (20, y0) in time t.
1
S x = ux t + ax t 2
2
1
Þ 20 = 0 + ´ 10 ´ t 2 Þ t = 2 s
2
1
S y = u y ´ t + ayt 2
2
1
y0 = 5 ´ 2 + ´ 4 ´ 22 = 18 m
2
12.
®
(d) r = 15t 2iˆ + (4 - 20t 2 ) ˆj
®
®
d r
= 30tiˆ - 40tjˆ
v =
dt
®
®
Acceleration, a = d v = 30iˆ - 40 ˆj
dt
13.
\ a = 302 + 402 = 50 m/s 2
r r 1r
(b) As S = ut + at 2
2
r
1
ˆ + (4iˆ+ 4 ˆj)4
S = (5iˆ + 4j)2
2
= 10iˆ + 8jˆ + 8iˆ+ 8 ˆj
r r
rf - ri = 18iˆ + 16jˆ
r
r r
[as s = change in position = rf - ri ]
r
rr = 20iˆ + 20ˆj
r
| rr |= 20 2
)
(
\ h1 = d
[Q ux = 0]
dx
dy
= ky and
= kx
dt
dt
dy
dt = x = dy
Now dx y dx ,Þ ydy = xdx
dt
Integrating both side
y2 = x2 + c
1 2
15. (c) Using S = ut + at
2
1
y = u y t + a y t 2 (along y Axis)
2
1
Þ 32 = 0 ´ t + (4)t 2
2
1
2
Þ
´ 4 ´ t = 32
2
Þt=4s
1
S x = u xt + a xt 2
(Along x Axis)
2
1
Þ x = 3 ´ 4 + ´ 6 ´ 4 2 = 60
2
16. (580)
For pariticle ‘A’
For particle ‘B’
XA = –3t + 8t + 10
r
VA = (8 – 6t )iˆ
r
aA = –6iˆ
YB = 5 – 8t3
r
VB = –24t 2 ˆj
r
aB = -48tjˆ
2
At t = 1 sec
r
r
VA = (8 – 6t )iˆ = 2iˆ and vB = –24 ˆj
r
r
r
\ V B / A = – v A + vB = –2iˆ – 24 ˆj
\ Speed of B w.r.t. A,
v = 22 + 242
= 4 + 576 = 580
\ v = 580 (m/s)
17. (d) Given, Position vector,
r
r = cos wtiˆ + sin wt ˆj
r
r dr
v
=
= w (– sin wtiˆ + cos wt ˆj )
Velocity,
dt
Acceleration,
r
r dv
a=
= - w 2 (cos wtiˆ + sin wt ˆj )
dt
r
r
a = -w 2 r
r
\ a is antiparallel to rr
r r
r r
Also v . r = 0 \v ^ r
Thus, the particle is performing uniform circular motion.
P-32
18.
Physics
(d) v = k(yi + xj)
v = kyi + kxj
We have,
tan q = 2 or cos q =
dx
dy
= ky,
= kx
dt
dt
dy
dy dt
´
=
dx
dt dx
dy
kx
=
dx
ky
ydy = xdx
Integrating equation (i)
and
...(i)
y2 = x2 + c
r
r
(a) Given u = 3iˆ + 4 ˆj , a = 0.4iˆ + 0.3 ˆj , t = 10 s
From 1st equatoin of motion.
v–u
a=
t
\ v = at tu
Þ v = ( 0.4iˆ + 0.3 ˆj ) ´ 10 + ( 3iˆ + 4 ˆj )
4iˆ + 3 ˆj + 3 ˆj + 4 ˆj
v = 7iˆ + 7 ˆj
Þ
Þ
20.
r
Þ v = 72 + 72 = 7 2 unit.
(b) Coordinates of moving particle at time ‘t’ are
x = at3 and y = bt3
vx =
2u cos q
\
dx
dy
= 3at 2 and v y =
= 3bt 2
dt
dt
= 3t
21.
2
a +b
= 9 or
=9
2u (1/ 5) 2
u = 5/3 m/s
t1 =
2u sin q
and
g
t2 =
2u sin(90° - q) 2u cos q
=
g
g
æ 2u sin q ö æ 2u cos q ö
Now, t1t2 = ç g ÷ ç
g ÷ø
è
øè
2 æ u 2 sin 2q ö 2 R
÷=
= gç
g
g
è
ø
24. (b) For same range, the angle of projections are :
q and 90° – q. So,
h1 =
u 2sin 2q
and
2g
h2 =
u 2sin 2 (90° - q) u 2 cos 2 q
=
2g
2g
Also, R =
h1 h2 =
2
(a) Using principal of conservation of linear momentum
for horizontal motion, we have
2mvx = mu + mu cos 60°
3u
vx =
4
For vertical motion
2h
1 2
gT Þ T =
g
2
Let R is the horizontal distance travelled by the body.
1
R = v xT + (0)(T ) 2 (For horizontal motion)
2
3u
2h
´
R = v xT =
4
g
h = 0+
3 3u 2
8g
(c) Given, y = 2x – 9x2
On comparing with,
=
u 2sin 2q
g
u 2sin 2q u 2 cos2 q
×
2g
2g
u 2 u 2 (2sin q cos q) 2
16
g2
R2
16
or R2 = 16 h1 h2
25. (a) On an inclined plane, time of flight (T) is given by
=
2u sin q
g cos a
Substituting the values, we get
T=
T=
(2)(2sin15°)
4sin15°
=
g cos 30°
10 cos30°
Distance, S = (2cos15°)T -
Þ R=
22.
y = x tan q -
gx
2
2u cos q
1
g sin 30°(T ) 2
2
x
y
2 m/s
q =15°
2
2
10
2
Time of flights:
\ v = v2x + v2y = 9a 2t 4 + 9b2t 4
2
2
5
23. (d) R will be same for q and 90° – q.
ò ydy = ò x × dx
19.
g
2
1
,
a = 30°
30
gsin
g gcos30
P-33
Motion in a Plane
= (2cos15°)
2
4 sin15°
æ1
ö 16sin 15°
- ç ´ 10sin 30°÷
ø 100cos 2 30°
10 10cos 30° è 2
16 3 - 16
; 0.1952m ; 20cm
60
(b)
=
26.
10 m/s
q
g
o
ux =
v
gcosq
60
5
g
(10 - 5 3)
Horizontal component of velocity
vx = 10cos 60° = 5 m/s
vertical component of velocity
vy = 10cos 30° = 5 3 m/s
After t = 1 sec.
Horizontal component of velocity vx = 5 m/s
Vertical component of velocity
(
)
vy = | 5 3 –10 | m / s = 10 – 5 3
Centripetal, acceleration an =
v +v
2
x
v2
R
2
y
25 +100 + 75 –100 3
...(i)
=
an
10cos q
From figure (using (i))
ÞR=
tan q=
R=
27.
(
100 2 – 3
10cos15
) = 2.8m
(a) As we know, range R =
Vertical velocity (initial), 50 = uy t +
u 2 sin 2q
g
4
A1 u14 é 1 ù
1
=
=
=
A 2 u 42 êë 2 úû 16
(c) Let ‘t’ be the time taken by the bullet to hit the target.
\ 700 m = 630 ms–1 t
\
700m
630ms -1
=
10
sec
9
1
(–10) ×4
2
or, 50 = 2uy – 20
70
= 35m / s
or, uy =
2
u y 35 7
=
=
\ tan q =
u x 20 4
7
4
r ˆ
30. (b) From equation, v = i + 2 ˆj
Þ x=t
Þ Angle q = tan–1
1
y = 2t - (10t 2 )
2
From (i) and (ii), y = 2x – 5x2
… (ii)
P
wall
45°
O
4m A
6m
As ball is projected at an angle 45° to the horizontal
therefore Range = 4H
10
= 2.5 m
or 10 = 4H Þ H =
4
(Q Range = 4 m + 6 m = 10m)
Maximum height, H =
\ u2 =
H ´ 2g
2
sin q
=
u 2 sin 2 q
2g
2.5 ´ 2 ´10
æ 1 ö
ç
÷
è 2ø
or, u = 100 = 10 ms -1
Height of wall PA
1 2
\ h = gt
2
= OA tan q 2
… (i)
31. (b)
For vertical motion,
Here, u = 0
1
æ 10 ö
= ´ 10 ´ ç ÷
è 9ø
2
1 2
gt
2
Þ uy × 2 +
32. (b)
\ A µ R2 or, A µ u4
Þ t=
40
= 20m/s
2
10 – 5 3
= 2 – 3 Þq= 15°
5
and, area A = p R2
28.
500
m = 6.1 m
81
Therefore, the rifle must be aimed 6.1 m above the centre
of the target to hit the target.
29. (c) From question,
Horizontal velocity (initial),
=
2
= 100
1 g(OA) 2
2 u 2 cos2 q
1
10 ´ 16
= 4- ´
= 2.4 m
1
1
2
10 ´ 10 ´
´
2
2
P-34
33.
Physics
(d) R =
u 2 sin2 q
u 2 sin 2 q
,H=
g
2g
vr
Hmax at 2q = 90°
vr
u2
Hmax =
2g
u 2 sin 2q
u2
Þ Rmax =
g
g
tan 60° =
v 2 sin 2q v 2 sin 90° v 2
=
=
Where Rmax =
g
g
g
tan 45° =
...(i)
35.
v4
g2
(a) A projectile have same range for two angle
Let one angle be q, then other is 90° – q
T1 =
vr
v
...(i)
2u sin q
2u cos q
, T2 =
g
g
vr
(b + 1)v
3v = (b + 1)v Þ b = 3 - 1 = 0.732.
40. (c)
sin q =
u
2 1
= =
v
4 2
37.
38.
36.
q
with respect to flow,
ĵ (North)
39.
B
41. (b)
rBA
iˆ (East)
A
r
vA = 30iˆ + 50 ˆj km/hr
r
vB = (-10iˆ) km/hr
rBA = (80iˆ + 150 ˆj ) km
r
r
r
vBA = vB - v A = -10iˆ - 30iˆ - 50iˆ = 40iˆ - 50 ˆj
u 2 sin 2q (10)2 sin(2 ´ 30°)
=
= 5 3 = 8.66 m
g
10
(a) Here, R = 0.1 m
tminimum =
2p 2 p
=
= 0.105 rad /s
T
60
Acceleration of the tip of the clock second's hand,
=
a = w 2 R = (0.105)2 (0.1) = 0.0011 = 1.1 ´ 10 -3 m/s2
Hence, average acceleration is of the order of 10–3.
(d) The given situation is shown in the diagram. Here vr
be the velocity of rain drop.
u
= 90° + 30° = 120°
R=
w=
v
or q = 30°
4u 2 sin q cos q
then, T1T2 =
= 2R
g
u 2 sin 2 q
)
(Q R =
g
Thus, it is proportional to R. (Range)
(c) Yes, Man will catch the ball, if the horizontal
component of velocity becomes equal to the constant
speed of man.
vo
= vo cos q
2
or q = 60°
(d) Horizontal range is required
...(ii)
Dividing (i) by (ii) we get,
...(ii)
From equation (i) and (ii)
A=p
–vcar = (b + 1)v
When car is moving with speed (1 + b)v ,
10 ´ g ´ 2
= 20 metre
g
(a) Let, total area around fountain
2
A = pRmax
45°
When car is moving with speed v,
Rmax =
34.
60°
–vcar = v
v
u2
= 10 Þ u2 = 10 g ´ 2
2g
R=
vr
( rrBA )(· vrBA )
2
( vrBA )
(80iˆ + 150 ˆj )( -40iˆ - 50 ˆj )
\t=
(10 41) 2
10700
10 41 ´ 10 41
=
107
= 2.6 hrs.
41
P-35
Motion in a Plane
42.
Change in velocity,
p p
=
2w 2
So, both have completed quater circle
(c) From, q = wt = w
| Dv | = v12 + v 22 + 2v1 v 2 cos ( p – q )
= 2vsin
wR1 A
r
r
(Q| v1 | = | v 2 |) = v
= (2 × 10) × sin(30°) = 2 × 10 ×
wR2 B
V2
r
ra = constant
Hence graph (c) correctly describes relation between
acceleration and radius.
a=
Relative velocity,
( )
v A – v B =wR1 –iˆ - wR 2 ( –i ) =w ( R 2 – R1 ) i
v2
(d)
q
1
2
= 10 m/s
44. (c) Speed, V = constant (from question)
Centripetal acceleration,
X
43.
q
2
v1
v2
v1
(p - q)
- v1
P-36
4
Physics
Laws of Motion
TOPIC 1
1.
2.
Ist, Ind & IIIrd Laws of
Motion
A particle moving in the xy plane experiences a velocity
r
dependent force F = k ( v y i$ + v x $j ) , where vx and vy are x
r
r
and y components of its velocity v . if a is the acceleration of the particle, then which of the following statements
is true for the particle?
[Sep. 06, 2020 (II)]
r r
(a) quantity v ´ a is constant in time
r
(b) F arises due to a magnetic field
(c) kinetic energy of particle is constant in time
r r
(d) quantity v × a is constant in time
A spaceship in space sweeps stationary interplanetary dust.
dM (t )
= bv 2 (t ),
As a result, its mass increases at a rate
dt
where v (t) is its instantaneous velocity. The instantaneous
acceleration of the satellite is :
[Sep. 05, 2020 (II)]
(a) -bv3 (t )
(b) -
5.
6.
bv 3
M (t )
2bv 3
bv 3
(d) M (t )
2 M (t )
A small ball of mass m is thrown upward with velocity u
from the ground. The ball experiences a resistive force
mkv2 where v is its speed. The maximum height attained
by the ball is :
[Sep. 04, 2020 (II)]
(c) -
3.
(a)
(b)
1 æ ku 2 ö
ln 1 +
2 g ÷ø
k çè
1 æ ku
ln 1 +
2 k çè
g ÷ø
A ball is thrown upward with an initial velocity V0 from the
surface of the earth. The motion of the ball is affected by a
drag force equal to mgv2 (where m is mass of the ball, v is
its instantaneous velocity and g is a constant). Time taken
by the ball to rise to its zenith is :
[10 April 2019 I]
(c)
4.
ku 2
1
tan -1
2k
g
1
ku 2
tan -1
k
2g
(d)
æ g ö
æ g ö
1
tan -1 ç
V0 ÷
sin -1 ç
V
(b)
ç
÷
ç g 0 ÷÷
gg
gg
è g ø
è
ø
æ
ö
æ
1
g
1
2g ö
l n ç1 +
V0 ÷ (d)
tan -1 ç
V
(c)
ç g 0 ÷÷
g ÷ø
g g çè
2g g
è
ø
A ball is thrown vertically up (taken as + z-axis) from the
ground. The correct momentum-height (p-h) diagram is:
[9 April 2019 I]
(a)
2ö
1
(a)
(b)
(c)
(d)
A particle of mass m is moving in a straight line with
momentum p. Starting at time t = 0, a force F = kt acts in the
same direction on the moving particle during time interval
T so that its momentum changes from p to 3p. Here k is a
constant. The value of T is :
[11 Jan. 2019 II]
(a)
2
k
p
(b) 2
p
k
2k
2p
(d)
p
k
A particle of mass m is acted upon by a force F given by
(c)
7.
R
the empirical law F = 2 v(t). If this law is to be tested
t
experimentally by observing the motion starting from rest,
the best way is to plot :
[Online April 10, 2016]
1
(a) log v(t) against
(b) v(t) against t2
t
1
(d) log v(t) against t
(c) log v(t) against 2
t
P-37
Laws of Motion
8.
A large number (n) of identical beads, each of mass m
and radius r are strung on a thin smooth rigid horizontal
rod of length L (L >> r) and are at rest at random
positions. The rod is mounted between two rigid
supports (see figure). If one of the beads is now given
a speed v, the average force experienced by each support
after a long time is (assume all collisions are elastic):
[Online April 11, 2015]
(a) Statement 1 is true, Statement 2 is true, Statement 2 is
the correct explanation of Statement 1.
(b) Statement 1 is false, Statement 2 is true.
(c) Statement 1 is true, Statement 2 is false.
(d) Statement 1 is true, Statement 2 is true, Statement 2 is
not the correct explanation of Statement 1.
12. Two fixed frictionless inclined planes making an angle 30°
and 60° with the vertical are shown in the figure. Two
blocks A and B are placed on the two planes. What is the
relative vertical acceleration of A with respect to B ? [2010]
A
L
B
(a)
mv 2
2(L - nr)
(b)
mv 2
L - 2nr
mv 2
(d) zero
L - nr
A body of mass 5 kg under the action of constant force
r
r
F = Fxˆi + Fy ˆj has velocity at t = 0 s as v = 6iˆ - 2ˆj m/s
(c)
9.
(
)
r
r
and at t = 10s as v = +6ˆj m / s . The force F is:
[Online April 11, 2014]
æ 3 ˆ 4 ˆö
(a)
(b) ç - i + j ÷ N
è 5 5 ø
æ 3ˆ 4 ˆö
(d) ç i - j ÷ N
(c) 3iˆ - 4ˆj N
è5 5 ø
10. A particle of mass m is at rest at the origin at time
t = 0. It is subjected to a force F(t) = F0e–bt in the x direction.
Its speed v(t) is depicted by which of the following
curves?
[2012]
(
)
F0
mb
F0
mb
v (t )
15.
16.
(a)
(b)
(c) Zero
(d) 4.9 ms–2 in vertical direction
A ball of mass 0.2 kg is thrown vertically upwards by applying
a force by hand. If the hand moves 0.2 m while applying the
force and the ball goes upto 2 m height further, find the
magnitude of the force. (Consider g = 10 m/s2).
[2006]
(a) 4 N
(b) 16 N
(c) 20 N
(d) 22 N
A player caught a cricket ball of mass 150 g moving at a
rate of 20 m/s. If the catching process is completed in 0.1s,
the force of the blow exerted by the ball on the hand of the
player is equal to
[2006]
(a) 150 N (b) 3 N
(c) 30 N
(d) 300 N
A particle of mass 0.3 kg subject to a force F = – kx with
k = 15 N/m . What will be its initial acceleration if it is
released from a point 20 cm away from the origin ?[2005]
(a) 15 m/s2 (b) 3 m/s2 (c) 10 m/s2 (d) 5 m/s2
A block is kept on a frictionless inclined surface with angle
of inclination ‘a’. The incline is given an acceleration ‘a’
to keep the block stationary. Then a is equal to [2005]
(b) v (t )
v (t )
t
t
(c)
14.
)
F0
mb
(a)
13.
-3jˆ + 4ˆj N
(
30°
60°
4.9 ms–2 in horizontal direction
9.8 ms–2 in vertical direction
F0
mb
(d) v (t )
t
t
11. This question has Statement 1 and Statement 2. Of the
four choices given after the Statements, choose the one
that best describes the two Statements.
Statement 1: If you push on a cart being pulled by a horse
so that it does not move, the cart pushes you back with an
equal and opposite force.
Statement 2: The cart does not move because the force
described in statement 1 cancel each other.
[Online May 26, 2012]
a
a
(a) g cosec a
(b) g / tan a
(c) g tan a
(d) g
17. A rocket with a lift-off mass 3.5 × 104 kg is blasted upwards
with an initial acceleration of 10m/s2. Then the initial thrust
of the blast is
[2003]
(a) 3.5 ´ 10 5 N
(b) 7.0 ´ 10 5 N
(c) 14.0 ´ 10 5 N
(d) 1.75 ´ 10 5 N
18. Three forces start acting simultaneously on a particle
r
moving with velocity, v . These forces are represented
in magnitude and direction by the three sides of a triangle
ABC. The particle will now move with velocity [2003]
P-38
Physics
C
r
(a) less than v
A
(a)
2g
3
(b)
g
2
(c)
5g
6
B
r
(b) greater than v
r
(c) v in the direction of the largest force BC
(d) vr , remaining unchanged
19. A solid sphere, a hollow sphere and a ring are released
from top of an inclined plane (frictionless) so that they
slide down the plane. Then maximum acceleration down
the plane is for (no rolling)
[2002]
(a) solid sphere
(b) hollow sphere
(c) ring
(d) all same
Motion of Connected Bodies,
TOPIC 2 Pulley & Equilibrium of
Forces
20. A mass of 10 kg is suspended by a rope of length 4 m, from
the ceiling. A force F is applied horizontally at the midpoint of the rope such that the top half of the rope makes
an angle of 45° with the vertical. Then F equals:
(Take g = 10 ms–2 and the rope to be massless)
[7 Jan. 2020 II]
(a) 100 N
(b) 90 N
(c) 70 N
(d) 75 N
21. An elevator in a building can carry a maximum of 10
persons, with the average mass of each person being 68
kg. The mass of the elevator itself is 920 kg and it moves
with a constant speed of 3 m/s. The frictional force
opposing the motion is 6000 N. If the elevator is moving
up with its full capacity, the power delivered by the motor
to the elevator (g =10 m/s2) must be at least:
[7 Jan. 2020 II]
(a) 56300 W
(b) 62360 W
(c) 48000 W
(d) 66000 W
22. A mass of 10 kg is suspended vertically by a rope from
the roof. When a horizontal force is applied on the rope
at some point, the rope deviated at an angle of 45°at the
roof point. If the suspended mass is at equilibrium, the
magnitude of the force applied is (g = 10 ms–2)
[9 Jan. 2019 II]
(a) 200 N
(b) 140 N
(c) 70 N
(d) 100 N
23. A mass ‘m’ is supported by a massless string wound around
a uniform hollow cylinder of mass m and radius R. If the
str ing does not slip on the cylinder, with what
acceleration will the mass fall or release?
[2014]
R
m
m
(d) g
24. Two blocks of mass M1 = 20 kg and M2 = 12 kg are
connected by a metal rod of mass 8 kg. The system is
pulled vertically up by applying a force of 480 N as shown.
The tension at the mid-point of the rod is :
[Online April 22, 2013]
480 N
(a) 144 N
M1
(b) 96 N
(c) 240 N
(d) 192 N
M2
25. A uniform sphere of weight W and radius 5 cm is being
held by a string as shown in the figure. The tension in the
string will be :
[Online April 9, 2013]
8 cm
W
W
W
W
(b) 5
(c) 13
(d) 13
5
12
5
12
26. A spring is compressed between two blocks of masses m1
and m2 placed on a horizontal frictionless surface as shown
in the figure. When the blocks are released, they have
initial velocity of v1 and v2 as shown. The blocks travel
distances x1 and x2 respectively before coming to rest.
(a)
12
æx ö
The ratio ç 1 ÷ is
è x2 ø
[Online May 12, 2012]
m2
m1
v2
v1
(a)
m2
m1
(b)
m1
m2
(c)
m2
m1
(d)
m1
m2
P-39
Laws of Motion
27. A block of mass m is connected to another block of mass
M by a spring (massless) of spring constant k. The block
are kept on a smooth horizontal plane. Initially the blocks
are at rest and the spring is unstretched. Then a constant
force F starts acting on the block of mass M to pull it.
Find the force of the block of mass m.
[2007]
mF
MF
(a)
(b)
M
(m + M )
mF
(c) ( M + m) F
(d)
(m + M )
m
28. Two masses m1 = 5g and m2 = 4.8 kg tied to a string
are hanging over a light frictionless pulley. What is the
acceleration of the masses when left free to move ?
[2004]
( g = 9.8m / s 2 )
33. When forces F1, F2, F3 are acting on a particle of mass m
such that F2 and F3 are mutually perpendicular, then the
particle remains stationary. If the force F1 is now removed
then the acceleration of the particle is
[2002]
(a) F1/m
(b) F2F3 /mF1
(c) (F2 - F3)/m
(d) F2 /m.
34. Two forces are such that the sum of their magnitudes is
18 N and their resultant is 12 N which is perpendicular
to the smaller force. Then the magnitudes of the forces
are
[2002]
(a) 12 N, 6 N
(b) 13 N, 5 N
(c) 10 N, 8 N
(d) 16N, 2N.
35. A light string passing over a smooth light pulley connects
two blocks of masses m1 and m2 (vertically). If the
acceleration of the system is g/8, then the ratio of the
masses is
[2002]
(a) 8 : 1
(b) 9 : 7
(c) 4 : 3
(d) 5 : 3
36. Three identical blocks of masses m = 2 kg are drawn by a
force F = 10. 2 N with an acceleration of 0. 6 ms-2 on a
frictionless surface, then what is the tension (in N) in
the string between the blocks B and C?
[2002]
C
(a) 5 m/s2
(b) 9.8 m/s2
(c) 0.2 m/s2
(d) 4.8 m/s2
29. A spring balance is attached to the ceiling of a lift. A man
hangs his bag on the spring and the spring reads 49 N,
when the lift is stationary. If the lift moves downward with
an acceleration of 5 m/s2, the reading of the spring balance
will be
[2003]
(a) 24 N
(b) 74 N
(c) 15 N
(d) 49 N
30. A block of mass M is pulled along a horizontal frictionless
surface by a rope of mass m. If a force P is applied at the
free end of the rope, the force exerted by the rope on the
block is
[2003]
Pm
Pm
PM
(a)
(b)
(c) P
(d)
M +m
M -m
M +m
31. A light spring balance hangs from the hook of the other
light spring balance and a block of mass M kg hangs from
the former one. Then the true statement about the scale
reading is
[2003]
(a) both the scales read M kg each
(b) the scale of the lower one reads M kg and of the
upper one zero
(c) the reading of the two scales can be anything but the
sum of the reading will be M kg
(d) both the scales read M/2 kg each
32. A lift is moving down with acceleration a. A man in the lift
drops a ball inside the lift. The acceleration of the ball as
observed by the man in the lift and a man standing
stationary on the ground are respectively
[2002]
(a) g, g
(b) g – a, g – a
(c) g – a, g
(d) a, g
B
A
F
(a) 9.2
(b) 3.4
(c) 4
(d) 9.8
37. One end of a massless rope, which passes over a massless
and frictionless pulley P is tied to a hook C while the other
end is free. Maximum tension that the rope can bear is 360
N. With what value of maximum safe acceleration (in ms-2)
can a man of 60 kg climb on the rope?
[2002]
P
C
(a) 16
(b) 6
(c) 4
(d) 8
TOPIC 3 Friction
38. An insect is at the bottom of a hemispherical ditch of
radius 1 m. It crawls up the ditch but starts slipping
after it is at height h from the bottom. If the coefficient
of friction between the ground and the insect is 0.75,
then h is : (g = 10 ms–2)
[Sep. 06, 2020 (I)]
(a) 0.20 m
(b) 0.45 m
(c) 0.60 m
(d) 0.80 m
39. A block starts moving up an inclined plane of inclination
30° with an initial velocity of v0. It comes back to its
v
initial position with velocity 0 . The value of the
2
coefficient of kinetic friction between the block and the
I
. The nearest integer to I
inclined plane is close to
1000
is _________.
[NA Sep. 03, 2020 (II)]
P-40
Physics
40. A block of mass 5 kg is (i) pushed in case (A) and (ii) pulled
in case (B), by a force F=20 N, making an angle of 30 o with
the horizontal, as shown in the figures. The coefficient of
friction between the block and floor is m = 0.2. The
difference between the accelerations of the block, in case
(B) and case (A) will be : (g =10 ms–2)
[12 April 2019 II]
44. Two masses m1 = 5 kg and m2 = 10 kg, connected by an
inextensible string over a frictionless pulley, are moving
as shown in the figure. The coefficient of friction of
horizontal surface is 0.15. The minimum weight m that
should be put on top of m2 to stop the motion is: [2018]
T
m
m2
(a) 18.3 kg
m
2
(b) 27.3 kg
T
(c) 43.3 kg
m1
(d) 10.3 kg
(a) 0.4 ms–2
(b) 3.2 ms–2
–2
(c) 0.8 ms
(d) 0 ms–2
41. Two blocks A and B masses mA=1 kg and mB = 3 kg are kept
on the table as shown in figure. The coefficient of friction
between A and B is 0.2 and between B and the surface of
the table is also 0.2. The maximum force F that can be
applied on B horizontally, so that the block A does not
slide over the block B is : [Take g = 10 m/s2]
[10 April 2019 II]
(a) 8 N
(b) 16 N (c) 40 N
(d) 12 N
42. A block kept on a rough inclined plane, as shown in the
figure, remains at rest upto a maximum force 2 N down
the inclined plane. The maximum external force up the
inclined plane that does not move the block is 10 N. The
coefficient of static friction between the block and the
plane is : [Take g = 10 m/s2]
[12 Jan. 2019 II]
10
N
2N
30°
3
3
(b)
2
4
2
1
(c)
(d)
3
2
43. A block of mass 10 kg is kept on a rough inclined plane as
shown in the figure. A force of 3 N is applied on the block.
The coefficient of static friction between the plane and
the block is 0.6. What should be the minimum value of
force P, such that the block doesnot move downward?
(take g = 10 ms–2)
[9 Jan. 2019 I]
(a)
m1g
45. A given object takes n times more time to slide down a 45°
rough inclined plane as it takes to slide down a perfectly
smooth 45° incline. The coefficient of kinetic friction
between the object and the incline is :
(a)
(c)
1-
n2
[Online April 15, 2018]
1
(b) 1 - 2
n
2
(d)
1
1
2-n
1
1 - n2
46. A body of mass 2kg slides down with an acceleration of
3m/s2 on a rough inclined plane having a slope of 30°.
The external force required to take the same body up the
plane with the same acceleration will be: (g = 10m/s2)
[Online April 15, 2018]
(a) 4N
(b) 14N
(c) 6N
(d) 20N
47. A rocket is fired vertically from the earth with an acceleration
of 2g, where g is the gravitational acceleration. On an
inclined plane inside the rocket, making an angle q with
the horizontal, a point object of mass m is kept. The
minimum coefficient of friction mmin between the mass and
the inclined surface such that the mass does not move is :
[Online April 9, 2016]
(a) tan2q
(b) tanq
(c) 3 tanq
(d) 2 tan q
48. Given in the figure are two blocks A and B of weight 20 N
and 100 N, respectively. These are being pressed against a
wall by a force F as shown. If the coefficient of friction
between the blocks is 0.1 and between block B and the
wall is 0.15, the frictional force applied by the wall on
block B is:
[2015]
F
A
B
P
3
N
k
10
(a) 32 N
g
45°
(b) 18 N
(c) 23 N
(d)
25 N
(a) 120 N
(c) 100 N
(b) 150 N
(d) 80 N
P-41
Laws of Motion
49. A block of mass m = 10 kg rests on a horizontal table. The
coefficient of friction between the block and the table is
0.05. When hit by a bullet of mass 50 g moving with speed
n, that gets embedded in it, the block moves and comes to
stop after moving a distance of 2 m on the table. If a freely
n
falling object were to acquire speed
after being dropped
10
from height H, then neglecting energy losses and taking g
= 10 ms–2, the value of H is close to:
[Online April 10, 2015]
(a) 0.05 km
(b) 0.02 km
(c) 0.03 km
(d) 0.04 km
50. A block of mass m is placed on a surface with a vertical
x3
. If the coefficient of friction
6
is 0.5, the maximum height above the ground at which the
block can be placed without slipping is:
[2014]
cross section given by y =
1
2
1
1
m
m
m
m
(b)
(c)
(d)
6
3
3
2
51. Consider a cylinder of mass M resting on a rough horizontal
rug that is pulled out from under it with acceleration ‘a’
perpendicular to the axis of the cylinder. What is Ffriction
at point P? It is assumed that the cylinder does not slip.
[Online April 19, 2014]
w
v
O
(a)
A
h
vo2
L
C
2
vo
(a)
2h
+
m 2mg
(b)
h
+
m 2mg
(c)
h v2o
+
2m mg
(d)
v2
h
+ o
2m 2mg
54. A block A of mass 4 kg is placed on another block B of
mass 5 kg, and the block B rests on a smooth horizontal
table. If the minimum force that can be applied on A so
that both the blocks move together is 12 N, the maximum
force that can be applied to B for the blocks to move
together will be:
[Online April 9, 2014]
(a) 30 N
(b) 25 N
(c) 27 N
(d) 48 N
55. A block is placed on a rough horizontal plane. A time
dependent horizontal force F = kt acts on the block, where
k is a positive constant. The acceleration - time graph of
the block is :
[Online April 25, 2013]
a
a
(a)
P
B
(b)
O
a
a
O
a
t
t
F friction
Ma
Ma
(d)
3
2
52. A heavy box is to dragged along a rough horizontal floor.
To do so, person A pushes it at an angle 30° from the
horizontal and requires a minimum force FA, while person
B pulls the box at an angle 60° from the horizontal and
needs minimum force FB. If the coefficient of friction
(a) Mg
(b) Ma
(c)
between the box and the floor is
(a)
3
(b)
FA
3
, the ratio
is
FB
5
[Online April 19, 2014]
5
3
2
3
(d)
3
2
53. A small ball of mass m starts at a point A with speed vo
and moves along a frictionless track AB as shown. The
track BC has coefficient of friction m. The ball comes to
stop at C after travelling a distance L which is:
[Online April 11, 2014]
(c)
(c)
(d)
O
O
t
t
56. A body starts from rest on a long inclined plane of slope
45°. The coefficient of friction between the body and
the plane varies as m = 0.3 x, where x is distance travelled
down the plane. The body will have maximum speed
(for g = 10 m/s2) when x =
[Online April 22, 2013]
(a) 9.8 m
(b) 27 m
(c) 12 m
(d) 3.33 m
57. A block of weight W rests on a horizontal floor with
coefficient of static friction m. It is desired to make the
block move by applying minimum amount of force. The
angle q from the horizontal at which the force should be
applied and magnitude of the force F are respectively.
[Online May 19, 2012]
mW
(a)
q = tan -1 ( m) , F =
(b)
æ 1ö
mW
q = tan -1 ç ÷ , F =
è mø
1 + m2
1 + m2
P-42
Physics
(c)
q = 0, F = mW
æ m ö
mW
,F =
(d) q = tan -1 ç
÷
1+ m
è 1 + mø
58. An insect crawls up a hemispherical surface very slowly.
The coefficient of friction between the insect and the
surface is 1/3. If the line joining the centre of the
hemispherical surface to the insect makes an angle a with
the vertical, the maximum possible value of a so that the
insect does not slip is given by [Online May 12, 2012]
64. A block rests on a rough inclined plane making an angle of
30° with the horizontal. The coefficient of static friction
between the block and the plane is 0.8. If the frictional
force on the block is 10 N, the mass of the block (in kg) is
2
(take g = 10 m / s )
[2004]
(a) 1.6
(b) 4.0
(c) 2.0
(d) 2.5
65. A horizontal force of 10 N is necessary to just hold a block
stationary against a wall. The coefficient of friction between
the block and the wall is 0.2. The weight of the block is
[2003]
a
10N
(a) cot a = 3
(b) sec a = 3
(c) cosec a = 3
(d) cos a = 3
59. The minimum force required to start pushing a body up
rough (frictional coefficient m) inclined plane is F1 while
the minimum force needed to prevent it from sliding down
is F2. If the inclined plane makes an angle q from the
horizontal such that tan q = 2m then the ratio
F1
is
F2
[2011 RS]
(a) 1
(b) 2
(c) 3
(d) 4
60. If a spring of stiffness ‘k’ is cut into parts ‘A’ and ‘B’ of
length l A : l B = 2 : 3, then the stiffness of spring ‘A’ is
given by
(a)
[2011 RS]
3k
5
(b)
2k
5
5k
2
61. A smooth block is released at rest on a 45° incline and then
slides a distance ‘d’. The time taken to slide is ‘n’ times as
much to slide on rough incline than on a smooth incline.
The coefficient of friction is
[2005]
(c) k
(a)
(c)
(d)
1
mk = 1 – 2
n
ms = 1 -
1
n
2
(b)
(d)
mk = 1ms = 1-
(c) 400 m
TOPIC 4
Circular Motion, Banking of
Road
67. A disc rotates about its axis of symmetry in a hoizontal
plane at a steady rate of 3.5 revolutions per second. A coin
placed at a distance of 1.25cm from the axis of rotation
remains at rest on the disc. The coefficient of friction
between the coin and the disc is (g = 10m/s2)
[Online April 15, 2018]
(a) 0.5
(b) 0.7
(c) 0.3
(d) 0.6
68. A conical pendulum of length 1 m makes an angle q = 45°
w.r.t. Z-axis and moves in a circle in the XY plane.The
radius of the circle is 0.4 m and its centre is vertically below O. The speed of the pendulum, in its circular path, will
be :
(Take g = 10 ms–2)
[Online April 9, 2017]
Z
1
(a) 0.4 m/s
n2
(b) 4 m/s
1
(c) 0.2 m/s
n2
62. The upper half of an inclined plane with inclination f is
perfectly smooth while the lower half is rough. A body
starting from rest at the top will again come to rest at the
bottom if the coefficient of friction for the lower half is
given by
[2005]
(a) 2 cos f (b) 2 sin f (c) tan f (d) 2 tan f
63. Consider a car moving on a straight road with a speed of
100 m/s . The distance at which car can be stopped is
[ m k = 0.5 ]
(a) 1000 m (b) 800 m
(a) 20 N
(b) 50 N
(c) 100 N (d) 2 N
66. A marble block of mass 2 kg lying on ice when given a
velocity of 6 m/s is stopped by friction in 10 s. Then the
coefficient of friction is
[2003]
(a) 0.02
(b) 0.03
(c) 0.04
(d) 0.06
O
q
C
(d) 2 m/s
69. A particle is released on a vertical smooth semicircular
track from point X so that OX makes angle q from the
vertical (see figure). The normal reaction of the track on
the particle vanishes at point Y where OY makes angle f
with the horizontal. Then:
[Online April 19, 2014]
X
Y
q
[2005]
(d) 100 m
f
O
P-43
Laws of Motion
1
cos q
2
2
3
(c) sin f = cos q
(d) sin f = cos q
3
4
70. A body of mass ‘m’ is tied to one end of a spring and
whirled round in a horizontal plane with a constant angular
velocity. The elongation in the spring is 1 cm. If the
angular velocity is doubled, the elongation in the spring
is 5 cm. The original length of the spring is :
[Online April 23, 2013]
(a) 15 cm
(b) 12 cm
(c) 16 cm
(d) 10 cm
71. A point P moves in counter-clockwise direction on a
circular path as shown in the figure. The movement of 'P' is
such that it sweeps out a length s = t3 + 5, where s is in
metres and t is in seconds. The radius of the path is 20 m.
The acceleration of 'P' when t = 2 s is nearly.
[2010]
y
(a) sin f = cos f
(b)
sin f =
B
P(x,y)
m
20
O
A
x
(a) 13m/s2
(b) 12 m/s2
(c) 7.2 ms2
(d) 14m/s2
72. For a particle in uniform circular motion, the acceleration
r
a at a point P(R,q) on the circle of radius R is (Here q is
measured from the x-axis)
[2010]
(a)
-
n2
n2
cos q iˆ +
sin q ˆj
R
R
(b)
-
n2
n2
sin q iˆ +
cos q ˆj
R
R
(c) -
n2
n2
cos q iˆ sin q ˆj
R
R
n2 ˆ n2 ˆ
i+
j
R
R
73. An annular ring with inner and outer radii R1 and R2 is
rolling without slipping with a uniform angular speed. The
ratio of the forces experienced by the two particles situated
F1
on the inner and outer parts of the ring , F is [2005]
2
(d)
(a)
æ R1 ö
çè R ÷ø
2
2
(b)
R2
R1
R1
(d) 1
R2
74. Which of the following statements is FALSE for a particle
moving in a circle with a constant angular speed ? [2004]
(a) The acceleration vector points to the centre of the
circle
(b) The acceleration vector is tangent to the circle
(c) The velocity vector is tangent to the circle
(d) The velocity and acceleration vectors are
perpendicular to each other.
75. The minimum velocity (in ms-1) with which a car driver must
traverse a flat curve of radius 150 m and coefficient of friction
0.6 to avoid skidding is
[2002]
(a) 60
(b) 30
(c) 15
(d) 25
(c)
P-44
1.
Physics
(a) Given
r
F = k ( v y iˆ + vx ˆj )
\ Fx = kv y iˆ, Fy = kv x ˆj
4.
dv
= a = – (g + gv2)
dt
Let time t required to rise to its zenith (v = 0) so,
v 2y = v x2 + C
\t =
0
ò
dp d (mv)
æ dm ö
=
= vç
÷
dt
dt
è dt ø
...(i)
dM (t )
= bv 2 (t )
dt
Thrust on the satellite,
...(ii)
We have given,
5.
r
F = mkv 2 - mg
r
r F
a=
= -[ kv 2 + g ]
m
Þ v×
0
Þò
u
dv
= -[kv 2 + g ]
dh
v × dv
kv 2 + g
h
= ò dh
0
0
1
Þ
ln éëkv 2 + g ùû = - h
u
2k
g + gv2
[for Hmax, v = 0]
0
æ g v0 ö
tan -1 ç
÷
gg
è g ø
(d) v2 = u2 – 2gh
u 2 - 2 gh
Momentum, P = mv = m u 2 - 2 gh
u2
,P = 0
f
upward direction is positive and downward direction
is negative.
(b) From Newton’s second law
At h = 0, P = mu and at h =
6.
dp
= F = kt
dt
Integrating both sides we get,
3p
òp
T
dp = ò kt dt Þ [ p ]
0
3p
p
7.
8.
T
é t2 ù
=k ê ú
êë 2 úû 0
kT 2
p
ÞT =2
2
k
R
dv R
(a) From F < 2 v(t) Þ m < 2 v(t)
dt t
t
dv
Rdt
<ò
Integrating both sides ò
dt
mt 2
R
In v < ,
mt
1
[ ln v ×
t
(b) Space between the supports for motion of beads is
L–2nr
Average force experienced by each support,
Þ 2p =
(Q mg and mkv2 act
opposite to each other)
dv ö
æ
çèQ a = v ÷ø
dh
= ò dt
1
or v =
æ dm ö
2
3
F = -v ç
÷ = -v (bv ) = -bv [Using (i) and (ii)]
è dt ø
-bv3
Þ F = M (t ) a = -bv3 Þ a =
M (t )
v
=
0
(d)
2
(g+kv ) = a (acceleration)
H
u
t
-dv
v0
k
r r
v ´ a = (v x iˆ + v y ˆj ) ´ (v y iˆ + vx ˆj )
m
k
2ˆ
2ˆ k
= (vx k - v y k ) = (vx2 - v 2y ) kˆ = constant
m
m
(b) From the Newton's second law,
F=
3.
(a) Net acceleration
mdvx
dv
k
= kv y Þ x = v y
dt
dt
m
dv y k
Similarly,
= vx
dt
m
dv y vx
=
Þ ò v y dvy = ò vx dvx
dvx v y
v 2y - v x2 = constant
2.
1 é ku 2 + g ù
ln ê
ú= h
2k ë g û
Þ
F=
2mV
mV 2
=
2( L – 2nr ) L – 2 nr
V
P-45
Laws of Motion
Þ 40 - 0 = 2 (a) 0.2 Þ a = 100 m/s2
\ F = ma = 0.2 ´ 100 = 20 N
Þ N - mg = 20 Þ N = 20 + 2 = 22 N
Note :
Whand + Wgravity = DK
mv
L–2nr
mv
9.
(a) From question,
Mass of body, m = 5 kg
Velocity at t = 0,
Þ F (0.2) + (0.2)(10)(2.2) = 0 Þ F = 22 N
14. (c) Given, mass of cricket ball, m = 150 g = 0.15 kg
Initial velocity, u = 20 m/s
Force,
u = (6iˆ - 2 ˆj) m/s
Velocity at t = 10s,
m(v - u ) 0.15(0 - 20)
=
= 30 N
t
0.1
15. (c) Mass (m) = 0.3 kg
Force, F = m.a = –kx
Þ ma = –15x
Þ 0.3a = –15x
15
-150
x=
x = - 50 x
Þ a= –
0.3
3
a = –50 × 0.2 = 10m/s2
16. (c) When the incline is given an acceleration a towards
the right, the block receives a reaction ma towards left.
F=
v = + 6 ĵ m/s
Force, F = ?
Acceleration, a =
v -u
t
-3iˆ + 4 ˆj
6 ˆj - (6iˆ - 2 ˆj )
=
m/s2
10
5
Force, F = ma
( -3iˆ + 4 ˆj )
= 5´
= ( -3iˆ + 4 ˆj ) N
5
10. (c) Given that F(t) = F0e–bt
dv
Þ m
= F0e–bt
dt
dv
F0 -bt
e
=
dt
m
=
v
ò dv =
0
ma
g cos
a
mg cosa
+ ma sina mg
F0 -bt
e dt
mò
0
F0 é
F é e - bt ù
- e - bt - e -0 ù
v= 0ê
ú =
û
mb ë
m ëê -b ûú
(
)
0
F0 é
1 - e -bt ù
û
mb ë
11. (a) According to newton third law of motion i.e. every
action is associated with equal and opposite reaction.
12. (d) mg sin q = ma
\ a = g sin q
\ Vertical component of acceleration
= g sin2 q
\ Relative vertical acceleration of A with respect to B is
Þ v=
g (sin 2 60 - sin 2 30]
æ3 1ö g
= g ç – ÷ = = 4.9 m/s2
è4 4ø 2
in vertical direction
13. (d) For the motion of ball, just after the throwing
v = 0, s = 2m, a = –g = –10ms–2
v2 – u2 = 2as for upward journey
Þ -u 2 = 2( -10) ´ 2 Þ u 2 = 40
When the ball is in the hands of the thrower
u = 0, v = 40 ms–1
s = 0.2 m
v2 – u2 = 2as
a
a
t
t
N
a
mg sin a
For block to remain stationary, Net force along the incline
should be zero.
mg sin a = ma cos a Þ a = g tan a
17. (b) In the absence of air resistance, if
Thrust (F)
the rocket moves up with an acceleration
a, then thrust
F = mg + ma
a
\ F = m ( g + a) = 3.5 × 104 ( 10 + 10)
= 7 × 105 N
mg
18. (d) Resultant force is zero, as three forces are represented
by the sides of a triangle taken in the same order. From
r
r
Newton’s second law, Fnet = ma.
Therefore, acceleration is also zero i.e., velocity remains
unchanged.
19. (d) This is a case of sliding (if plane is friction less) and
therefore the acceleration of all the bodies is same.
20. (a) From the free body diagram
P-46
Physics
T cos45° = 100 N
T sin45° = F
On dividing (i) by (ii) we get
...(i)
...(ii)
480
= 12 ms -2
20 + 12 + 8
Tension at the mid point
=
T cos 45° 100
=
T sin 45°
F
Þ F = 100 N
Mass of rod ö
æ
T = ç M2 +
÷a
2
è
ø
= (12 + 4) × 12 = 192 N
25. (d) P
21. (d) Net force on the elevator = force on elevator
+ frictional force
q
Þ F = (10 m + M)g + f
where, m = mass of person, M = mass of elevator,
8 cm
T
f = frictional force
Þ F = (10 × 68 + 920) × 9.8 + 600
Q 5 cm O
Þ F = 22000 N
Þ P = FV = 22000 × 3 = 66000 W
22. (d)
w wcosq
PQ = OP 2 + OQ2
o
45
= 132 + 52 = 12
Tension in the string
o
45
T = w cos q =
F
26. (a)
27. (d) Writing free body-diagrams for m & M,
100 N
At equilibrium,
mg 100
=
tan 45° =
F
F
\ F = 100 N
23. (b) From figure,
m
M
K
F
N
N
m
R
mg
a
T
T
m a
mg
Acceleration a = Ra
and mg – T = ma
From equation (i) and (ii)
æ aö
T × R = mR2a = mR2 çè ÷ø
R
or T = ma
Þ mg – ma = ma
g
Þ a=
2
24. (d) Acceleration produced in upward direction
a=
13
W
12
F
M1 + M 2 + Mass of metal rod
…(i)
…(ii)
a
T T
M
F
Mg
we get T = ma and F – T = Ma
where T is force due to spring
Þ F – ma = Ma or,, F = Ma + ma
\ Acceleration of the system
F
a=
.
M +m
Now, force acting on the block of mass m is
æ F ö = mF
ma = m ç
.
è M + m ÷ø m + M
If a is the acceleration along the inclined plane, then
28. (c) Here, m1 = 5kg and m2 = 4.8 kg.
If a is the acceleration of the masses,
m1a = m1g – T ...(i)
m2a = T – m2g ...(ii)
Solving (i) and (ii) we get
æ m - m2 ö
a=ç 1
÷g
è m1 + m2 ø
Þa=
(5 - 4.8) ´ 9.8
m / s2 = 0.2 m/s2
(5 + 4.8)
P-47
Laws of Motion
29. (a) When lift is stationary, W1 = mg ...(i)
When the lift descends with acceleration, a
W2 = m(g – a)
49
(10 – 5) = 24.5 N
W2 =
10
34. (b) Let the two forces be F1 and F2 and let F2 < F1. R is
the resultant force.
Given F1 + F2 = 18
...(i)
From the figure F22 + R 2 = F12
F12 - F22 = R 2
\ F12 - F22 = 144
Only option (b) follows equation (i) and (ii).
F1
T
a
mg
R
F2
30. (d) Taking the rope and the block as a system
a
M
T
m
P
we get P = (m + M)a
P
\ Acceleration produced, a =
m+M
Taking the block as a system,
Force on the block, F = Ma
MP
m+M
31. (a) The Earth exerts a pulling force Mg. The block in turn
exerts a reaction force Mg on the spring of spring balance
S1 which therefore shows a reading of M kgf.
As both the springs are massless. Therefore, it exerts a
force of Mg on the spring of spring balance S2 which
shows the reading of M kgf.
35. (b) For mass m1
m1g – T = m1a
For mass m2
T–m2g = m2a
\ F=
s2
Mkgf
Mkgf
s1
M
Mg
32. (c) Case - I: For the man standing in the lift, the
acceleration of the ball
r
r r
abm = ab - am Þ abm = g – a
Case - II: The man standing on the ground, the acceleration
of the ball
r
r r
abm = ab - am Þ abm = g – 0 = g
33. (a) When forces F1, F2 and F3 are acting on the particle,
it remains in equilibrium. Force F2 and F3 are perpendicular
to each other,
F1 = F2 + F3
F22 + F32
The force F1 is now removed, so, resultant of F2 and F3
will now make the particle move with force equal to F1.
F
Then, acceleration, a = 1
m
\ F1 =
F1
...(i)
...(ii)
T
a m2
T
m1
m2g
a
m1g
Adding the equations we get
a=
(m1 - m2 ) g
m1 + m2
g
8
m1
-1
m
m
m
9
1 m2
Þ 1 +1 = 8 1 - 8 Þ 1 =
\
=
m
m
m
m
7
8
1 +1
2
2
2
m2
36. (b) Force = mass × acceleration
\ F = (m + m + m) × a
F = 3m × a
F
a=
3m
10.2
m / s2
\a =
6
10.2
\ T2 = ma = 2 ´
= 3.4N
6
Here a =
2 kg
C
T2
T2
2 kg
2 kg
B
A
T1
T1
37. (c) Tension, T = 360 N
Mass of a man m = 60 kg
mg – T = ma
F
...(ii)
P-48
Physics
\ a=g-
T
m
Þ
360
= 4m / s 2
60
38. (a) For balancing, mg sin q = f = mmg cos q
= 10 -
Þ 5 + 5 3m = 4(5 - 5 3m)
s2)
3
346
= 0.346 =
5
1000
I
346
So,
=
1000 1000
40. (c) A : N = 5g + 20 sin30°
1
= 50 + 20 ×
= 60 N
2
N
Þm=
f = mgcosq
R
h
h q
mgsinq
Rcosq
q
R
20 cos 30°
5
5g
4
20 sin 30°
F-f
20 cos 30° - µN
=
Accelaration, a1 =
m
5
æ 4ö R
h = R - R cos q = R - R ç ÷ =
è 5ø 5
R
= 0.2 m
5
f
3
q
h
\h =
[Q radius, R = 1m]
é
ù
3
- 0.2 ´ 60 ú
ê 20 ´
2
ú = 1.06 m/s2
= ê
5
ê
ú
êë
úû
39. (346)
Acceleration of block while moving up an inclined plane,
a1 = g sin q + mg cos q
Þ a1 = g sin 30° + mg cos30°
g mg 3
= +
2
2
Using v 2 - u 2 = 2a( s )
..(i)
Þ v02 - 02 = 2a1 ( s)
Þ
v02
N
(Q q =
20 cos 30°
f
(Q u = 0)
5g
v02
...(ii)
a1
Acceleration while moving down an inclined plane
a2 = g sin q - mg cos q
Þs=
Þ a2 = g sin 30° - mg cos 30°
g m 3
g
...(iii)
2
2
Using again v 2 - u 2 = 2as for downward motion
Þ a2 =
2
v2
æv ö
Þ ç 0 ÷ = 2a2 ( s ) Þ s = 0
4a2
è 2ø
Equating equation (ii) and (iv)
20 sin 30°
30o)
- 2a1 ( s ) = 0
v02
v2
= 0 Þ a1 = 4a2
a1 4a2
(Substituting, g = 10 m/
Þ 5 + 5 3m = 20 - 20 3m Þ 25 3m = 15
3
Þ tan q = m = = 0.75
4
Rcosq q
æg m 3ö
g mg 3
+
= 4ç ÷
ç2
2
2
2 ÷ø
è
...(iv)
B : N = 5g – 20 sin 30°
= 50 – 20 ×
1
= 40 N
2
F - f é 20cos 30° - 0.2 ´ 40 ù
=ê
ú = 1.86 m/s2
m
5
ë
û
2
Now a2 – a1 = 1.86 – 1.06 = 0.8 m/s
41. (b) Taking (A + B) as system
F – m(M + m)g
= (M + m)a
a2 =
Þa=
a=
F – m ( M + m) g
( M + m)
F - (0.2)4 ´10 æ F - 8 ö
=ç
÷ ...(i)
4
è 4 ø
P-49
Laws of Motion
45. (b) The coefficients of kinetic friction between the object
and the incline
But, amax = mg = 0.2 × 10 = 2
F -8
=2
4
Þ F = 16 N
\
1
1ö
æ
m = tan q ç1 - 2 ÷ Þ m = 1 - 2
è n ø
n
42. (a) From figure, 2 + mg sin 30° = mmg cos 30° and
10 = mg sin 30° + m mg cos 30°
= 2mmg cos 30° – 2
Þ 6 = mmg cos 30° and
4 = mg cos 30°
By dividing above two
3
2
2
10
k
g
P
µ = 0.6
Ma
C
kg
q 30°
B
A
k
m3mgcosq
µmgcos?
3 + mgsing?
N
3
2
A
Þ F – 10 – 4 = 6
F 30°
q
F = 20 N
C
B
47. (b) Let m be the minimum coefficient of friction
P
43. (a)
46. (d) Equation of motion when the mass slides down
Mg sin q – f = Ma
Þ 10 – f = 6 (M = 2 kg, a = 3 m/s2, q = 30° given)
\ f = 4N
f
Equation of motion when the block is
pushed up
Let the external force required to take
the block up the plane with same
Ma
acceleration be F
f
F – Mg sin q – f = Ma
g
3
Þ =m´ 3
2
\ Coefficient of friction, m =
(Q q = 45°)
45°
mg sin 45° =
3mgsinq
100
2
= 50 2
[Qm = 10kg, g = 9.8 m s
mmg cosq = 0.6 × mg ×
-2
At equilibrium, mass does not move so,
3mg sinq = m3mg cosq
]
1
2
3 + mg sinq = P + mmg cosq
3g
[ μ min < tan θ
= 0.6 ´ 50 2
3 + 50 2 = P + 30 2
\ P = 31.28 = 32 N
44. (b) Given : m1 = 5kg; m2 = 10kg; m = 0.15
FBD for m1, m1g – T = m1a
Þ 50 – T = 5 × a
and, T – 0.15 (m + 10)g = (10 + m)a
For rest a = 0
or, 50 = 0.15 (m + 10) 10
N
m
T
m2
m(m+m2)g (m+m2)g
f2
f1
48. (a)
F
A
20N
B
f1
N
100N
Assuming both the blocks are stationary
N= F
f1 = 20 N
f2 = 100 + 20 = 120N
f
T
m1
m1g = 50N
Þ 5=
3
(m + 10)
20
100
= m + 10 \ m = 23.3kg; close to option (b)
3
120N
Considering the two blocks as one system and due to
equilibrium f = 120N
P-50
Physics
49. (d) f = µ(M + m) g
a=
52. (d) F (Push)
A
f
µ( M + m) g
=
= µg
M +m
( M + m)
R
q = 60°
30°
= 0.05 × 10 = 0.5 ms–2
V0 =
f
Initial momentum 0.05V
=
(M + m)
10.05
m = 50g
mg
M = 10 kg
n
V0
mmg
sin q - m cos q
Similarly,
FA =
FB =
mmg
FA sin q - m cos q
=
mmg
FB
sin q + m cos q
\
v2 – u2 = 2as
0 – u2 = 2as
u2 = 2as
2
æ 0.05v ö
ç
÷ = 2 ´ 0.5 ´ 2
è 10.05 ø
=
Solving we get v = 201 2
Object falling from height H.
V
= 2 gH
10
201 2
= 2 ´10 ´ H
10
H = 40 m = 0.04 km
50. (a) At limiting equilibrium,
mmg
3
sin 60° cos 60°
5
mmg
3
sin 30° +
cos 30°
5
é
ù
3
given ú
êm =
5
ë
û
3
cos30°
5
=
3
sin 60° cos 60°
5
sin 30° +
m = tan q
m
q
y
dy x 2
=
(from question)
dx
2
Q Coefficient of friction m = 0.5
tanq = m =
x2
2
Þ x = ±1
\ 0.5 =
x3 1
= m
6 6
51. (d) Force of friction at point P,
1
Ffriction = Ma sin q
3
1
= Ma sin 90° [ here q = 90°]
3
Ma
=
3
Now, y =
mmg
sin q + m cos q
1
3
3
+
´
2 5
2
=
3
3 1
´
2
5 2
1æ 3ö
1 8
´
ç1 + ÷
2è 5ø
= 2 5
=
3æ 1ö
3´4
ç1 - ÷
5 è 5ø
10
8
8
2
10
=
=
=
3´4
3´4
3
10
53. (b) Initial speed at point A, u = v0
Speed at point B, v = ?
v2 – u2 = 2gh
v2 = v20 + 2gh
Let ball travels distance ‘S’ before coming to rest
S=
=
v 2 + 2 gh
v2
= 0
2mg
2mg
v02
2 gh
h v2
+
= + 0
2mg 2mg
m 2mg
P-51
Laws of Motion
54. (c) Minimum force on A
= frictional force between the surfaces
= 12 N
A
B
1
cos q =
1 + m2
Hence, Fmin
4 kg
=
5 kg
Smooth table
mw
1
m
+
1 + m2
h
mg sin a
mg
The insect crawls up the bowl upto a certain height h only
till the component of its weight along the bowl is balanced
by limiting frictional force.
For limiting condition at point A
R = mg cosa ...(i)
F1 = mg sina ...(ii)
Dividing eq. (ii) by (i)
tan a =
F
1
= 1 = m [ As F1 = mR ]
cot a R
\ cot a = 3
59. (c)
N1
mg sin q
w
q
1é
1
ù
Q m = ( Given ) ú
ê
3ë
3
û
F1
f1
mg
N2
mg cos q
2
For horizontal equilibrium,
F cos q = µR
...(i)
For vertical equilibrium,
R + F sin q = mg
or, R = mg – F sinq
...(ii)
Substituting this value of R in eq. (i), we get
F cosq = µ (mg – F sinq)
= µ mg – µ Fsinq
or, F (cosq + µsinq) = µmg
µmg
or, F =
...(iii)
cosq + m sinq
For F to be minimum, the denominator (cosq + µ sinq)
should be maximum.
d
\
(cosq + m sinq ) = 0
dq
or, – sinq + µ cosq = 0
or, tanq = µ
or, q = tan–1(µ)
m
Then, sinq =
and
1 + m2
Bowl
mg cos a
Þ tan a = m =
F cosq
r
F
F = µR
1 + m2
a
R y
A
R
F sinq
mw
O
F1
F
=
1 + m2
58. (a)
Therefore maximum acceleration
12N
= 3m / s 2
amax =
4kg
Hence maximum force,
Fmax = total mass × amax
= 9 × 3 = 27 N
55. (b) Graph (b) correctly dipicts the acceleration-time
graph of the block.
56. (d) When the body has maximum speed then
m = 0.3x = tan 45°
\ x = 3.33 m
57. (a) Let the force F is applied at an angle q with the
horizontal.
2
f2
mg sin q
q
mg
mg cos q
When the body slides up the inclined plane, then
mg sin q + f1 = F1
or, F1 = mg sin q + mmg cos q
When the body slides down the inclined plane, then
mg sin q – f 2 = F2
or
\
F2 = mg sin q – mmg cos q
F1
sin q + m cos q
=
sin q - m cos q
F2
F1 tan q + m 2m + m 3m
Þ F = tan q - m = 2m - m = m = 3
2
P-52
Physics
60. (d) It is given lA : lB = 2 : 3
lA
2l
æ 3l ö
= , lB = ç ÷
è 5ø
5
1
l
If initial spring constant is k, then
\ We know that k µ
k l = k Al A = k B l B
æ 2l ö
kl = kA ç ÷
è 5ø
5k
kA =
2
61. (b)
a = g sin q - mg cos q
d q
sin d
g
=
a 45°
45°
smooth
rough
On smooth inclined plane, acceleration of the body = g
sin q. Let d be the distance travelled
1
\ d = ( g sin q)t12 ,
2
2d
,
t1 =
g sin q
On rough inclined plane,
mg sin q – mR
a=
m
mg sin q – mmg cos q
Þ a=
m
Þ a = g sin q – mkg cos q
1
ˆ cos q) t 2
\ d = ( g sin q - mkg
2
2
2d
ˆ cos q
g sin q - mkg
According to question, t2 = nt1
g sin f = -( g sin f - mg cos f)
Þ m = 2 tan f
NOTE
According to work-energy theorem, W = DK = 0
(Since initial and final speeds are zero)
\ Workdone by friction + Work done by gravity = 0
l
i.e., -( µ mg cos f ) + mg l sin f = 0
2
µ
cos f = sin f or µ = 2 tan f
or
2
63. (a) Given, initial velocity, u = 100m/s.
Final velocity, v = 0.
Acceleration, a = mkg = 0.5 × 10
v2 – u2 = 2as or
Þ 02 – u2 = 2(–mkg)s
1
Þ -1002 = 2 ´ - ´ 10 ´ s
2
Þ s = 1000 m
64. (c)
fs
N
m
in
gs
2d
ˆ cos q
g sin q - mkg
2d
=
g sin q
Here, m is coefficient of kinetic friction as the block
moves over the inclined plane.
\ sin q = (sin q – mk̂ cos q)n2
1
1
2
Þ n =
Þ n=
1 - mk
1 - mk
Þ mk = 1 -
1
n2
62. (d) For first half
acceleration = g sin f;
For second half
acceleration = – ( g sin f - mg cos f)
For the block to come to rest at the bottom, acceleration
in I half = retardation in II half.
sq
co
30°
Since the body is at rest on the inclined plane,
mg sin 30° = Force of friction
Þ m ´ 10 ´ sin 30° = 10
Þ m ´ 5 = 10 Þ m = 2.0 kg
65. (d) Horizontal force, N = 10 N.
Coefficient of friction m = 0.2.
f = mN
10N
t2 =
n
°
30
mg
mg
10N
10N
W
The block will be stationary so long as
Force of friction = weight of block
\ mN = W
Þ 0.2 × 10 = W
Þ W = 2N
66. (d) u = 6 m/s, v = 0, t = 10s, a = ?
Acceleration a =
v–u
t
0–6
10
-6
= -0.6m / s2
Þ a=
10
Þ a=
mg
f = mN N
The retardation is due to the frictional force
\ f = ma Þ mN = ma
P-53
Laws of Motion
Þ mmg = ma
Þ
m=
ma
mg
At t = 2s, at = 6 ´ 2 = 12 m/s2
9 ´ 16
= 7.2 m/s2
20
\ Resultant acceleration
ac =
a 0.6
Þm= =
= 0.06
g 10
mv 2
= mrw 2
r
w = 2pn = 2p × 3.5 = 7p rad/sec
Radius, r = 1.25 cm = 1.25 × 10–2 m
Coefficient of friction, µ = ?
67. (d) Using, mmg =
mmg =
m( rw ) 2
(Q v = rw)
r
=
at2 + ac2
=
(12) 2 + (7.2) 2 = 144 + 51.84
195.84 = 14 m/s2
r
72. (c ) Clearly a = ac cos q(-iˆ) +ac sin q(- ˆj )
=
=
-v 2
v2
cos q iˆ - sin q ˆj
R
R
Y
O
P( R, q)
1.25 cm
R
Disc
q
X
O
µmg = mrw2
Þ
22 ö
æ
1.25 ´ 10 -2 ´ ç 7 ´ ÷
è
rw
7ø
m=
=
g
10
2
73. (c)
a2
2
1.25 ´ 10-2 ´ 222
= 0.6
10
68. (d) Given, q = 45°, r = 0.4 m, g = 10 m/s2
mv 2
...... (i)
T sin q =
r
T cos q = mg
...... (ii)
From equation (i) & (ii) we have,
a1
R2
R1
v 2 = wR 2
v1 = wR 1
=
tan q =
T
v2
rg
q
v = rg = 0.4 ´ 10 = 2 m/s
69. (c)
70. (a)
s = t3 + 5
ds
= 3t 2
dt
dv
= 6t
Tangential acceleration at =
dt
Þ velocity, v =
Radial acceleration ac =
v 2 w 2 R12
a1 = 1 =
= w 2 R1
R1
R1
v2
a2 = 2 = w 2 R2
R2
Taking particle masses equal
v2 = rg
Q q = 45°
Hence, speed of the pendulum in its circular path,
71. (d)
Let m is the mass of each particle and w is the angular
speed of the annular ring.
v 2 9t 4
=
R
R
F1 ma1 mR1w2 R1
=
=
=
F2 ma2 mR2w2 R2
NOTE :
The force experienced by any particle is only along radial
direction.
Force experienced by the particle, F = mw2R
\
F1 R1
=
F2 R2
74. (b) Only option (b) is false since acceleration vector is
always radial (i.e. towards the center) for uniform circular
motion.
75. (b) The maximum velocity of the car is
mrg
Here m = 0.6, r = 150 m, g = 9.8
vmax =
vmax =
0.6 ≥150 ≥9.8 ; 30m / s
P-54
5
Physics
Work, Energy and
Power
1.
2.
3.
A person pushes a box on a rough horizontal platform
surface. He applies a force of 200 N over a distance of
15 m. Thereafter, he gets progressively tired and his applied
force reduces linearly with distance to 100 N. The total
distance through which the box has been moved is 30 m.
What is the work done by the person during the total
movement of the box ?
[4 Sep. 2020 (II)]
(a) 3280 J
(b) 2780 J
(c) 5690 J
(d) 5250 J
C
1
3
J
(c) 1J
(d)
J
2
2
A block of mass m is kept on a platform which starts
from rest with constant acceleration g/2 upward, as
shown in fig. work done by normal reaction on block in
time t is:
[10 Jan. 2019 I]
(a) 2J
TOPIC 1 Work
4.
(a) -
B
q
A
A small block starts slipping down from a point B on an
inclined plane AB, which is making an angle q with the
horizontal section BC is smooth and the remaining section
CA is rough with a coefficient of friction m. It is found that
the block comes to rest as it reaches the bottom (point A)
of the inclined plane. If BC = 2AC, the coefficient of friction
is given by m = k tanq. The value of k is _________.
5.
[NA 2 Sep. 2020 (I)]
r
Consider a force F = - xiˆ + yjˆ . The work done by this
6.
force in moving a particle from point A(1, 0) to B(0, 1)
along the line segment is: (all quantities are in SI units)
[9 Jan. 2020 I]
m g2 t 2
8
(b)
m g2 t2
8
3m g 2 t 2
8
A body of mass starts moving from rest along x-axis so
that its velocity varies as v = a s where a is a constant s
and is the distance covered by the body. The total work
done by all the forces acting on the body in the first second
after the start of the motion is: [Online April 16, 2018]
1 4 2
ma t
(a)
(b) 4ma 4 t 2
8
1
ma 4 t 2
(c) 8ma 4 t 2
(d)
4
When a rubber-band is stretched by a distance x, it exerts
restoring force of magnitude F = ax + bx2 where a and b are
constants. The work done in stretching the unstretched
rubber-band by L is:
[2014]
(c) 0
(d)
(a) aL2 + bL3
(b)
(c)
7.
(b)
aL2 bL3
+
2
3
(
1
aL2 + bL3
2
)
1 æ aL2 bL3 ö
(d) 2 çç 2 + 3 ÷÷
è
ø
A uniform chain of length 2 m is kept on a table such that
a length of 60 cm hangs freely from the edge of the table.
The total mass of the chain is 4 kg. What is the work done
in pulling the entire chain on the table ?
[2004]
(a) 12 J
(b) 3.6 J
(c) 7.2 J
(d) 1200 J
P-55
Work, Energy & Power
8.
r
r
r
r
A force F = (5i + 3 j + 2k ) N is applied over a particle
which displaces it from its origin to the point
r r
r
r = (2i - j )m. The work done on the particle in joules is
[2004]
(a) +10
(b) +7
(c) –7
(d) +13
9. A spring of spring constant 5 × 103 N/m is stretched initially
by 5cm from the unstretched position. Then the work
required to stretch it further by another 5 cm is [2003]
(a) 12.50 N-m
(b) 18.75 N-m
(c) 25.00 N-m
(d) 6.25 N-m
10. A spring of force constant 800 N/m has an extension of 5 cm.
The work done in extending it from 5 cm to 15 cm is
[2002]
(a) 16 J
(b) 8 J
(c) 32 J
(d) 24 J
th
TOPIC 2 Energy
11. A cricket ball of mass 0.15 kg is thrown vertically up by a
bowling machine so that it rises to a maximum height of 20
m after leaving the machine. If the part pushing the ball
applies a constant force F on the ball and moves
horizontally a distance of 0.2 m while launching the ball,
the value of F (in N) is (g = 10 ms–2) __________.
[NA 3 Sep. 2020 (I)]
12. A particle (m = l kg) slides down a frictionless track
(AOC) starting from rest at a point A (height 2 m). After
reaching C, the particle continues to move freely in air
as a projectile. When it reaching its highest point P
(height 1 m), the kinetic energy of the particle (in J) is:
(Figure drawn is schematic and not to scale; take g = 10
ms–2) ¾¾¾ .
[NA 7 Jan. 2020 I]
Height
A
14. A spring whose unstretched length is l has a force
constant k. The spring is cut into two pieces of
unstretched lengths 11 and l2 where, l1 = nl2 and n is an
integer. The ratio k 1/k 2 of the corresponding force
constants, k1 and k2 will be:
[12 April 2019 II]
1
1
(c)
(d) n 2
(a) n
(b) 2
n
n
15. A body of mass 1 kg falls freely from a height of 100m, on
a platform of mass 3 kg which is mounted on a spring
having spring constant k = 1.25 × 106 N/m. The body sticks
to the platform and the spring’s maximum compression is
found to be x. Given that g = 10 ms–2, the value of x will be
close to :
[11 April 2019 I]
(a) 40 cm (b) 4 cm
(c) 80 cm
(d) 8 cm
16. A uniform cable of mass ‘M’ and length ‘L’ is placed on a
P
C
2m
æ 1ö
horizontal surface such that its ç ÷ part is hanging
è nø
below the edge of the surface. To lift the hanging part of
the cable upto the surface, the work done should be:
[9 April 2019 I]
2MgL
MgL
MgL
(a)
(b)
(c)
(d) nMgL
2
2
2n
n
n2
17. A wedge of mass M = 4m lies on a frictionless plane. A
particle of mass m approaches the wedge with speed v.
There is no friction between the particle and the plane
or between the particle and the wedge. The maximum
height climbed by the particle on the wedge is given by:
[9 April 2019 II]
(a)
v2
g
(b)
2v 2
7g
2v 2
v2
(d)
5g
2g
18. A particle moves in one dimension from rest under the
influence of a force that varies with the distance travelled
by the particle as shown in the figure. The kinetic energy
of the particle after it has travelled 3 m is :
[8 April 2019 I]
(c)
O
13. A particle moves in one dimension from rest under the
influence of a force that varies with the distance travelled
by the particle as shown in the figure. The kinetic energy
of the particle after it has travelled 3 m is :
[7 Jan. 2020 II]
(a) 4 J
(c) 6.5 J
(b) 2.5 J
(d) 5 J
(a) 4 J
(b) 2.5 J
(c) 6.5 J
(d) 5 J
19. A particle which is experiencing a force, given by
r
r
r
r
r
F = 3i - 12 j, undergoes a displacement of d = 4i. If
the particle had a kinetic energy of 3 J at the beginning
of the displacement, what is its kinetic energy at the end
of the displacement?
[10 Jan. 2019 II]
(a) 9 J
(b) 12 J
(c) 10 J
(d) 15 J
P-56
Physics
20. A block of mass m, lying on a smooth horizontal surface,
is attached to a spring (of negligible mass) of spring
constant k. The other end of the spring is fixed, as shown
in the figure. The block is initally at rest in its equilibrium
position. If now the block is pulled with a constant force
F, the maximum speed of the block is: [9 Jan. 2019 I]
v (m/s)
-1
50 ms
(0,0)
m
(a)
2F
mk
(b)
F
p mk
(c)
F
pF
mk
F
(d)
mk
21. A force acts on a 2 kg object so that its position is given
as a function of time as x = 3t2 + 5. What is the work
done by this force in first 5 seconds?
[9 Jan. 2019 II]
(a) 850 J (b) 950 J
(c) 875 J
(d) 900 J
22. A particle is moving in a circular path of radius a under the
action of an attractive potential U = (a) -
k
(b)
4a 2
k
2r 2
. Its total energy is:
[2018]
(a) 3h
(b)
¥
(c)
5
h
3
(d)
8
h
3
26. A time dependent force F = 6t acts on a particle of mass
1 kg. If the particle starts from rest, the work done by the
force during the first 1 second will be
[2017]
(a) 9 J
(b) 18 J
(c) 4.5 J
(d) 22 J
27. Velocity–time graph for a body of mass 10 kg is shown in
figure. Work–done on the body in first two seconds of
the motion is :
[Online April 10, 2016]
t(s)
P
k
2a 2
3 k
(c) zero
(d) - 2
2a
23. Two particles of the same mass m are moving in circular
-16 3
-r
orbits because of force, given by F(r) =
r
The first particle is at a distance r = 1, and the second, at
r = 4. The best estimate for the ratio of kinetic energies
of the first and the second particle is closest to
[Online April 16, 2018]
(a) 10–1
(b) 6 × 10–2 (c) 6 × 102 (d) 3 × 10–3
24. A body of mass m = 10–2 kg is moving in a medium and
experiences a frictional force F = –kv2. Its intial speed is v0 =
1 2
10 ms–1. If, after 10 s, its energy is mv0 , the value of k will
8
be:
[2017]
(a) 10–4 kg m–1
(b) 10–1 kg m–1 s–1
(c) 10–3 kg m–1
(d) 10–3 kg s–1
25. An object is dropped from a height h from the ground.
Every time it hits the ground it looses 50% of its kinetic
energy. The total distance covered as t ® ¥ is
[Online April 8, 2017]
10s
(a) – 9300 J
(b) 12000 J
(c) –4500 J
(d) –12000 J
28. A point particle of mass m, moves long the uniformly
rough track PQR as shown in the figure. The coefficient
of friction, between the particle and the rough track
equals m. The particle is released, from rest from the
point P and it comes to rest at a point R. The energies,
lost by the ball, over the parts, PQ and QR, of the track,
are equal to each other, and no energy is lost when particle
changes direction from PQ to QR.
The value of the coefficient of friction m and the distance
x (= QR), are, respectively close to :
[2016]
h=2m
30°
Horizontal
Surface
R
Q
(a) 0.29 and 3.5 m
(b) 0.29 and 6.5 m
(c) 0.2 and 6.5 m
(d) 0.2 and 3.5 m
29. A person trying to lose weight by burning fat lifts a mass
of 10 kg upto a height of 1 m 1000 times. Assume that the
potential energy lost each time he lowers the mass is
dissipated. How much fat will he use up considering the
work done only when the weight is lifted up? Fat supplies
3.8 × 107 J of energy per kg which is converted to
mechanical energy with a 20% efficiency rate. Take g = 9.8
ms–2 :
[2016]
(a) 9.89 × 10–3 kg
(b) 12.89 × 10–3 kg
(c) 2.45 × 10–3 kg
(d) 6.45 × 10–3 kg
30. A particle is moving in a circle of radius r under the action
of a force F = ar2 which is directed towards centre of the
circle. Total mechanical energy (kinetic energy + potential
energy) of the particle is (take potential energy = 0 for r = 0) :
[Online April 11, 2015]
5 3
4 3
1 3
αr
ar (b) ar
(c)
(d) ar3
6
3
2
31. A block of mass m = 0.1 kg is connected to a spring of
unknown spring constant k. It is compressed to a distance
x from its equilibrium position and released from rest. After
approaching half the distance æç x ö÷ from equilibrium
è 2ø
position, it hits another block and comes to rest
momentarily, while the other block moves with a velocity
3 ms–1.
The total initial energy of the spring is :
[Online April 10, 2015]
(a)
P-57
Work, Energy & Power
(a) 0.3 J
(c) 0.8 J
(a) constant
(b) 0.6 J
(d) 1.5 J
(c)
th
æ1ö
32. A bullet looses ç ÷ of its velocity passing through
ènø
one plank. The number of such planks that are required
to stop the bullet can be:
[Online April 19, 2014]
(c)
1 2
mv
3
(d)
1
mv 2
6
34. Two springs of force constants 300 N/m
(Spring A) and 400 N/m (Spring B) are joined together in
series. The combination is compressed by 8.75 cm. The
E
E
ratio of energy stored in A and B is A . Then A is
EB
EB
equal
to :
[Online April 9, 2013]
(a)
4
3
(b)
16
9
(c)
3
4
(d)
3
F (in N)
2
1
x (in m)
0
–1
–2
0 1 2 3 4 5 6 78
(a) 34 J
(b) 34.5 J
(c) 4.5 J
(d) 29.4 J
36. A particle gets displaced by
D r = 2iˆ + 3 ˆj + 4kˆ m under the action of a force
r
F = 7iˆ + 4 ˆj + 3kˆ . The change in its kinetic energy is
(
(
)
)
[Online May 7, 2012]
(a) 38 J
(b) 70 J
(c) 52.5 J
(d) 126 J
37. At time t = 0 a particle starts moving along the x-axis. If
its kinetic energy increases uniformly with time ‘t’, the
net force acting on it must be proportional to [2011 RS]
t
D = éëU ( x = ¥) - U at equilibrium ùû , D is
[2010]
b2
b2
b2
b2
(b)
(c)
(d)
2a
12a
4a
6a
An athlete in the olympic games covers a distance of 100
m in 10 s. His kinetic energy can be estimated to be in the
range
[2008]
5
5
(a) 200 J - 500 J
(b) 2 × 10 J - 3 × 10 J
(c) 20, 000 J - 50,000 J
(d) 2,000 J - 5, 000 J
A 2 kg block slides on a horizontal floor with a speed of 4m/
s. It strikes a uncompressed spring, and compresses it till
the block is motionless. The kinetic friction force is 15N and
spring constant is 10,000 N/m. The spring compresses by
[2007]
(a) 8.5 cm
(b) 5.5 cm
(c) 2.5 cm
(d) 11.0 cm
A particle is projected at 60o to the horizontal with a kinetic
energy K. The kinetic energy at the highest point is
(a) K /2
(b) K
[2007]
(c) Zero
(d) K /4
A particle of mass 100g is thrown vertically upwards with
a speed of 5 m/s. The work done by the force of gravity
during the time the particle goes up is
[2006]
(a)
39.
40.
9
16
r
35. The force F = Fiˆ on a particle of mass 2 kg, moving along
the x-axis is given in the figure as a function of its position
x. The particle is moving with a velocity of 5 m/s along the
x-axis at x = 0. What is the kinetic energy of the particle at
x = 8 m?
[Online May 26, 2012]
(d)
t
atoms in a diatomic molecule is approximately given by
a
b
U(x) = 12 - 6 , where a and b are constants and x is
x
x
the distance between the atoms. If the dissociation energy
of the molecule is
n2
2n 2
(b)
(c) infinite (d) n
2n - 1
n -1
33. A spring of unstretched length l has a mass m with one
end fixed to a rigid support. Assuming spring to be made
of a uniform wire, the kinetic energy possessed by it if its
free end is pulled with uniform velocity v is:
[Online April 12, 2014]
1
mv 2 (b) mv2
2
1
38. The potential energy function for the force between two
(a)
(a)
(b) t
41.
42.
(a) –0.5 J
(b) –1.25 J
(c) 1.25 J
(d) 0.5 J
43. The potential energy of a 1 kg particle free to move along
æ x 4 x2 ö
the x-axis is given by V ( x) = ç
- ÷ J.
2ø
è 4
The total mechanical energy of the particle is 2 J. Then,
the maximum speed (in m/s) is
[2006]
(a)
3
(b)
1
(d) 2
2
2
44. A mass of M kg is suspended by a weightless string. The
horizontal force that is required to displace it until the
string makes an angle of 45° with the initial vertical
direction is
[2006]
(a) Mg ( 2 + 1)
Mg
(c)
2
2
(c)
(b) Mg 2
(d) Mg ( 2 - 1)
P-58
Physics
45. A spherical ball of mass 20 kg is stationary at the top of
a hill of height 100 m. It rolls down a smooth surface to
the ground, then climbs up another hill of height 30 m and
finally rolls down to a horizontal base at a height of 20 m
above the ground. The velocity attained by the ball is
[2005]
(a) 20 m/s
(b) 40 m/s
46.
47.
48.
49.
(c) 10 30 m/s
(d) 10 m/s
A particle moves in a straight line with retardation
proportional to its displacement. Its loss of kinetic energy
for any displacement x is proportional to
[2004]
(a) x
(b) e x
(c) x2
(d) loge x
A particle is acted upon by a force of constant magnitude
which is always perpendicular to the velocity of the particle,
the motion of the particles takes place in a plane. It follows
that
[2004]
(a) its kinetic energy is constant
(b) its acceleration is constant
(c) its velocity is constant
(d) it moves in a straight line
A wire suspended vertically from one of its ends is
stretched by attaching a weight of 200N to the lower end.
The weight stretches the wire by 1 mm. Then the elastic
energy stored in the wire is
[2003]
(a) 0.2 J
(b) 10 J
(c) 20 J
(d) 0.1 J
A ball whose kinetic energy is E, is projected at an angle of
45° to the horizontal. The kinetic energy of the ball at the
highest point of its flight will be
[2002]
(a) E
(b) E / 2
(c) E/2
(d) zero
TOPIC 3 Power
50. A body of mass 2 kg is driven by an engine delivering a
constant power of 1 J/s. The body starts from rest and
moves in a straight line. After 9 seconds, the body has
moved a distance (in m) ________. [5 Sep. 2020 (II)]
51. A particle is moving unidirectionally on a horizontal plane
under the action of a constant power supplying energy
source. The displacement (s) - time (t) graph that describes
the motion of the particle is (graphs are drawn schematically
and are not to scale) :
[3 Sep. 2020 (II)]
S
S
(a)
(b)
t
t
S
S
(c)
(d)
t
t
52. A 60 HP electric motor lifts an elevator having a
maximum total load capacity of 2000 kg. If the frictional
force on the elevator is 4000 N, the speed of the elevator
at full load is close to: (1 HP = 746 W, g = 10 ms–2)
[7 Jan. 2020 I]
(a) 1.7 ms–1
(b) 1.9 ms–1
(c) 1.5 ms–1
(d) 2.0 ms–1
53. A particle of mass M is moving in a circle of fixed radius
R in such a way that its centripetal acceleration at time t
is given by n2 R t2 where n is a constant. The power
delivered to the particle by the force acting on it, is :
[Online April 10, 2016]
1
M n2 R2t2
(b) M n2R2t
2
(c) M n R2t2
(d) M n R2t
54. A car of weight W is on an inclined road that rises by 100
m over a distance of 1 Km and applies a constant frictional
(a)
W
on the car. While moving uphill on the road at
20
a speed of 10 ms–1, the car needs power P. If it needs
force
P
while moving downhill at speed v then value of
2
v is:
[Online April 9, 2016]
(a) 20 ms–1 (b) 5 ms–1 (c) 15 ms–1 (d) 10 ms–1
55. A wind-powered generator converts wind energy into electrical
energy. Assume that the generator converts a fixed fraction
of the wind energy intercepted by its blades into electrical
energy. For wind speed n, the electrical power output will be
most likely proportional to
[Online April 25, 2013]
4
2
(a) n
(b) n
(c) n
(d) n
56. A 70 kg man leaps vertically into the air from a crouching
position. To take the leap the man pushes the ground with
a constant force F to raise himself. The center of gravity
rises by 0.5 m before he leaps. After the leap the c.g. rises
by another 1 m. The maximum power delivered by the
muscles is : (Take g = 10 ms–2) [Online April 23, 2013]
(a) 6.26 × 103 Watts at the start
(b) 6.26 × 103 Watts at take off
(c) 6.26 × 104 Watts at the start
(d) 6.26 × 104 Watts at take off
57. A body of mass ‘m’, accelerates uniformly from rest to ‘v1’
in time ‘t1’. The instantaneous power delivered to the body
as a function
of time ‘t’ is
[2004]
mv12t
mv1t 2
(a)
(b)
t1
t12
mv1t
mv12t
(c) t
(d)
t1
1
power
58. A body is moved along a straight line by a machine
delivering a constant power. The distance moved by the
body in time ‘t’ is proportional to
[2003]
(a) t3/4
(b) t3/2
(c) t1/4
(d) t1/2
P-59
Work, Energy & Power
TOPIC 4 Collisions
59. Two bodies of the same mass are moving with the same
speed, but in different directions in a plane. They have a
completely inelastic collision and move together
thereafter with a final speed which is half of their initial
speed. The angle between the initial velocities of the
two bodies (in degree) is ______.[NA 6 Sep. 2020 (I)]
60. Particle A of mass m moving with velocity ( 3$i + $j ) ms-1
1
collides with another particle B of mass m2 which is at rest
r
r
initially. Let V1 and V2 be the velocities of particles A and
B after collision respectively. If m1 = 2m2 and after collision
r
r
r
V1 = ($i + 3 $j ) ms -1 , the angle between V1 and V2 is :
64. A particle of mass m is moving along the x-axis with initial
velocity uiˆ. It collides elastically with a particle of mass
10 m at rest and then moves with half its initial kinetic
energy (see figure). If sin q1 = n sin q2 , then value of n
is ___________.
[NA 2 Sep. 2020 (II)]
m
q1
q2
m
10
m
ui$
10 m
65. Two particles of equal mass m have respective initial
æ iˆ + ˆj ö
velocities uiˆ and u ç 2 ÷ . They collide completely
è
ø
inelastically. The energy lost in the process is: [9 Jan. 2020 I]
(a)
1
mu2
3
(b)
(c)
3
mu2
4
(d)
[6 Sep. 2020 (II)]
(a) 15º
(b) 60º
(c) – 45º
(d) 105º
61. Blocks of masses m, 2m, 4m and 8m are arranged in a line on
a frictionless floor. Another block of mass m, moving with
speed v along the same line (see figure) collides with mass
m in perfectly inelastic manner. All the subsequent collisions
are also perfectly inelastic. By the time the last block of
mass 8m starts moving the total energy loss is p% of the
original energy. Value of ‘p’ is close to :
[4 Sep. 2020 (I)]
v
m
m
2m
4m
8m
(a) 77
(b) 94
(c) 37
(d) 87
62. A block of mass 1.9 kg is at rest at the edge of a table, of
height 1 m. A bullet of mass 0.1 kg collides with the block
and sticks to it. If the velocity of the bullet is 20 m/s in
the horizontal direction just before the collision then the
kinetic energy just before the combined system strikes
the floor, is [Take g = 10 m/s2 . Assume there is no
rotational motion and losss of energy after the collision
is negligiable.]
[3 Sep. 2020 (II)]
(a) 20 J
(b) 21 J
(c) 19 J
(d) 23 J
63. A particle of mass m with an initial velocity u iˆ collides
perfectly elastically with a mass 3 m at rest. It moves
with a velocity v ˆj after collision, then, v is given by :
[2 Sep. 2020 (I)]
(a) v =
(c) v =
2
u
3
u
2
(b) v =
(d) v =
u
3
1
6
u
1
mu2
8
2
mu2
3
66. A body A, of mass m = 0.1 kg has an initial velocity of 3 iˆ ms–1.
It collides elastically with another body, B of the same
mass which has an initial velocity of 5 ĵ ms–1. After
r
collision, A moves with a velocity v = 4 iˆ + ˆj . The
(
)
x
J. The value of
10
x is _______.
[NA 8 Jan. 2020 I]
67. A particle of mass m is dropped from a height h above the
ground. At the same time another particle of the same
mass is thrown vertically upwards from the ground with a
energy of B after collision is written as
speed of
2 gh . If they collide head-on completely
inelastically, the time taken for the combined mass to reach
the ground, in units of
(a)
1
2
h
is:
g
(b)
[8 Jan. 2020 II]
3
4
1
3
(d)
2
2
68. A man (mass = 50 kg) and his son (mass = 20 kg) are
standing on a frictionless surface facing each other. The
man pushes his son so that he starts moving at a speed of
0.70 ms–1 with respect to the man. The speed of the man
with respect to the surface is :
[12 April 2019 I]
(a) 0.28 ms–1
(b) 0.20 ms–1
(c) 0.47 ms–1
(d) 0.14 ms–1
(c)
69. Two particles, of masses M and 2M, moving, as shown,
with speeds of 10 m/s and 5 m/s, collide elastically at the
origin. After the collision, they move along the indicated
directions with speeds v1 and v2, respectively. The values
of v1 and v2 are nearly :
[10 April 2019 I]
P-60
Physics
5
th of the initial kinetic
6
energy is lost in whole process. What is value of M/m?
[9 Jan. 2019 I]
C, also perfectly inelastically
A
m
(a) 6.5 m/s and 6.3 m/s (b) 3.2 m/s and 6.3 m/s
(c) 6.5 m/s and 3.2 m/s (d) 3.2 m/s and 12.6 m/s
70. A body of mass 2 kg makes an elastic collision with a
second body at rest and continues to move in the original
direction but with one fourth of its original speed. What
is the mass of the second body?
[9 April 2019 I]
(a) 1.0 kg (b) 1.5 kg
(c) 1.8 kg
(d)
1.2 kg
71. A particle of mass ‘m’ is moving with speed ‘2v’ and
collides with a mass ‘2m’ moving with speed ‘v’ in the
same direction. After collision, the first mass is stopped
completely while the second one splits into two particles
each of mass ‘m’, which move at angle 45° with respect to
the original direction.
[9 April 2019 II]
The speed of each of the moving particle will be:
(a)
2 v
(b) 2 2 v
(c) v / (2 2)
(d) v/ 2
72. A body of mass m1 moving with an unknown velocity of
v1 iˆ , undergoes a collinear collision with a body of mass
m2 moving with a velocity v2 iˆ . After collision, m1 and m2
move with velocities of v3 iˆ and v4 iˆ , respectively..
If m2 = 0.5 m1 and v3 = 0.5 v1, then v1 is : [8 April 2019 II]
v
v
(a) v4 – 2 (b) v4 – v2 (c) v4 – 2 (d) v4 + v2
2
4
73. An alpha-particle of mass m suffers 1-dimensional elastic
collision with a nucleus at rest of unknown mass. It is
scattered directly backwards losing, 64% of its initial
kinetic energy. The mass of the nucleus is :
[12 Jan. 2019 II]
(a) 2 m
(b) 3.5 m
(c) 1.5 m
(d) 4 m
74. A piece of wood of mass 0.03 kg is dropped from the
top of a 100 m height building. At the same time, a bullet
of mass 0.02 kg is fired vertically upward, with a velocity
100 ms–1, from the ground. The bullet gets embedded in
the wood. Then the maximum height to which the
combined system reaches above the top of the building
before falling below is: (g = 10 ms–2) [10 Jan. 2019 I]
(a) 20 m
(b) 30 m
(c) 40 m
(d)
10 m
75. There block A, B and C are lying on a smooth horizontal
surface, as shown in the figure. A and B have equal
masses, m while C has mass M. Block A is given an inital
speed v towards B due to which it collides with B
perfectly inelastically. The combined mass collides with
(a) 5
(b) 2
B
m
C
M
(c) 4
(d)
3
76. In a collinear collision, a particle with an initial speed n0
strikes a stationary particle of the same mass. If the final
total kinetic energy is 50% greater than the original
kinetic energy, the magnitude of the relative velocity
between the two particles, after collision, is:
[2018]
n0
n0
n0
(a)
(b) 2n0
(c)
(d)
2
4
2
77. The mass of a hydrogen molecule is 3.32×10–27 kg. If 1023
hydrogen molecules strike, per second, a fixed wall of area
2 cm2 at an angle of 45° to the normal, and rebound
elastically with a speed of 103 m/s, then the pressure on
the wall is nearly:
[2018]
(a) 2.35 × 103 N/m2
(b) 4.70 × 103 N/m2
(c) 2.35 × 102 N/m2
(d) 4.70 × 102 N/m2
78. It is found that if a neutron suffers an elastic collinear
collision with deuterium at rest, fractional loss of its energy
is pd; while for its similar collision with carbon nucleus at
rest, fractional loss of energy is Pc. The values of Pd and
Pc are respectively:
[2018]
(a)
( ×89, ×28) (b) ( ×28, ×89 ) (c)
(0, 0)
(d) (0, 1)
79. A proton of mass m collides elastically with a particle of
unknown mass at rest. After the collision, the proton and
the unknown particle are seen moving at an angle of 90°
with respect to each other. The mass of unknown particle
is:
[Online April 15, 2018]
m
3
m
(c) 2m
(d) m
2
80. Two particles A and B of equal mass M are moving with
the same speed v as shown in the figure. They collide
completely inelastically and move as a single particle C.
The angle q that the path of C makes with the X-axis is
given by:
[Online April 9, 2017]
(a)
(a) tanq =
(b) tanq =
(c) tanq =
(d) tanq =
(b)
3+ 2
Y
1- 2
C
3- 2
1- 2
1- 2
2(1 + 3)
1- 3
1+ 2
q
A
30°
X
45°
B
P-61
Work, Energy & Power
81. A neutron moving with a speed 'v' makes a head on
collision with a stationary hydrogen atom in ground state.
The minimum kinetic energy of the neutron for which
inelastic collision will take place is :
[Online April 10, 2016]
(a) 20.4 eV (b) 10.2 eV (c) 12.1 eV (d) 16.8 eV
82. A particle of mass m moving in the x direction with speed
2v is hit by another particle of mass 2m moving in the y
direction with speed v. If the collision is perfectly inelastic,
the percentage loss in the energy during the collision is
close to :
[2015]
(a) 56%
(b) 62%
(c) 44%
(d) 50%
83. A bullet of mass 4g is fired horizontally with a speed of
300 m/s into 0.8 kg block of wood at rest on a table. If the
coefficient of friction between the block and the table is
0.3, how far will the block slide approximately?
[Online April 12, 2014]
(a) 0.19 m (b) 0.379 m (c) 0.569 m (d) 0.758 m
84. Three masses m, 2m and 3m are moving in x-y plane with
speed 3u, 2u and u respectively as shown in figure. The
three masses collide at the same point at P and stick
together. The velocity of resulting mass will be:
[Online April 12, 2014]
y
2m, 2u
60°
m, 3u
P
60°
(
)
(
x
)
u ˆ
u ˆ
i + 3jˆ
i - 3jˆ
(b)
12
12
u ˆ
u ˆ
-i + 3jˆ
-i - 3jˆ
(c)
(d)
12
12
85. This question has statement I and statement II. Of the four
choices given after the statements, choose the one that
best describes the two statements.
Statement - I: Apoint particle of mass m moving with
speed u collides with stationary point particle of mass
M. If the maximum energy loss possible is given as
æ1
ö
f ç mv2 ÷ then f = æç m ö÷ .
è2
ø
èM + mø
(
)
(
(a) m
v02
x02
v0
(b) m 2 x
0
2
2 æ v0 ö
m
(d)
3 çè x0 ÷ø
89. A projectile moving vertically upwards with a velocity of
200 ms–1 breaks into two equal parts at a height of 490 m.
One part starts moving vertically upwards with a velocity
of 400 ms–1. How much time it will take, after the break
up with the other part to hit the ground?
[Online May 12, 2012]
v0
(c) 2m x
0
3m, u
(a)
86. A projectile of mass M is fired so that the horizontal
range is 4 km. At the highest point the projectile explodes
in two parts of masses M/4 and 3M/4 respectively and
the heavier part starts falling down vertically with zero
initial speed. The horizontal range (distance from point
of firing) of the lighter part is :
[Online April 23, 2013]
(a) 16 km (b) 1 km
(c) 10 km (d) 2 km
87. A moving particle of mass m, makes a head on elastic
collision with another particle of mass 2m, which is initially
at rest. The percentage loss in energy of the colliding
particle on collision, is close to
[Online May 19, 2012]
(a) 33%
(b) 67%
(c) 90%
(d) 10%
88. Two bodies A and B of mass m and 2m respectively are
placed on a smooth floor. They are connected by a spring
of negligible mass. A third body C of mass m is placed
on the floor. The body C moves with a velocity v0 along
the line joining A and B and collides elastically with A.
At a certain time after the collision it is found that the
instantaneous velocities of A and B are same and the
compression of the spring is x0. The spring constant k
will be
[Online May 12, 2012]
)
Statement - II: Maximum energy loss occurs when the
particles get stuck together as a result of the collision.
[2013]
(a) Statement - I is true, Statment - II is true, Statement
- II is the correct explanation of Statement - I.
(b) Statement-I is true, Statment - II is true, Statement II is not the correct explanation of Statement - II.
(c) Statement - I is true, Statment - II is false.
(d) Statement - I is false, Statment - II is true.
(a) 2 10 s
(b) 5 s
(c) 10 s
(d)
10 s
90. Statement -1: Two particles moving in the same direction
do not lose all their energy in a completely inelastic
collision.
Statement -2 : Principle of conservation of momentum
holds true for all kinds of collisions.
[2010]
(a) Statement -1 is true, Statement -2 is true ; Statement
-2 is the correct explanation of Statement -1.
(b) Statement -1 is true, Statement -2 is true; Statement -2
is not the correct explanation of Statement -1
(c) Statement -1 is false, Statement -2 is true.
(d) Statement -1 is true, Statement -2 is false.
91. A block of mass 0.50 kg is moving with a speed of 2.00
ms–1 on a smooth surface. It strikes another mass of 1.00
kg and then they move together as a single body. The
energy loss during the collision is
[2008]
(a) 0.16 J (b) 1.00 J (c) 0.67 J (d) 0.34 J
P-62
Physics
(a) 0.16 J (b) 1.00 J (c) 0.67 J (d) 0.34 J
92. A bomb of mass 16kg at rest explodes into two pieces of
masses 4 kg and 12 kg. The velolcity of the 12 kg mass is
4 ms–1. The kinetic energy of the other mass is [2006]
(a) 144 J (b) 288 J
(c) 192 J
(d) 96 J
93. The block of mass M moving on the frictionless horizontal
surface collides with the spring of spring constant k and
compresses it by length L. The maximum momentum of
the block after collision is
[2005]
M
kL2
ML2
(b) Mk L (c)
(d) zero
2M
k
94. A mass ‘m’ moves with a velocity ‘v’ and collides
inelastically with another identical mass. After collision
(a)
the l st mass moves with velocity
v
in a direction
3
perpendicular to the initial direction of motion. Find the
speed of the 2 nd mass after collision.
m
A before
collision
m
v
[2005]
3
Aafter
collision
v
2
v
3
3
95. Consider the following two statements :
[2003]
A. Linear momentum of a system of particles is zero
B. Kinetic energy of a system of particles is zero.
Then
(a) A does not imply B and B does not imply A
(b) A implies B but B does not imply A
(c) A does not imply B but B implies A
(d) A implies B and B implies A
(a)
3v
(b) v
(c)
(d)
P-63
Work, Energy & Power
1.
(d) The given situation can be drawn graphically as
shown in figure.
Work done = Area under F-x graph
= Area of rectangle ABCD + Area of trapezium BCFE
200N
F
A
5.
B
a 2t2
4
a2
2
F=m´
F
Þ W = 5250 J
ma 2 a 2 t 2 1 4 2
´
= ma t
2
4
8
(c) Work done in stretching the rubber-band by a
distance dx is
dW = F dx = (ax + bx2)dx
Integrating both sides,
(3) If AC = l then according to question, BC = 2l and AB =
3l.
W = ò axdx + ò bx 2 dx =
C
15m
B
Smooth
Rough
m
C
Work done =
30m
1
W = (200 ´ 15) + (100 + 200) ´ 15 = 3000 + 2250
2
6.
8.
\k = 3
(
)(
0
1
1
0
)
9.
(
1ö 1
æ
= ç 0 + ÷ + = 1J
è
2ø 2
mg
3mg
ÞN=
2
2
N = normal reaction
Now, work done by normal reaction ‘N’ on
r r æ 3mg ö æ 1
2ö
block in time t, W=NS= çè
÷ø çè g/ 2 t ÷ø
2
2
)
Displacement, x = ( 2iˆ – ˆj )
Work done,
r r
W = F × x = (5iˆ + 3 ˆj + 2kˆ ) × (2iˆ - ˆj )
= 10 – 3 = 7 joules
(b) Spring constant, k = 5 × 103 N/m
Let x1 and x2 be the initial and final stretched position of
the spring, then
(
Þ W = – ò xdx + ò ydy
(d) Here, N – mg = ma =
0
4
´ (0.6)kg = 1.2 kg
2
Weight of hanging part of the chain
= 1.2 × 10 = 12 N
C.M. of hanging part = 0.3 m below the table
Workdone in putting the entire cha in on the table = 12 ×
0.30 = 3.6 J.
r
(b) Given, Force, F = 5iˆ + 3 ˆj + 2kˆ
Þ m = 3tan q = k tan q
= - xi$ + yjˆ × d ´ i$ + dyjˆ
0
aL2 bL3
+
2
3
m¢ =
Þ mmg cos ql = 3mgl sin q
r uur
(c) Work done, W = ò F × ds
L
(b) Mass of over hanging part of the chain
3lsinq
mg (3l )sin q - mmg cos q(l ) = 0
L
7.
q
A
Here, work done by all the forces is zero.
Wfriction + Wmg = 0
4.
2 s = at Þ S =
ds
dt
E
D
3.
3mg 2 t 2
8
(a) From question, v = a s =
or,
100N
2.
or, W =
)
1
Work done, W = k x22 - x12
2
1
= ´ 5 ´ 103 é(0.1)2 - (0.05) 2 ù
ë
û
2
5000
=
´ 0.15 ´ 0.05 = 18.75 Nm
2
10. (b) Small amount of work done in extending the spring
by dx is
dW = k x dx
0.15
\W= k
ò
0.05
x dx
P-64
Physics
solving x » 4 cm
16. (a) W = uf – ui
æ mg L ö MgL .
= 0-ç´
=
è n 2n ÷ø
2n 2
17. (c) mv = (m + M) V’
800 é
(0.15) 2 - (0.05) 2 ù
û
2 ë
= 400 [(0.15 + 0.05)(0.15 – 0.05)]
= 400 × 0.2 × 0.1 = 8 J
11. (150.00) From work energy theorem,
=
W = F × s = DKE =
1 2
mv
2
mv
mv
v
=
=
m + M m + 4m 5
Using conservation of ME, we have
or v =
Here V 2 = 2 gh
2 1 15
= ´
´ 2 ´ 10 ´ 20
10 2 100
\ F = 150 N.
2
1 2 1
æ vö
mv = ( m + 4m) ç ÷ + mgh
è 5ø
2
2
\F ×s = F ´
12. (10.00) Kinetic energy = change in potential energy of
the particle,
KE = mgDh
Given, m = 1 kg,
1
´ (3 + 2) ´ (3 - 2) + 2 ´ 2
2
= 2.5 + 4
= 6.5 J
Using work energy theorem,
D K.E = work done
\ D K.E = 6.5 J
r
r
r r
r
19. (d) Work done = F.d = 3i –12J . 4i = 12J
\W=
Dh = h2 – h1 = 2 – 1 = 1m
\ KE = 1 × 10 × 1 = 10 J
13. (c) We know area under F-x graph gives the work done
by the body
1
´ (3 + 2) ´ (3 - 2) + 2 ´ 2 = 2.5 + 4 = 6.5 J
2
Using work energy theorem,
D K.E = work done
\ D K.E = 6.5 J
\W=
(
14. (c) l1 + l2 = l and l1 = nl2
nl
l
and l2 =
n +1
n +1
1
As k µ
,
l
k1
l / (n + 1)
1
\ k = (nl ) / (n + 1) = n
2
15. (b) Velocity of 1 kg block just before it collides with 3 kg
\
2 v2
5 g
18. (c) We know area under F-x graph gives the work done
by the body
or h =
l1 =
From work energy theorem,
wnet = DK.E. = kf – ki
Þ 12 = kf – 3
\ Kf = 15J
20. (d) Maximum speed is at mean position or equilibrium
At equilibrium Position
F
k
From work-energy theorem,
F = kx Þ x =
block = 2gh = 2000 m/s
WF + Wsp = DKE
Using principle of conservation of linear momentum just
before and just after collision, we get
F(x) –
2000
1 ´ 2000 = 4v Þ v =
m/s
4
Initial compression of spring
1.25 × 106 x0 = 30 Þ x0 » 0
using work energy theorem,
Wg + Wsp = DKE
Þ 40 ´ x +
1
´ 1.25 ´ 106 (02 - x 2 )
2
1
= 0 - ´ 4 ´ v2
2
)( )
1 2 1
kx = mv 2 - 0
2
2
2
4 kg
v
1
æ F ö 1 æ Fö
F ç ÷ - k ç ÷ = mv2
è kø 2 è kø
2
1 F2 1
= mv 2
2 K 2
F
or, v max =
mk
21. (d) Position, x = 3t2 + 5
Þ
\ Velocity, v =
Þ v= 6t +0
d ( 3t 2 + 5 )
dx
Þv =
dt
dt
P-65
Work, Energy & Power
At t = 0
v= 0
And, at t = 5 sec
v = 30 m/s
According to work-energy theorem, w = DKE
1
1
2
2
or, W = mv – 0 = (2)(30) = 900J
2
2
¶u
K
rˆ = 3 rˆ
22. (c) F = ¶r
r
Since particle is moving in circular path
F=
2
mv
K
K
=
Þ mv 2 =
3
r
r
r2
1
K
\ K.E. = mv2 = 2
2
2r
Total energy = P.E. + K.E.
K
K
K
(Q P.E. = - 2 given)
= - 2 + 2 = Zero
2r
2r
2r
23. (b) As the particles moving in circular orbits, So
mv2 16 2
= +r
r
r
dV
ò V2
10
6t = 1.
dV
dt
dV
dt
[Q m = 1 kg given]
1
v
ét 2 ù
ò dV = ò 6t dt V = 6 ê 2 ú = 3 ms –1
ë û0
0
From work-energy theorem,
W = DKE =
(
[Q t = 1 sec given]
)
1
1
m V 2 - u 2 = ´ 1 ´ 9 = 4.5 J
2
2
27. (c) Acceleration (a) =
v – 4 (0 , 50) < ,5 m / s 2
= (10 , 0)
t
u = 50 m/s
\ v = u + at = 50 – 5t
Veocity in first two seconds t = 2
v(at t <2) < 40 m / s
From work-energy theorem,
1
1
Kinetic energy, KE0 = mv2 = [16 + r 4 ]
2
2
1
For first particle, r = 1, K1 = m(16 + 1)
2
1
Similarly, for second particle, r = 4, K 2 = m (16 + 256)
2
16 + 1
K1
17
2
\
=
=
; 6 ´ 10-2
K 2 16 + 256 272
2
24. (a) Let Vf is the final speed of the body.
From questions,
V
1
1
mV f2 = mV02
Þ V f = 0 = 5 m/s
2
8
2
dV
æ dV ö
2
F = mç
= –kV2
÷ = -kV \ (10–2)
dt
dt
è
ø
5
26. (c) Using, F = ma = m
10
= -100 K ò dt
1
(40 2 , 50 2 ) ´10 < , 4500 J
2
28. (a) Work done by friction at QR = µmgx
1 2
In triangle, sin 30° = =
2 PQ
Þ PQ = 4m
Work done by friction at PQ = µmg × Cos 30° × 4
ΧK.E. < W <
3
´ 4 = 2 3 µmg
2
Since work done by friction on parts PQ and QR are equal,
µmg x = 2 3 umg
= µmg ×
Þ x = 2 3 @ 3.5m
Using work energy theorem mg sin 30° × 4 = 2 3 µmg + µmgx
Þ2= 4 3 µ
Þ µ = 0.29
29. (b) n =
0
1 1
- = 100K (10)
or, K = 10–4 kgm–1
5 10
25. (a) (K.E.)' = 50% of K.E. after hit i.e.,
v
1
50 1
mv '2 =
´ mv 2 Þ v ' =
2
100 2
2
1
Coefficient of restitution =
2
Now, total distance travelled by object is
1ö
æ
1+
æ 1 + e2 ö
ç
2 ÷ = 3h
= hç
H = hç
÷
2÷
1
è1- e ø
çè 1 - ÷ø
2
Input =
W
mgh ´ 1000 10 ´ 9.8 ´ 1 ´ 1000
=
=
input
input
input
98000
= 49 × 104J
0.2
49 ´ 10 4
= 12.89 × 10–3kg.
3.8 ´ 107
30. (b) As we know, dU = F.dr
Fat used =
r
U = ò ar 2 dr =
0
ar 3
3
...(i)
1 3
ar
2
...(ii)
mv 2
= ar 2
r
m2v2 = mar3
As,
or, 2m(KE) =
P-66
Physics
Total energy = Potential energy + kinetic energy
Now, from eqn (i) and (ii)
Total energy = K.E. + P.E.
ar 3 ar 3 5 3
+
= ar
3
2
6
31. (b) Applying momentum conservation
m1u1 + m2u2 = m1v1 + m2v2
0.1u + m(0) = 0.1(0) + m(3)
0.1u = 3m
1
k (x )2
EA 2 A A
300 ´ (5)2
4
=
=
=
2
EB 1
3
k B (x B )2 400 ´ (3.75)
2
35. (d)
36. (a) According to work-energy theorem,
Change in kinetic energy = work done
1
1
0.1u 2 = m(3) 2
2
2
Solving we get, u = 3
2
1 2 1 æ xö 1
kx = K ç ÷ + (0.1)32
2
2 è2ø
2
3 2
kx = 0.9
Þ
4
3 1 2
Þ
´ kx = 0.9
2 2
1 2
\
Kx = 0.6 J (total initial energy of the spring)
2
32. (a) Let u be the initial velocity of the bullet of mass m.
After passing through a plank of width x, its velocity
decreases to v.
4
4 u(n - 1)
or, v = u - =
n
n
n
If F be the retarding force applied by each plank, then
using work – energy theorem,
u–v=
1
1
1
1
( n - 1)
Fx = mu2 – mv2 = mu2 – mu2
2
2
2
2
n2
2ù
é
1
2 ê1 - ( n - 1) ú
mu
= 2
ê
ú
n2
ë
û
1
æ 2n - 1 ö
Fx = mu 2 ç
÷
2
è n2 ø
Let P be the number of planks required to stop the bullet.
Total distance travelled by the bullet before coming to
rest = Px
Using work-energy theorem again,
2
1
mu 2 - 0
2
é1
2 ( 2n - 1) ù 1
2
ú = mu
or, P ( Fx ) = P ê 2 mu
2
2
n
ë
û
F ( Px ) =
\ P=
33. (d)
F = 300 ´ x = 400(8.75 - x)
Solving we get, x = 5 cm
x B = 8.75 - 5 = 3.75cm
=
\
x B = (8.75 - x) cm
n2
2n - 1
34. (a) Given : k A = 300 N / m, k B = 400 N / m
Let when the combination of springs is compressed by
force F. Spring A is compressed by x. Therefore compression
in spring B
)(
(
® ®
= F .D r = 7iˆ + 4 ˆj + 3kˆ . 2iˆ + 3 ˆj + 4kˆ
= 14 + 12 + 12 = 38 J
37. (c) K.E. µ t
K.E. = ct
[Here, c = constant]
1 2
mv = ct
Þ
2
(mv )2
Þ
= ct
2m
Þ
p2
= ct (Q p = mv)
2m
Þ
p = 2ctm
)
dp
d ( 2 ctm )
=
dt
dt
1
Þ F = 2 cm ´
2 t
1
Þ Fµ
t
– dU ( x )
38. (d) At equilibrium : F =
dx
–d é a
bù
–
Þ F=
dx êë x12 x 6 úû
é 12a 6b ù
Þ F = – ê 13 + 7 ú
x û
ëx
Þ F=
Þ
12a
x13
6b
= 7
x
1
æ 2a ö 6
x=ç ÷
è b ø
Þ
\ U at equilibrium =
a
æ 2a ö
ç ÷
è b ø
æ 2 ö b2
\ D = 0 -ç- b ÷ =
è 4a ø 4a
2
-
b
b2
= and U(x=¥) = 0
4a
æ 2a ö
ç ÷
è b ø
39. (d) The average speed of the athelete
5 100
v= =
= 10 m / s
t 10
1
\ K.E. = mv 2
2
P-67
Work, Energy & Power
Assuming the mass of athelet to 40 kg his average K.E
would be
1
K.E = ´ 40 ´ (10)2 = 2000J
2
Assuming mass to 100 kg average kinetic energy
1
K.E. = ´ 100 ´ (10) 2 = 5000 J
2
40. (b) Suppose the spring gets compressed by x before
stopping.
kinetic energy of the block = P.E. stored in the spring +
work done against friction.
1
1
2
2
Þ ´ 2 ´ (4) = ´ 10,000 ´ x + 15 ´ x
2
2
Þ 10,000 x2 + 30x – 32 = 0
Þ 5000 x 2 + 15 x - 16 = 0
-15 ± (15)2 - 4 ´ (5000)(-16)
2 ´ 5000
= 0.055m = 5.5cm.
41. (d) Let u be the velocity with which the particle is thrown
and m be the mass of the particle. Then
\ x=
1
K = mu 2 .
... (1)
2
At the highest point the velocity is u cos 60° (only the
horizontal component remains, the vertical component
being zero at the top-most point). Therefore kinetic energy
at the highest point.
1
K
1
K ' = m(u cos 60°) 2 = mu 2 cos2 60° =
[From 1]
2
2
4
42. (b) Given, Mass of the particle, m = 100g
Initial speed of the particle, m = 5 m/s
Final speed of the particle, v = 0
Work done by the force of gravity
= Loss in kinetic energy of the body.
1
1 100
= m (v2 – u2) = ´
(02 – 52)
2
2 1000
= –1.25 J
43. (a) Potential energy
x4 – x2
joule
4 2
For maxima of minima
dV
= 0 Þ x 3 - x = 0 Þ x = ±1
dx
1 1
1
Þ Min. P.E. = - = - J
4 2
4
K.E.(max.) + P.E.(min.) = 2 (Given)
1 9
\ K.E.(max.) = 2 + =
4 4
1 2
K.E.max . = mvmax .
2
1
9
3
2
Þ ´ 1 ´ vmax
. = Þ vmax. =
2
4
2
V(x) =
44. (d) Work done by tension + Work done by force (applied)
+ Work done by gravitational force = change in kinetic
energy
Work done by tension is zero
O
45°
l
B
A
F
C
F
Þ 0 + F ´ AB - Mg ´ AC = 0
1
é
ê1 æ AC ö
2
Þ F = Mgç
÷ = Mg ê
ê 1
è AB ø
êë 2
l
[Q AB = l sin 45° =
and
2
ù
ú
ú
ú
úû
1 ö
æ
AC = OC - OA = l - l cos 45° = l ç1 ÷
è
2ø
where l = length of the string.]
Þ F = Mg ( 2 - 1)
45. (b)
100
30
mgH
Using conservation of energy,
Total energy at 100 m height
= Total energy at 20m height
20
1
mv 2 + mgh
2
æ1 2
ö
m (10 × 100) = m çè v + 10 ´ 20÷ø
2
1 2
or
v = 800 or v = 1600 = 40 m/s
2
Note :
Loss in potential energy = gain in kinetic energy
1
m ´ g ´ 80 = mv 2
2
1 2
10 ´ 80 = v
2
v2 = 1600 or v = 40 m/s
46. (c) Given : retardation µ displacement
i.e., a = –kx [Here, k = constant]
dv
But a = v
dx
vdv
\
= - kx Þ
dx
v2
x
v1
0
ò v dv = - ò kxdx
P-68
Physics
v
Þ
x
v
é v2 ù 2
é x2 ù
ê ú = –k ê ú
ë 2 û v1
ë 2 û0
Þ ò v dv =
0
2
t
P
dt
m ò0
Þ
(v22 - v12 ) = - kx2
Þ
v 2 Pt
æ 2 Pt ö
=
Þv=ç
è m ÷ø
m
2
Þ
1
1 æ - kx 2 ö
m v22 - v12 = m ç
÷
2
2 çè 2 ÷ø
Þ
dx
=
dt
(
)
\ Loss in kinetic energy, \ DK µ x2
47. (a) Work done by such force is always zero since force is
acting in a direction perpendicular to velocity.
\ From work-energy theorem = DK = 0
K remains constant.
48. (d) The elastic potential energy
1
= ´ Force ´ extension
2
1
= ´F ´x
2
1
= ´ 200 ´ 0.001 = 0.1 J
2
49. (c) Let u be the speed with which the ball of mass m is
projected. Then the kinetic energy (E) at the point of
projection is
u
u
x
Þ ò dx =
0
t
2P 1/ 2
t dt
m ò0
\ Distance, x =
But F = mav = m
\ P = mv
dv
Þ P dt = mv dv
dt
t
Integrating both sides
(Remember that the horizontal
component of velocity does not change during a
projectile motion).
\ The kinetic energy at the highest point
2
1 æ u ö
E
1 mu 2
mç
=
=
[From (i)]
÷
2 è 2ø
2
2 2
50. (18) Given, Mass of the body, m = 2 kg
Power delivered by engine, P = 1 J/s
Time, t = 9 seconds
Power, P = Fv
[Q F = ma]
Þ P = mav
=
Þm
dv
v=P
dt
P
dt
m
Integrating both sides we get
Þ v dv =
dv ö
æ
çèQ a = ÷ø
dt
ò
v
ò
P dt = m v dv
0
0
1 2 Þ v = æ 2P ö t1/ 2
çç
÷÷
mv
2
è m ø
ò
Distance, s = v dt =
0
E=
2
dv
v
dt
t
1
mu 2
...(i)
2
When the ball is at the highest point of its flight, the speed
2 P 2 3/ 2
´ t
m 3
2 ´ 1 2 3/ 2 2
´ ´ 9 = ´ 27 = 18.
2
3
3
51. (b) We know that
Power, P = Fv
P. t =
2
u
2 P t 3/ 2
=
m 3/ 2
Þx=
u
of the ball is
dx ö
æ
çèQ v = ÷ø
dt
2 P 1/ 2
t
m
2
45°
1/ 2
t
2P 1/ 2
t dt =
m
ò
0
2 P t 3/ 2
×
m 3/2
8P 3/ 2
×t
Þ s µ t3/2
9m
So, graph (b) is correct.
52. (b) Total force required to lift maximum load capacity
against frictional force = 400 N
Ftotal = Mg + friction
= 2000 × 10 + 4000
= 20,000 + 4000 = 24000 N
Using power, P = F × v
60 × 746 = 24000 × v
Þ v = 1.86 m/s » 1.9 m/s
Hence speed of the elevator at full load is close to 1.9 ms–1
53. (b) Centripetel acceleration ac = n2Rt2
Þs=
ac <
v2
< n 2 Rt 2
R
v2 < n2 R 2 t2
v = nRt
dv
< nR
dt
Power = matv = m nR nRt = Mn 2R2t.
ac <
P-69
Work, Energy & Power
59. (120)
54. (c) While moving downhill power
m
wö
æ
P = ç w sin q + ÷10
20
è
ø
æ w w ö÷
3w
P < çç ∗ ÷÷10 <
èç10 20 ø
2
v0
2
æv ö
æ mv öæ v t ö
= ç 1 ÷ç 1 ÷ = m ç 1 ÷ t
è t1 ø
è t1 øè t1 ø
58. (b) Power, P = Fv = ma.v
mdv
Þ P=
v = c = contant
dt
mdv ö
æ
çQ F = ma =
÷
è
dt ø
mv0v = cdt
Integrating both sides, we get
t
0
0
m ò vdv = c ò dt
e
Velocity of particle A, u1 = ( 3iˆ + ˆj ) m/s
Velocity of particle B, u2 = 0
After collision,
Velocity of particle A, v1 = (iˆ + 3 ˆj)
Velocity of particle B, v2 = 0
Using principal of conservation of angular momentum
r
r
r
r
m1u1 + m2 u2 = m1v1 + m2 v2
r
Þ 2m2 ( 3iˆ + ˆj ) + m2 ´ 0 = 2m2 (iˆ + 3 ˆj ) + m2 ´ v2
r
Þ 2 3iˆ + 2 ˆj = 2iˆ + 2 3 ˆj + v2
r
Þ v2 = ( 3 - 1)iˆ - ( 3 - 1) ˆj
r
Þ v1 = iˆ + 3 ˆj
r
r
For angle between v1 and v2 ,
r r
v ×v
2( 3 - 1)(1 - 3) 1 - 3
=
cos q = 1r r 2 =
v1v2
2 ´ 2 2( 3 - 1)
2 2
v2 = ( 3 - 1)$
i - ( 3 - 1) $
j
dx
2c 12
=
´t
dt
m
ò dx =
= q + q = 60° + 60° = 120°.
60. (d) Before collision,
60°
45°
2c 1 2
´t
Þ v=
m
Þ
v0
1
Þ cos q = or q = 60°
2
2
Hence angle between the initial velocities of the two bodies
2mv0 cos q = 2m
v1 = $
i+ 3$
j
v 2 c.t
=
2
m
x
Momentum conservation along x direction,
r
r
Angle between v1 and v2 is 105°
2c.t
Þ v2 =
m
Þ
v0
m
Þ q = 105°
1
Þ mv 2 = ct
2
Þ
where v =
61. (b) According to the question, all collisions are perfectly
inelastic, so after the final collision, all blocks are moving
together.
v
m
dx
dt
t
1
2c
´ t 2 dt
m ò
m
2m
4m
8m
Let the final velocity be v', using momentum conservation
0
3
Þ x=
v0/2
q
sq
co
P 3w æç w w ÷ö
1
w
<
< ç , ÷÷ V
tanq = 10
çè10 20 ø
2
4
3 v
=
Þ v = 15 m/s
4 20
\ Speed of car while moving downhill v = 15 m/s.
55. (d)
56. (b)
57. (b) Let a be the acceleration of body
Using, v = u + at
v
v1 = 0 + at1 Þ a = 1
t1
Velocity of the body at instant t,
v = at
v1t
Þ v=
t1
r r
r r
Instantaneous powr, P = F × v = ( ma ) × v
v
2m
q
2c 2t 2
´
m
3
Þ xµt
3
2
mv = 16mv ' Þ v ' =
v
16
P-70
Physics
1
Now initial energy Ei = mv 2
2
2
2
or, mu = mv +
2
1
1 mv 2
æ v ö
Final energy : E f = ´ 16m ´ ç ÷ =
2
2 16
è 16 ø
1 2 1 v2
mv - m
2
2 16
1
1ù
1
é
é15 ù
Þ mv 2 ê1 - ú Þ mv2 ê ú
2
2
ë 16 û
ë16 û
The total energy loss is P% of the original energy.
\ %P =
1 2 é15 ù
mv ê ú
2
ë16 û ´ 100 = 93.75%
=
1 2
mv
2
Hence, value of P is close to 94.
62. (b) Given,
Mass of block, m1 = 1.9 kg
Mass of bullet, m2 = 0.1 kg
Velocity of bullet, v2 = 20 m/s
Let v be the velocity of the combined system. It is an
inelastic collision.
Using conservation of linear momentum
m1 ´ 0 + m2 ´ v2 = (m1 + m2 )v
Þ 0.1 ´ 20 = (0.1 + 1.9) ´ v
Þ v = 1 m/s
Using work energy theorem
Work done = Change in Kinetic energy
Let K be the Kinetic energy of combined system.
(m1 + m2 )gh
1
= K - (m1 + m2 )V2
2
1
Þ 2 ´ g ´1 = K - ´ 2 ´ 12 Þ K = 21 J
2
63. (c) From conservation of linear momentum
ur
muiˆ + 0 = mvjˆ + 3mv '
v
3m
m
u
2
64. (10.00)
From momentum conservation in perpendicular direction
of initial motion.
mu1 sin q1 = 10mv1 sin q2
...(i)
It is given that energy of m reduced by half. If u1 be
velocity of m after collision, then
æ 1 2ö 1 1
2
çè mu ÷ø = mu1
2
2 2
Energy loss
´ 100
Original energy
ur u
v
v ' = iˆ - ˆj
3
3
y
u
\v =
Energy loss : Ei - E f =
mu 2 mv 2
+
3
3
m
3m
v'
x
Before
collision
From kinetic energy conservation,
æ æ u ö2 æ v ö2 ö
1
1
1
mu 2 = mv 2 + (3m) ç ç ÷ + ç ÷ ÷
çè 3 ø è 3 ø ÷
2
2
2
è
ø
After
collision
Þ u1 =
u
2
If v1 be the velocity of mass 10 m after collision, then
1
1 u2
´ 10 m ´ v12 = m
Þ v1 =
2
2 2
From equation (i), we have
sin q1 = 10 sin q2
65. (b)
m
u
20
u/2
u
m
y
u/2
d
x-direction
mu
3u
mu +
= 2mv'x Þ Vx' =
2
4
mu
u
'
= 2mv y Þ v 'y =
y-direction 0 +
2
4
2
2ù
é
1
1
æ uö
æ uö
K .E.i = m u 2 + m ê ç ÷ + ç ÷ ú
è 2ø ú
2
2 ëê è 2 ø
û
1 2 mu 2 3mu 2
mu +
=
2
4
4
2
2
1
1
K .E . f = ( 2m) vx' + ( 2m) v 'y
2
2
2
2
é
1
æ 3u ö
æ uö ù 5
= 2m êç ÷ + ç ÷ ú = mu 2
è 4ø ú 8
2
êëè 4 ø
û
\ Loss in KE = KEf – KEi
=
( )
( )
2
æ 6 5 ö mu
= mu 2 ç - ÷ =
è 8 8ø
8
66. (a) For elastic collision KEi = KEf
1
1
1
1
m ´ 25 + ´ m ´ 9 = m ´ 32 + mvB2
2
2
2
2
34 = 32 + VB2 Þ VB = 2
KEB =
\x=1
1 2 1
1
mvB = ´ 0.1 ´ 2 = 0.1J = J
2
2
10
O
x
P-71
Work, Energy & Power
67. (d) Let t be the time taken by the particle dropped from
height h to collide with particle thrown upward.
Using,
h
4
S1
2
1
1
1
æ uö
´ 2 ´ u 2 + 0 = ´ 2 ´ ç ÷ + mv22
è
ø
2
2
4
2
On solving, we get
m = 1.5 kg
71. (b) m (2v) + 2mv = 0 + 2mv’ cos 45°
and
1 2 1 h
h
gt = g.
=
2
2 2g 4
Distance of collision point from ground
h 3h
s2 = h – =
4 4
Speed of (A) just before collision
S1 =
or v’ = 2 2v
72. (b) m1v1 + m2v2 = m1v2 + m2v1
or m1v1 + (0.5m1)v2 = m10.5v1 + (0.5m1)v4
On solving, v1 = v4 – v2
73. (d) Using conservation of momentum,
mv0 = mv2 – mv1
a
V0 M
gh
2
And speed of(B) just before collision
v1 = gt =
After collision
a
gh
2
V1
Using principle of conservation of linear momentum
mv1 + mv2 = 2mvf
æ
gh ö
gh
Þ v f = m ç 2 gh –
–m
=0
2 ÷ø
2
è
2m
After collision, time taken (t1) for combined mass to
reach the ground is
M
V2
1
1
mv12 = 0.36 ´ mv02
2
2
Þ v1 = 0.6v0
The collision is elastic. So,
1
1
MV22 = 0.64´ mv20 [\ M = mass of nucleus]
2
2
m
Þ V2 =
´ 0.8V0
M
mV0 = mM ´ 0.8V0 – m ´ 0.6V0
3h 1 2
= gt1
4 2
Þ1.6 m = 0.8 mM
3h
Þ t1 =
2g
68. (b) Pi = Pf
or 0 = 20(0.7 – v) = 50v
or v = 0.2 m/s
69. (a) Apply concervation of linear momentum in X and Y
direction for the system then
M (10cos30°) + 2M (5cos45°) = 2M (v1cos30°)
+ M(v2cos45°)
v
5 3 + 5 2 = 3 v1 + 2
2
)
3 + 1 v1 = 5 3 + 10 2 - 5 Þ v1 = 6.5m/s
æ uö
70. (b) 2u + 0 = 2 ç ÷ + mv2
è 4ø
Þ v = 2 gh
Downward distance travelled
Þ
....(2)
v2 = 6.3 m/s
v2 –u2 = 2gh
Þ v2 – 02 =2gh
v2 = 2 gh –
v
5 2 - 5 = v1 - 2
2
Solving equation (1) and (2)
(
3h
4
S2
Also
2M(5sin45°) – M(10sin30°) = 2Mv1sin30° – Mv2sin45°
....(1)
Þ 4m 2 = mM
\ M = 4m
74. (c)
0.03 kg
100 m
100 m/s
0.02 kg
Time taken for the particles to collide,
d 100
t = V = 100 =1sec
rel
Speed of wood just before collision = gt = 10 m/s and
speed of bullet just before collision = v – gt
P-72
Physics
= 100 – 10 = 90 m/s
1
× 10 × 1 = 95 m
2
Now, using conservation of linear momentum just before
and after the collision
– (0.03) (10) + (0.02) (90) = (0.05)v
Þ 150 = 5v
\ v = 30 m/s
Max. height reached by body
77. (a) Change in momentum
j^
S = 100 × 1 –
30 ´ 30
h = = 2 ´ 10 = 45m
Before
0.03 kg 10 m/s
0.02 kg
After
v
90 m/s
2m + M
M
=6\ =4
m
m
76. (b) Before Collision
m
V0
After Collision
m
45°
– P J^
2
P
P^
45° 2 i
P ^i
2
i^
P
P ˆ P ˆ P ˆ P ˆ
J+
J+
ii
2
2
2
2
2P ˆ
DP =
J = IH molecule
2
2P ˆ
Þ Iwall = J
2
Pressure, P
DP =
0.05 kg
\ Height above tower = 40 m
75. (c) Kinetic energy of block A
1
k1 = mv 20
2
\ From principle of linear momentum conservation
mv0
mv0 = ( 2m + M ) vf Þ v f =
2m + M
5
According to question, of th the initial kinetic energy
6
is lost in whole process.
1
mv02
2
Þ
=6
k
2
\ i =6
mv
1
æ
ö
0
kf
( 2m + M ) çè
2
2m + M ÷ø
Þ
P ^J
2
Þ
m
V1
m
V2
Stationary
1 2 1 2
3æ1
ö
mv + mv = ç mv20 ÷
2 1 2 2
2è2
ø
3
Þ v12 + v 22 = v 20
....(i)
2
From momentum conservation
mv0 = m(v1 + v2)
....(ii)
Squarring both sides,
(v1 + v2)2 = v02
Þ v12 + v22 + 2v1v2 = v02
v2
2v1v2 = - 0
2
(v1 - v 2 ) 2 = v 21 + v 22 - 2v1v 2 =
v2
3 2
v0 + 0
2
2
Solving we get relative velocity between the two particles
v1 - v2 = 2v0
F
2P
=
n (Q n = no.of particles)
A
A
2 ´ 3.32 ´ 10 -27 ´ 103 ´ 1023
=
=2.35 × 103N/m2
2 ´ 10 -4
78. (a) For collision of neutron with deuterium:
v
v1
v2
m
m
2m
2m
Applying conservation of momentum :
mv + 0 = mv1 + 2mv2
.....(i)
v2 – v1 = v
.....(ii)
Q Collision is elastic, e = 1
v
From eqn (i) and eqn (ii) v1 = 3
1
2 1
2
mv - mv1
8
2
= = 0.89
Pd = 2
1
9
2
mv
2
Now, For collision of neutron with carbon nucleus
v
v1
v2
m
m
12m
12m
Applying Conservation of momentum
mv + 0 = mv1+ 12mv2
....(iii)
v = v2 – v1
....(iv)
From eqn (iii) and eqn (iv)
11
v1 = - v
13
2
1
1 æ 11 ö
mv2 - m ç v ÷
48
2
2 è 13 ø
=
» 0.28
Pc =
1
169
mv2
2
79. (d) Apply principle of conservation of momentum along
x-direction,
mu = mv1 cos 45° + Mv2 cos 45°
1
mu =
(mv1 + Mv2 )
..... (i)
2
Along y-direction,
o = mv1 sin 45° – Mv2 sin 45°
=
P-73
Work, Energy & Power
1
1
mv 2 < 2´10.2 < 20.4 eV
2
Y
pf = 3 m V
..... (ii)
90°
M, u2 = 0
,v2
M
m, u1 = u
Proton
m
,v
1
o = (mv1 – Mv2)
2
Unknown mass
Before collision
After collision
v2 - v1 cos90
u cos 45
(Q Collision is elastic)
m
82. (a)
2v
pi
Coefficient of restution e = 1 =
v2
=1
u
2
..... (iii)
Þ u = 2v2
Solving eqs (i), (ii), & (iii), we get mass of unknown
particle, M = m
80. (a) For particle C,
According to law of conservation of linear momentum,
verticle component,
2 mv' sin q = mv sin 60° + mv sin 45°
Þ
mv mv 3 ...... (i)
+
2
2
Horizontal component,
2 mv' cos q = mv sin 60° – mv cos 45°
mv mv
...... (ii)
2mv'cos q =
+
2
2
Y
A
B
v sin 60°
2mv 'sin q =
Y'
v sin 45°
30°
60°
45°
X'
– v cos 45°
For particle B
X
v cos 60°
For particle A
1
v
<
v<
(1 ∗ 1)
2
n ↑ v(H)
(n)(H) ↑
v
2
(m1 ∗ m 2 )
v
Before
After
æ v ö2 1
1
1
mv 2 , (2m) çç ÷÷÷ < mv2
èç 2 ø
2
2
4
K.E. lost is used to jump from 1st orbit to 2nd orbit
DK.E. = 10.2ev
Minimum K.E. of neutron for inelastic collision
Loss in K.E. =
v
2m
Initial momentum of the system
pi =
[m(2V) 2 ´ 2m(2V) 2 ]
= 2m ´ 2V
Final momentum of the system = 3mV
By the law of conservation of momentum
2 2mv = 3mV Þ
2 2v
= Vcombined
3
Loss in energy
1
1
1
2
DE = m1V12 + m2V22 - (m1 + m2 )Vcombined
2
2
2
4
5
DE = 3mv2 - mv 2 = mv 2 = 55.55%
3
3
Percentage loss in energy during the collision ; 56%
83. (b) Given, m1 = 4g, u1 = 300 m/s
m2 = 0.8 kg = 800 g, u2 = 0 m/s
From law of conservation of momentum,
m1u1 + m2u2 = m1v1 + m2v2
Let the velocity of combined system = v m/s
then,
4 × 300 + 800 × 0 = (800 + 4) × v
1200
= 1.49 m / s
804
Now, m = 0.3 (given)
a = mg
a = 0.3 × 10
= 3 m/s2
then, from v2 = u2 + 2as
(1.49)2 = 0 + 2 × 3 × s
1
3
+
2+ 3
2
tan q = 2
=
1 1
1
- 2
2
2
m1
X
v=
Dividing eqn (i) by eqn (ii),
81. (a) For inelastic collision v ' <
45°
s=
(take g = 10 m/s2)
(1.49 )2
6
2.22
s=
6
= 0.379 m
84. (d) From the law of conservation of momentum we know
that,
m1u1 + m2u2 + .... = m1v1 + m2v2 + ....
Given m1 = m, m2 = 2m and m3 = 3m
and u1 = 3u, u2 = 2u and u3 = u
®
Let the velocity when they stick = v
P-74
Physics
Initial momentum of the system block (C)
= mv0. After striking with A, the block C comes to rest
and now both block A and B moves with velocity v when
compression in spring is x0.
By the law of conservation of linear momentum
v0
mv0 = (m + 2m) v Þ v =
3
By the law of conservation of energy
K.E. of block C = K.E. of system + P.E. of system
Then, according to question,
Y
2m, 2u
j sin 60°
j
m, 3u
P
60°
60° i
X
i cos 60°
(–j) sin 60°
2
1 2 1
æv ö
1
mv0 = ( 3m ) ç 0 ÷ + kx02
è 3ø
2
2
2
–3m, u
()
(
m × 3u î + 2m × 2u -ˆi cos 60° - ˆjsin 60°
(
)
)
®
+ 3 m × u -ˆi cos 60° + ˆjsin 60° = (m + 2m + 3m) v
ˆ
ˆ
æ
ö
Þ 3muiˆ - 4mu i - 4mu ç 3 ˆj÷ - 3mu i
2
2
è 2 ø
®
æ 3 ˆö
+ 3mu çç
j ÷÷ = 6m v
è 2 ø
®
3
3 ˆ
muj = 6m v
Þ muiˆ - muiˆ 2
2
®
1
3 ˆ
muj = 6m v
Þ - muiˆ 2
2
®
u ˆ
- i - 3jˆ
Þ v=
12
P2
P2
85. (d) Maximum energy loss =
2m 2(m + M)
(
)
é
P2 1 2 ù
= mv ú
êQ K.E. =
2m 2
êë
úû
P2 é M ù 1
ì M ü
= mv 2 í
ý
ê
ú
2m ë (m + M) û 2
îm + M þ
Statement II is a case of perfectly inelastic collision.
By comparing the equation given in statement I with above
equation, we get\
=
æ M ö
æ m ö
f =ç
è m + M ÷ø instead of çè M + m ÷ø
Hence statement I is wrong and statement II is correct.
86. (c)
87. (c) Fractional decrease in kinetic energy of mass m
æ m2 - m1 ö
= 1- ç
è m2 + m1 ÷ø
2
æ 2 - 1ö
= 1- ç
è 2 + 1÷ø
2
2
1 8
æ 1ö
= 1 – ç ÷ = 1- =
è 3ø
9 9
Percentage loss in energy
8
= ´ 100 ; 90%
9
A
C
88. (d)
m
m
v0
B
2m
Þ
1 2 1 2 1 2
mv0 = mv0 + kx0
2
6
2
Þ
2
1 2 1 2 1 2 mv0
kx0 = mv0 - mv0 =
2
2
6
3
\
2 æv ö
k = mç 0 ÷
3 è x0 ø
89. (c)
Y
2
m/2, + v = 400 m/s
Mass before explosion = m
490 m and velocity v = 200 m/s (vertically)
X
O
Momentum before explosion = Momentum after explosion
m
m
m ´ 200 ˆj = ´ 400 ˆj + v
2
2
m
ˆ
400 j + v
=
2
Þ 400 ˆj - 400 ˆj = v
(
)
\ v=0
i.e., the velocity of the other part of the mass, v = 0
Let time taken to reach the earth by this part be t
1
Applying formula, h = ut + gt 2
2
1
490 = 0 + × 9.8 × t2
2
980
2
= 100
Þ t =
9.8
\ t = 100 = 10sec
90. (a) In completely inelastic collision, all initial kinetic
energy is not lost but loss in kinetic energy 15 as large
as it can be. Linear momentum remain conserved in all
types of collision. Statement -2 explains statement -1
correctly because applying the principle of conservation
of momentum, we can get the common velocity and hence
the kinetic energy of the combined body.
91. (c) Initial kinetic energy of the system
K.Ei =
1 2 1
mu + M (0) 2
2
2
P-75
Work, Energy & Power
Momentum of the block, = M × v
1
´ 0.5 ´ 2 ´ 2 + 0 = 1J
2
Momentum before collision
= Momentum after collision
m1u1 + m2u2 = (m + M) × v
=
k
× L = kM × L
M
94. (d) Considering conservation of momentum along x-direction,
mv = mv1 cos q
...(1)
where v1 is the velocity of second mass
In y-direction,
mv
0=
- mv1 sin q
3
mv
or m1v1 sin q =
...(2)
3
=M×
\ 0.5 × 2 + 1 × 0 = (0.5 + 1) × v Þ v =
2
m/s
3
Final kinetic energy of the system is
1
K.E f = (m + M )v 2
2
1
2 2 1
= (0.5 + 1) ´ ´ = J
2
3 3 3
\ Energy loss during collision
æ 1ö
= ç1 - ÷ J = 0.67J
è 3ø
92. (b) Let the velocity and mass of 4 kg piece be v1 and m1
and that of 12 kg piece be v2 and m2.
v/ 3
m
v
v
16 kg
Situation 1
4 kg = m1
v1
Initial momentum
=0
m2 = 12 kg Final momentum
v2
= m2v2 – m1v1
Situation 2
Applying conservation of linear momentum
16 × 0 = 4 × v1 + 12 × 4
12 ´ 4
= –12 ms -1
Þ v1 = –
4
Kinetic energy of 4 kg mass
1
1
\ K . E. = m1v12 = ´ 4 ´144 = 288 J
2
2
93. (b) When the spring gets compressed by length L.
K.E. lost by mass M = P.E. stored in the compressed spring.
1
1
Mv 2 = k L2
2
2
k
×L
Þ v=
M
M
v1 cosq
q
v1
v1 sinq
Squaring and adding eqns. (1) and (2) we get
v12 = v2 +
v2
Þ v1 =
2
v
3
3
95. (c) Kinetic energy of a system of particle is zero only
when the speed of each particles is zero. This implies
momentum of each particle is zero, thus linear momentum
of the system of particle has to be zero.
Also if linear momentum of the system is zero it does not
mean linear momentum of each particle is zero. This is because
linear momentum is a vector quantity. In this case the kinetic
energy of the system of particles will not be zero.
\ A does not imply B but B implies A.
Given, force, F = 200 N extension of wire, x = 1mm.
P-76
6
Physics
System of Particles and
Rotational Motion
Centre of Mass, Centre of
TOPIC 1 Gravity & Principle of
Moments
1.
2.
The centre of mass of a solid hemisphere of radius 8 cm is
x cm from the centre of the flat surface. Then value of x is
______.
[NA Sep. 06, 2020 (II)]
a
A square shaped hole of side l =
is carved out at a
2
a
distance d = from the centre ‘O’ of a uniform circular
2
disk of radius a. If the distance of the centre of mass of the
a
remaining portion from O is - , value of X (to the nearest
X
integer) is ___________.
[NA Sep. 02, 2020 (II)]
5.
(a) (1.25 m, 1.50 m)
(b) (0.75 m, 1.75 m)
(c) (0.75 m, 0.75 m)
(d) (1 m, 1.75 m)
As shown in fig. when a spherical cavity (centred at O) of
radius 1 is cut out of a uniform sphere of radius R (centred
at C), the centre of mass of remaining (shaded) part of
sphere is at G, i.e on the surface of the cavity. R can be
determined by the equation:
[8 Jan. 2020 II]
(a) (R2 + R + 1) (2 – R) = 1
(b) (R2 – R – l) (2 – R) = 1
a
O
(c) (R2 – R + l) (2 – R) = l
d
(d) (R2 + R – 1) (2 – R) = 1
l = a/2
6.
3.
A rod of length L has non-uniform linear mass density
2
æ xö
given by r(x) = a + b ç ÷ , where a and b are constants
èLø
and 0 £ x £ L. The value of x for the centre of mass of the
rod is at:
[9 Jan. 2020 II]
(a)
(b)
3 æ 2a + b ö
ç
÷L
4 è 3a + b ø
4æ a +b ö
3 æ 2a + b ö
(d) ç
ç
÷L
÷L
3 è 2a + 3b ø
2 è 3a + b ø
The coordinates of centre of mass of a uniform flag shaped
lamina (thin flat plale) of mass 4 kg. (The coordinates of
the same are shown in figure) are:
[8 Jan. 2020 I]
(c)
4.
3æ a +b ö
ç
÷L
2 è 2a + b ø
Three point particles of masses 1.0 kg, 1.5 kg and 2.5 kg
are placed at three corners of a right angle triangle of
sides 4.0 cm, 3.0 cm and 5.0 cm as shown in the figure.
The center of mass of the system is at a point:
[7 Jan. 2020 I]
(a)
(b)
(c)
(d)
0.6 cm right and 2.0 cm above 1 kg mass
1.5 cm right and 1.2 cm above 1 kg mass
2.0 cm right and 0.9 cm above 1 kg mass
0.9 cm right and 2.0 cm above 1 kg mass
P-77
System of Particles and Rotational Motion
7.
Three particles of masses 50 g, 100 g and 150 g are placed
at the vertices of an equilateral triangle of side 1 m (as
shown in the figure). The (x, y) coordinates of the centre of
mass will be :
[12 Apr. 2019 II]
10. The position vector of the centre of mass r cm of an
asymmetric uniform bar of negligible area of crosssection as shown in figure is:
[12 Jan. 2019 I]
L
L
8.
æ 3
5 ö
(a) çç 4 m, 12 m ÷÷
è
ø
æ 7
3 ö
(b) çç 12 m, 8 m ÷÷
è
ø
(a)
æ 7
3 ö
(c) çç 12 m, 4 m ÷÷
è
ø
æ 3
7 ö
(d) çç 8 m, 12 m ÷÷
è
ø
(c)
Four particles A, B, C and D with masses mA = m, mB =
2m, mC = 3m and mD = 4m are at the corners of a square.
They have accelerations of equal magnitude with
directions as shown. The acceleration of the centre of
mass of the particles is :
[8 April 2019 I]
2L
3L
r
13
5
rcm = L xˆ + L yˆ
8
8
(b)
r
5
13
rcm = L xˆ + L yˆ
8
8
r
3
11
rcm = L xˆ + L yˆ
8
8
(d)
r
11
3
rcm = L xˆ + L yˆ
8
8
11. A force of 40 N acts on a point B at the end of an L-shaped
object, as shown in the figure. The angle q that will produce
maximum moment of the force about point A is given by:
[Online April 15, 2018]
A
1
(a) tan q =
4
(b) tan q = 2
(c)
4m
1
tan q =
2
®
F
(d) tan q = 4
(a)
( )
a $ $
i– j
5
( )
a $ $
i+ j
5
A uniform rectangular thin sheet ABCD of mass M has
length a and breadth b, as shown in the figure. If the shaded
portion HBGO is cut-off, the coordinates of the centre of
mass of the remaining portion will be : [8 Apr. 2019 II]
(c) Zero
9.
(b) a
(d)
2m
q
B
12. In a physical balance working on the principle of moments,
when 5 mg weight is placed on the left pan, the beam
becomes horizontal. Both the empty pans of the balance
are of equal mass. Which of the following statements is
correct ?
[Online April 8, 2017]
(a) Left arm is longer than the right arm
(b) Both the arms are of same length
(c) Left arm is shorter than the right arm
(d) Every object that is weighed using this balance
appears lighter than its actual weight.
13. In the figure shown ABC is a uniform wire. If centre of
BC
is
AB
[Online April 10, 2016]
mass of wire lies vertically below point A, then
close to :
A
æ 3a 3b ö
(a) ç , ÷
è 4 4 ø
æ 5a 5b ö
(b) ç , ÷
è 3 3 ø
æ 2a 2b ö
(c) ç , ÷
è 3 3 ø
5a 5b
(d) æç , ö÷
12
12 ø
è
60°
(a)
(c)
B
1.85
1.37
C
(b) 1.5
(d) 3
P-78
Physics
14. Distance of the centre of mass of a solid uniform cone
from its vertex is z0. If the radius of its base is R and its
height is h then z0 is equal to :
[2015]
5h
3h
3h 2
h2
(b)
(c)
(d)
8
4
8R
4R
15. A uniform thin rod AB of length L has linear mass
bx
density m (x) = a +
, where x is measured from A. If
L
æ 7ö
the CM of the rod lies at a distance of ç ÷ L from A,
è 12 ø
then a and b are related as : [Online April 11, 2015]
(a) a = 2b
(b) 2a = b
(c) a = b
(d) 3a = 2b
16. A thin bar of length L has a mass per unit length l, that
increases linearly with distance from one end. If its total
mass is M and its mass per unit length at the lighter end is
lO, then the distance of the centre of mass from the lighter
end is:
[Online April 11, 2014]
(a)
2
L l o L2
L loL
+
(b)
3 8M
2 4M
L l o L2
2L l o L2
+
(c)
(d)
3
4M
3
6M
17. A boy of mass 20 kg is standing on a 80 kg free to move
long cart. There is negligible friction between cart and
ground. Initially, the boy is standing 25 m from a wall. If he
walks 10 m on the cart towards the wall, then the final
distance of the boy from the wall will be
[Online April 23, 2013]
(a) 15 m
(b) 12.5 m (c) 15.5 m (d) 17 m
18. A thin rod of length ‘L’ is lying along the x-axis with its
ends at x = 0 and x = L. Its linear density (mass/length)
n
æ xö
varies with x as k ç ÷ , where n can be zero or any
è Lø
positive number. If the position xCM of the centre of mass
of the rod is plotted against ‘n’, which of the following
graphs best approximates the dependence of xCM on n?
[2008]
xCM
xCM
(a)
(a)
(c)
L
L
2
O
xCM
L
L
2
O
(b)
n
(d)
n
L
2
O
xCM
L
L
2
O
n
20. Consider a two particle system with particles having masses
m1 and m2. If the first particle is pushed towards the centre
of mass through a distance d, by what distance should the
second particle is moved, so as to keep the centre of mass
at the same position?
[2006]
m2
m1
(a) m d
(b) m + m d
1
1
2
(c)
m1
d
m2
(d) d
21. A body A of mass M while falling vertically downwards
1
under gravity breaks into two parts; a body B of mass
3
2
M and a body C of mass
M. The centre of mass of
3
bodies B and C taken together shifts compared to that of
body A towards
[2005]
(a) does not shift
(b) depends on height of breaking
(c) body B
(d) body C
22. A ‘T’ shaped object with dimensions shown in the figure,
r
is lying on a smooth floor. A force ‘ F ’ is applied at the
point P parallel to AB, such that the object has only the
translational motion without rotation. Find the location of
P with respect to C.
[2005]
l
B
A
P
2l
F
C
(a)
3
l
2
TOPIC 2
(b)
2
l
3
(c) l
4
l
3
Angular Displacement,
Velocity and Aceleration
23. A bead of mass m stays at point P(a, b) on a wire bent in
the shape of a parabola y = 4Cx2 and rotating with angular
speed w (see figure). The value of w is (neglect friction) :
[Sep. 02, 2020 (I)]
y
w
P (a, b)
n
19. A circular disc of radius R is removed from a bigger circular
disc of radius 2R such that the circumferences of the discs
coincide. The centre of mass of the new disc is
a/R form the centre of the bigger disc. The value of a is
[2007]
(a) 1/4
(b) 1/3
(c) 1/2
(d) 1/6
(d)
x
O
(a) 2 2gC
(c)
2gC
ab
(b) 2 gC
(d)
2g
C
P-79
System of Particles and Rotational Motion
24. A cylindrical vessel containing a liquid is rotated about
its axis so that the liquid rises at its sides as shown in the
figure. The radius of vessel is 5 cm and the angular speed
of rotation is w rad s –1. The difference in the height,
h (in cm) of liquid at the centre of vessel and at the side
will be :
[Sep. 02, 2020 (I)]
(a)
w
(b)
h
10 cm
(a)
2w2
25g
(b)
25w2
(c)
2g
5w2
2g
2w2
(d)
5g
25. A spring mass system (mass m, spring constant k and
natural length l) rests in equilibrium on a horizontal disc.
The free end of the spring is fixed at the centre of the disc.
If the disc together with spring mass system, rotates about
it’s axis with an angular velocity w, (k >> mw2) the relative
change in the length of the spring is best given by the
option:
[9 Jan. 2020 II]
(a)
2 æ mw2 ö
ç
÷
3 çè k ÷ø
(b)
2mw2
k
mw2
mw2
(d)
k
3k
26. A particle of mass m is fixed to one end of a light spring
having force constant k and unstretched length l. The other
end is fixed. The system is given an angular speed w about
the fixed end of the spring such that it rotates in a circle in
gravity free space. Then the stretch in the spring is:
[8 Jan. 2020 I]
(c)
(a)
(c)
ml w2
k - wm
(b)
ml w 2
k + mw
(c)
2
(d)
ml w 2
k - mw 2
ml w 2
k + mw
27. A uniform rod of length l is being rotated in a horizontal
plane with a constant angular speed about an axis passing
through one of its ends. If the tension generated in the rod
due to rotation is T(x) at a distance x from the axis, then
which of the following graphs depicts it most closely?
[12 Apr. 2019 II]
(d)
28. A smooth wire of length 2pr is bent into a circle and kept
in a vertical plane. A bead can slide smoothly on the wire.
When the circle is rotating with angular speed w about
the vertical diameter AB, as shown in figure, the bead is
at rest with respect to the circular ring at position P as
shown. Then the value of w2 is equal to :
[12 Apr. 2019 II]
3g
(b) 2 g / (r 3)
2r
(d) 2g/r
(c) ( g 3) / r
29. A long cylindrical vessel is half filled with a liquid. When
the vessel is rotated about its own vertical axis, the liquid
rises up near the wall. If the radius of vessel is 5 cm and its
rotational speed is 2 rotations per second, then the
difference in the heights between the centre and the sides, in
cm, will be :
[12 Jan. 2019 II]
(a) 2.0
(b) 0.1
(c) 0.4
(d) 1.2
(a)
P-80
Physics
30. A particle is moving with a uniform speed in a circular
orbit of radius R in a central force inversely proportional
to the nth power of R. If the period of rotation of the
particle is T, then:
[2018]
(b) T µ R n /2+1
(c) T µ R (n +1)/2
(d) T µ R n /2
31. The machine as shown has 2 rods of length 1 m connected
by a pivot at the top. The end of one rod is connected to
the floor by a stationary pivot and the end of the other rod
has a roller that rolls along the floor in a slot.
As the roller goes back and forth, a 2 kg weight moves up
and down. If the roller is moving towards right at a constant
speed, the weight moves up with a : [Online April 9, 2017]
(a) T µ R3/2 for any n.
2 kg
F
Fixed pivot
x
34. A cubical block of side 30 cm is moving with velocity
2 ms–1 on a smooth horizontal surface. The surface has a
bump at a point O as shown in figure. The angular velocity
(in rad/s) of the block immediately after it hits the bump,
is :
[Online April 9, 2016]
a = 30 cm
O
(a) 13.3
(b) 5.0
(c) 9.4
(d) 6.7
35. Two point masses of mass m1 = fM and m2 = (1 – f) M (f
< 1) are in outer space (far from gravitational influence of
other objects) at a distance R from each other. They move
in circular orbits about their centre of mass with angular
velocities w1 for m1 and w2 for m2. In that case
[Online May 19, 2012]
(a) (1 – f) w1 = fw
(b) w1 = w2 and independent of f
(c) fw1 = (1 – f)w2
(d) w1 = w2 and depend on f
Movable roller
TOPIC 3
36. Four point masses, each of mass m, are fixed at the corners
of a square of side l. The square is rotating with angular
frequency w, about an axis passing through one of the
corners of the square and parallel to its diagonal, as
shown in the figure. The angular momentum of the square
about this axis is :
[Sep. 06, 2020 (I)]
(a) ml 2w
(a)
3g
cos q
2l
(b)
2g
cos q
3l
3g
2g
sin q
sin q
(d)
2l
2l
33. Concrete mixture is made by mixing cement, stone and
sand in a rotating cylindrical drum. If the drum rotates too
fast, the ingredients remain stuck to the wall of the drum
and proper mixing of ingredients does not take place. The
maximum rotational speed of the drum in revolutions per
minute (rpm) to ensure proper mixing is close to :
(Take the radius of the drum to be 1.25 m and its axle to
be horizontal):
[Online April 10, 2016]
(a) 27.0
(b) 0.4
(c) 1.3
(d) 8.0
(c)
Torque, Couple and
Angular Momentum
ax
is
(a) constant speed
(b) decreasing speed
(c) increasing speed
3
(d) speed which is th of that of the roller when the
4
weight is 0.4 m above the ground
32. A slender uniform rod of mass M and length l is pivoted
at one end so that it can rotate in a vertical plane (see
figure). There is negligible friction at the pivot. The free
end is held vertically above the pivot and then released.
The angular acceleration of the rod when it makes an angle
[2017]
q with the vertical is
(b) 4 ml2w
(c) 3 ml2w
(d) 2 ml2w
37. A thin rod of mass 0.9 kg and length 1 m is suspended, at
rest, from one end so that it can freely oscillate in the
vertical plane. A particle of move 0.1 kg moving in a straight
line with velocity 80 m/s hits the rod at its bottom most
point and sticks to it (see figure). The angular speed
(in rad/s) of the rod immediately after the collision will be
______________.
[NA Sep. 05, 2020 (II)]
38. A person of 80 kg mass is standing on the rim of a circular
platform of mass 200 kg rotating about its axis at 5
revolutions per minute (rpm). The person now starts
moving towards the centre of the platform. What will be
the rotational speed (in rpm) of the platform when the
person reaches its centre __________.
[NA Sep. 03, 2020 (I)]
P-81
System of Particles and Rotational Motion
39. A block of mass m = 1 kg slides with velocity v = 6 m/s
on a frictionless horizontal surface and collides with a
uniform vertical rod and sticks to it as shown. The rod is
pivoted about O and swings as a result of the collision
making angle q before momentarily coming to rest. If
the rod has mass M = 2 kg, and length l = 1 m, the value
of q is approximately: (take g = 10 m/s2)
42. A uniform cylinder of mass M and radius R is to be pulled
over a step of height a (a < R) by applying a force F at its
centre 'O' perpendicular to the plane through the axes of
the cylinder on the edge of the step (see figure). The
minimum value of F required is :
[Sep. 02, 2020 (I)]
F
[Sep. 03, 2020 (I)]
O
M, l
q
m v
(a) 63°
(c) 69°
40.
æ R-aö
(a) Mg 1 - ç
÷
è R ø
m
m
(b) 55°
(d) 49°
w
l
q
A uniform rod of length 'l' is pivoted at one of its ends on
a vertical shaft of negligible radius. When the shaft rotates
at angular speed w the rod makes an angle q with it (see
figure). To find q equate the rate of change of angular
momentum (direction going into the paper)
ml 2 2
w sin q cos q about the centre of mass (CM) to the
12
torque provided by the horizontal and vertical forces FH
and FV about the CM. The value of q is then such that :
[Sep. 03, 2020 (II)]
(a) cos q =
(c) cos q =
2g
(b) cos q =
3l w2
g
lw
(d) cos q =
2
g
2l w2
3g
2l w2
41.
A
25
50
75
2m
a
2
æ R ö
(b) Mg ç
÷ -1
è R-aø
a
a2
(d) Mg 1 - 2
R
R
43. Consider a uniform rod of mass M = 4m and length l pivoted
about its centre. A mass m moving with velocity v making
p
angle q = to the rod’s long axis collides with one end of
4
the rod and sticks to it. The angular speed of the rod-mass
system just after the collision is:
[8 Jan. 2020 I]
3 v
3v
(a)
(b)
7 2l
7l
4v
3 2v
(d)
7l
7 l
44. A particle of mass m is moving along a trajectory given by
x = x0 + a cos w1t
y = y0 + b cos w2 t
The torque, acting on the particle about the origin, at t = 0
is:
[10 Apr. 2019 I]
2$
2$
+
my
a
(a) m ( - x0 b + y0 a ) w1 k
(b)
0 w1 k
2
2 $
(c) zero
(d) -m( x0 bw2 - y0 aw1 )k
45. The time dependence of the position of a particle of mass
r
m = 2 is given by r (t ) = 2 t i$ - 3t 2 $j . Its angular momentum,
with respect to the origin, at time t = 2 is :
[10 Apr. 2019 II]
$
$
(a) 48 i + j
(b) 36k$
(c)
(
(
)
(c) -34 k$ - $i
0
2
R
(c) Mg
FV
FH
O
100
B
Shown in the figure is rigid and uniform one meter long
rod AB held in horizontal position by two strings tied to
its ends and attached to the ceiling. The rod is of mass 'm'
and has another weight of mass 2 m hung at a distance of
75 cm from A. The tension in the string at A is :
[Sep. 02, 2020 (I)]
(a) 0.5 mg
(b) 2 mg
(c) 0.75 mg
(d) 1 mg
)
(d) -48k$
46. A metal coin of mass 5 g and radius 1 cm is fixed to a thin
stick AB of negligible mass as shown in the figure The
system is initially at rest. The constant torque, that will
make the system rotate about AB at 25 rotations per second
in 5s, is close to :
[10 Apr. 2019 II]
P-82
Physics
(a) 4.0×10–6 Nm
(b) 1.6×10–5 Nm
(c) 7.9×10–6 Nm
(d) 2.0×10–5 Nm
47. A rectangular solid box of length 0.3 m is held
horizontally, with one of its sides on the edge of a
platform of height 5m. When released, it slips off the
table in a very short time t = 0.01 s, remaining essentially
horizontal. The angle by which it would rotate when it
hits the ground will be (in radians) close to :
[8 Apr. 2019 II]
(a) 0.5
(b) 0.3
(c) 0.02
(d) 0.28
48. A particle of mass 20 g is released with an initial velocity
5 m/s along the curve from the point A, as shown in the
figure. The point A is at height h from point B. The particle
slides along the frictionless surface. When the particle
reaches point B, its angular momentum about O will be :
(Take g = 10 m/s2)
[12 Jan. 2019 II]
O
a = 10 m
A
h = 10 m
B
(a) 2 kg-m2/s
(b) 8 kg-m2/s
50. The magnitude of torque on a particle of mass 1 kg is 2.5
Nm about the origin. If the force acting on it is 1 N, and the
distance of the particle from the origin is 5m, the angle
between the force and the position vector is (in radians):
[11 Jan. 2019 II]
p
p
p
p
(b)
(c)
(d)
6
3
8
4
51. To mop-clean a floor, a cleaning machine presses a
circular mop of radius R vertically down with a total
force F and rotates it with a constant angular speed
about its axis. If the force F is distributed uniformly
over the mop and if coefficient of friction between the
mop and the floor is m, the torque, applied by the
machine on the mop is:
[10 Jan. 2019 I]
(a) m FR/3
(b) m FR/6
2
μ mFR
(c) m FR/2
(d)
3
52. A rigid massless rod of length 3l has two masses attached
at each end as shown in the figure. The rod is pivoted at
point P on the horizontal axis (see figure). When released
from initial horizontal position, its instantaneous angular
acceleration will be:
[10 Jan. 2019 II]
2l
l
(a)
P
5 Mo
2 Mo
g
g
(b)
13l
3l
g
7g
(c)
(d)
2l
3l
53. An L-shaped object, made of thin rods of uniform mass
density, is suspended with a string as shown in figure. If
AB = BC, and the angle made by AB with downward
vertical is q, then:
[9 Jan. 2019 I]
(a)
(d) 3 kg-m2/s
uur
uur
49. A slab is subjected to two forces F1 and F2 of same
uur
magnitude F as shown in the figure. Force F2 is in XY(c) 6
kg-m2/s
plane while force F1 acts along z-axis at the point
r r
2i + 3 j . The moment of these forces about point O will
(
)
be :
[11 Jan. 2019 I]
Z
F1
F2
O
30º
6m
(a)
(c)
( 3$i - 2 $j + 3k$ ) F
( 3$i + 2 $j – 3k$ ) F
54.
(b)
(d)
tan q =
(c)
tan q =
y
4m
x
(a)
( 3$i - 2 $j – 3k$ ) F
( 3$i + 2 $j + 3k$ ) F
A
1
(b) tan q =
2 3
2
(d) tan q =
3
B
x
1
2
1
3
P-83
System of Particles and Rotational Motion
A uniform rod AB is suspended from a point X, at a variable
distance from x from A, as shown. To make the rod
horizontal, a mass m is suspended from its end A. A set of
(m, x) values is recorded. The appropriate variable that
give a straight line, when plotted, are:
[Online April 15, 2018]
1
1
(a) m,
(b) m, 2 (c) m, x
(d) m, x2
x
x
55. A thin uniform bar of length L and mass 8m lies on a smooth
horizontal table. Two point masses m and 2m moving in
the same horizontal plane from opposite sides of the bar
with speeds 2v and v respectively. The masses stick to the
L
L
bar after collision at a distance
and
respectively
3
6
from the centre of the bar. If the bar starts rotating about
its center of mass as a result of collision, the angular speed
of the bar will be:
[Online April 15, 2018]
L/6
v
L/3
2v
O
58. A bob of mass m attached to an inextensible string of
length l is suspended from a vertical support. The bob
rotates in a horizontal circle with an angular speed w rad/s
about the vertical. About the point of suspension: [2014]
(a) angular momentum is conserved.
(b) angular momentum changes in magnitude but not in
direction.
(c) angular momentum changes in direction but not in
magnitude.
(d) angular momentum changes both in direction and
magnitude.
59. A ball of mass 160 g is th rown up at an an gle
of 60° to the horizontal at a speed of 10 ms–1. The angular
momentum of the ball at the highest point of the trajectory
with respect to the point from which the ball is thrown is
nearly (g = 10 ms–2)
[Online April 19, 2014]
(a) 1.73 kg m2/s
(b) 3.0 kg m2/s
(c) 3.46 kg m2/s
(d) 6.0 kg m2/s
60. A particle is moving in a circular path of radius a, with a
constant velocity v as shown in the figure. The centre of
circle is marked by ‘C’. The angular momentum from the
origin O can be written as:
[Online April 12, 2014]
y
v
6v
3v
v
(b)
(c)
(d)
6L
5L
5L
5L
56. A particle of mass m is moving along the side of a square
of side 'a', with a uniform speed v in the x-y plane as shown
in the figure :
[2016]
(a)
O
C
y
D
O
a
V
C
a V
V a
V
A
B
a
V
45° R
61.
a
Which of the following statements is false for the angular
ur
momentum L about the origin?
ur
éR
ù
+ a ú k$ when the particle is moving from
(a) L = mv ê
ë 2
û
B to C.
ur mv
Rk$ when the particle is moving from D to A.
(b) L =
2
ur
mv $
Rk when the particle is moving from A to B.
(c) L = –
2
ur
éR
ù
- a ú k$ when the particle is moving from
(d) L = mv ê
ë 2
û
C to D.
57. A particle of mass 2 kg is on a smooth horizontal table
and moves in a circular path of radius 0.6 m. The height
of the table from the ground is 0.8 m. If the angular speed
of the particle is 12 rad s–1, the magnitude of its angular
momentum about a point on the ground right under the
centre of the circle is :
[Online April 11, 2015]
(a) 14.4 kg m2s–1
(b) 8.64 kg m2s–1
(c) 20.16 kg m2s–1
(d) 11.52 kg m2s–1
a
q
62.
63.
64.
x
(a) va (1 + cos 2q)
(b) va (1 + cos q)
(c) va cos 2q
(d) va
A particle of mass 2 kg is moving such that at time t, its
r
position, in meter, is given by r (t ) = 5iˆ - 2t 2 ˆj . The angular
momentum of the particle at t = 2s about the origin in
kg m–2 s–1 is :
[Online April 23, 2013]
(a) - 80 k$
(b) (10iˆ - 16 ˆj )
(c) -40k$
(d) 40k$
A bullet of mass 10 g and speed 500 m/s is fired into a door
and gets embedded exactly at the centre of the door. The
door is 1.0 m wide and weighs 12 kg. It is hinged at one end
and rotates about a vertical axis practically without friction.
The angular speed of the door just after the bullet embeds
into it will be :
[Online April 9, 2013]
(a) 6.25 rad/sec
(b) 0.625 rad/sec
(c) 3.35 rad/sec
(d) 0.335 rad/sec
A stone of mass m, tied to the end of a string, is whirled
around in a circle on a horizontal frictionless table. The
length of the string is reduced gradually keeping the
angular momentum of the stone about the centre of the
circle constant. Then, the tension in the string is given by
T = Arn, where A is a constant, r is the instantaneous
radius of the circle. The value of n is equal to
[Online May 26, 2012]
(a) – 1
(b) – 2
(c) – 4
(d) – 3
A thin horizontal circular disc is rotating about a vertical
axis passing through its centre. An insect is at rest at a
P-84
Physics
point near the rim of the disc. The insect now moves along
a diameter of the disc to reach its other end. During the
journey of the insect, the angular speed of the disc.[2011]
(a) continuously decreases
(b) continuously increases
(c) first increases and then decreases
(d) remains unchanged
65. A small particle of mass m is projected at an angle q with
the x-axis with an initial velocity n0 in the x-y plane as
n sin q
shown in the figure. At a time t < 0
, the angular
g
momentum of the particle is
[2010]
y
v0
(a) Angular velocity
(b) Angular momentum
(c) Moment of inertia
(d) Rotational
kinetic energy
r
70. Let F be the force
r acting on a particle having position
r
vector r , and T be the torque of this force about the
origin. Then
[2003]
r r
r r
(a) r .Tr = 0 and Fr .Tr ¹ 0
r
(b) r .T ¹ 0 and F .T = 0
r r
r r
(c) r .T ¹ 0 and F .T ¹ 0
r
r r
r
(d) r .T = 0 and F .T = 0
71. A particle of mass m moves along line PC with velocity v
as shown. What is the angular momentum of the particle
about P?
[2002]
C
L
q
P
x
(a)
-mg n0t 2 cos q ˆj
1
(c) - mg n 0 t 2 cos q kˆ
2
(b)
mg n 0t cos q kˆ
(d)
1
mgn 0 t 2 cos q iˆ
2
where iˆ, ˆj and k̂ are unit vectors along x, y and z-axis
respectively.
66. Angular momentum of the particle rotating with a central
force is constant due to
[2007]
(a) constant torque
(b) constant force
(c) constant linear momentum
(d) zero torque
67. A thin circular ring of mass m and radius R is rotating
about its axis with a constant angular velocity w. Two
objects each of mass M are attached gently to the
opposite ends of a diameter of the ring. The ring now
rotates with an angular velocity w' =
[2006]
w(m - 2M )
w (m + 2 M )
(b)
(a)
(m + 2M )
m
wm
wm
(c)
(d)
(m + M )
(m + 2M )
68. A force of – Fkˆ acts on O, the origin of the coordinate
system. The torque about the point (1, –1) is
[2006]
Z
O
Y
X
(a) F (iˆ - ˆj )
(b) - F (iˆ + ˆj )
ˆ
ˆ
(c) F (i + j )
(d) - F (iˆ - ˆj )
69. A solid sphere is rotating in free space. If the radius of
the sphere is increased keeping mass same, which one
of the following will not be affected ?
[2004]
l
(a) mvL
TOPIC 4
O
(b) mvl
(c) mvr
(d) zero.
Moment of Inertia and
Rotational K.E.
72. Shown in the figure is a hollow icecream cone (it is open at
the top). If its mass is M, radius of its top, R and height, H,
then its moment of inertia about its axis is :
[Sep. 06, 2020 (I)]
R
H
MR 2
M (R2 + H 2 )
(b)
2
4
MH 2
MR 2
(c)
(d)
3
3
73. The linear mass density of a thin rod AB of length L varies
xö
æ
from A to B as l ( x ) = l 0 ç1 + ÷ , where x is the distance
è Lø
from A. If M is the mass of the rod then its moment of inertia
about an axis passing through A and perpendicular to the
rod is :
[Sep. 06, 2020 (II)]
(a)
5
7
ML2
(b)
ML2
12
18
2
3
(c) ML2
(d) ML2
5
7
74. A wheel is rotating freely with an angular speed w on a
shaft. The moment of inertia of the wheel is I and the moment of inertia of the shaft is negligible. Another wheel
(a)
P-85
System of Particles and Rotational Motion
of moment of inertia 3I initially at rest is suddenly coupled
to the same shaft. The resultant fractional loss in the
kinetic energy of the system is :
[Sep. 05, 2020 (I)]
(a)
5
6
(b)
1
4
3
4
75. ABC is a plane lamina of the shape of an equilateral
triangle. D, E are mid points of AB, AC and G is the
centroid of the lamina. Moment of inertia of the lamina
about an axis passing through G and perpendicular to the
plane ABC is I0. If part ADE is removed, the moment of
(c) 0
(d)
inertia of the remaining part about the same axis is
NI0
16
where N is an integer. Value of N is _______.
[NA Sep. 04, 2020 (I)]
A
79. Moment of inertia of a cylinder of mass M, length L and
radius R about an axis passing through its centre and
perpendicular to the axis of the cylinder is
æ R 2 L2 ö
+ ÷÷ . If such a cylinder is to be made for a
I = M çç
è 4 12 ø
given mass of a material, the ratio L/R for it to have minimum
possible I is :
[Sep. 03, 2020 (I)]
(a)
2
3
(b)
3
2
(d)
2
3
80. An massless equilateral triangle EFG of side 'a' (As shown
in figure) has three particles of mass m situated at its
vertices. The moment of inertia of the system about the
N
ma 2
line EX perpendicular to EG in the plane of EFG is
20
where N is an integer. The value of N is _________.
[Sep. 03, 2020 (II)]
(c)
X
E
D
3
2
F
G
B
C
76. A circular disc of mass M and radius R is rotating about
its axis with angular speed w1. If another stationary disc
R
having radius
and same mass M is dropped co-axially
2
on to the rotating disc. Gradually both discs attain
constant angular speed w2. The energy lost in the process
is p% of the initial energy. Value of p is ___________.
[NA Sep. 04, 2020 (I)]
77. Consider two uniform discs of the same thickness and
different radii R1 = R and R2 = aR made of the same material.
If the ratio of their moments of inertia I1 and I2, respectively,
about their axes is I1 : I2 = 1 : 16 then the value of a is :
[Sep. 04, 2020 (II)]
(a) 2 2
(c) 2
78. y
O
(b) 2
(d) 4
O'
80 cm
x
60 cm
For a uniform rectangular sheet shown in the figure, the
ratio of moments of inertia about the axes perpendicular to
the sheet and passing through O (the centre of mass) and
O' (corner point) is :
[Sep. 04, 2020 (II)]
(a) 2/3
(b) 1/4
(c) 1/8
(d) 1/2
E
a
G
(b)
20
J
3
81. Two uniform circular discs are rotating independently in
the same direction around their common axis passing
through their centres. The moment of inertia and angular
velocity of the first disc are 0.1 kg-m2 and 10 rad s–1
respectively while those for the second one are 0.2 kg-m 2
and 5 rad s–1 respectively. At some instant they get stuck
together and start rotating as a single system about their
common axis with some angular speed. The kinetic energy
of the combined system is :
[Sep. 02, 2020 (II)]
(a)
10
J
3
5
2
J
J
(d)
3
3
82. Three solid spheres each of mass m and diameter d are
stuck together such that the lines connecting the centres
form an equilateral triangle of side of length d. The ratio
I0
of moment of inertia I0 of the system about an axis
IA
passing the centroid and about center of any of the spheres
IA and perpendicular to the plane of the triangle is:
[9 Jan. 2020 I]
13
(a)
23
15
(b)
13
23
(c)
13
13
(d)
15
(c)
P-86
Physics
83. One end of a straight uniform 1 m long bar is pivoted on
horizontal table. It is released from rest when it makes
an angle 30° from the horizontal (see figure). Its angular
speed when it hits the table is given as ns -1 , where n is
an integer. The value of n is ________ . [9 Jan. 2020 I]
84. A uniformly thick wheel with moment of inertia I and
radius R is free to rotate about its centre of mass (see
fig). A massless string is wrapped over its rim and two
blocks of masses m1 and m2 (m1 > m2) are attached to the
ends of the string. The system is released from rest. The
angular speed of the wheel when m1 descents by a distance
h is:
[9 Jan. 2020 II]
87. Mass per unit area of a circular disc of radius a depends
on the distance r from its centre as s(r) = A + Br. The
moment of inertia of the disc about the axis, perpendicular
to the plane and passing through its centre is:
[7 Jan. 2020 II]
aB ö
4æ A
(a) 2pa çè + ÷ø
4 5
Bö
4 æ A aB ö
4æA
(c) pa çè + ÷ø
(d) 2pa çè + ÷ø
4 5
4 5
88. A circular disc of radius b has a hole of radius a at its centre
(see figure). If the mass per unit area of the disc varies as
æ s0 ö
ç
÷ , then the radius of gyration of the disc about its axis
è r ø
passing through the centre is :
[12 Apr. 2019 I]
1/ 2
(a)
a2 + b2 + ab
2
(b)
a+b
2
1/ 2
(c)
a2 + b2 + ab
3
(d)
a+b
3
é 2(m1 - m2 ) gh ù
ú
(a) ê
2
ëê (m1 + m2 )R + 1 ûú
é 2(m1 + m2 ) gh ù
ú
(b) ê
2
êë (m1 + m2 )R + 1 úû
89. Two coaxial discs, having moments of inertia I1 and
1/2
é (m1 - m2 ) ù
ú
(c) ê
2
ëê (m1 + m2 )R + 1 ûú
gh
1/2
é
ù
m1 + m2
ú gh
(d) ê
2
êë (m1 + m2 )R + 1 úû
85. As shown in the figure, a bob of mass m is tied by a
massless string whose other end portion is wound on a
fly wheel (disc) of radius r and mass m. When released
from rest the bob starts falling vertically. When it has
covered a distance of h, the angular speed of the wheel
will be:
[7 Jan. 2020 I]
1 4 gh
r
3
(b) r
3
2 gh
w1
,
2
about their common axis. They are brought in contact with
each other and thereafter they rotate with a common angular
velocity. If Ef and Ei are the final and initial total energies,
then (Ef – Ei) is :
[10 Apr. 2019 I]
3 2
I w2
I1w12
I w2
(b) 1 1 (c) I1w1
(d) - 1 1
8
6
12
24
90. A thin disc of mass M and radius R has mass per unit area
s(r) = kr2 where r is the distance from its centre. Its moment
of inertia about an axis going through its centre of mass
and perpendicular to its plane is :
[10 Apr. 2019 I]
(a) -
3
1 2 gh
r
(d)
4 gh
r
3
86. The radius of gyration of a uniform rod of length l, about
l
an axis passing through a point
away from the centre
4
of the rod, and perpendicular to it, is: [7 Jan. 2020 I]
(a)
1
l
4
(b)
1
l
8
(c)
7
l
48
(d)
3
l
8
MR 2
3
(b)
MR 2
6
(d)
2MR 2
3
MR 2
2
91. A solid sphere of mass M and radius R is divided into two
(c)
(c)
I1
, are
2
rotating with respective angular velocities w1 and
(a)
(a)
Bö
4 æ aA
+ ÷
(b) 2pa çè
4 5ø
7M
and is
8
converted into a uniform disc of radius 2R. The second
part is converted into a uniform solid sphere. Let I1 be
the moment of inertia of the new sphere about its axis.
The ratio I1/I2 is given by :
[10 Apr. 2019 II]
(a) 185
(b) 140
(c) 285
(d) 65
unequal parts. The first part has a mass of
P-87
System of Particles and Rotational Motion
92. A stationary horizontal disc is free to rotate about its axis.
When a torque is applied on it, its kinetic energy as a
function of q, where q is the angle by which it has rotated,
is given as kq2. If its moment of inertia is I then the angular
acceleration of the disc is:
[9 April 2019 I]
k
k
k
2k
q
q (c) 2 I q
q
(a)
(b)
(d)
4I
I
I
93. Moment of inertia of a body about a given axis is 1.5 kg
m2. Initially the body is at rest. In order to produce a
rotational kinetic energy of 1200 J, the angular
acceleration of 20 rad/s2 must be applied about the axis
for a duration of:
[9 Apr. 2019 II]
(a) 2.5s
(b) 2s
(c) 5s
(d) 3s
94. A thin smooth rod of length L and mass M is rotating
freely with angular speed w0 about an axis perpendicular
to the rod and passing through its center. Two beads of
mass m and negligible size are at the center of the rod
initially. The beads are free to slide along the rod. The
angular speed of the system, when the beads reach the
opposite ends of the rod, will be:
[9 Apr. 2019 II]
M w0
M w0
(a)
(b)
M +m
M + 3m
M w0
M w0
(d)
(c)
M + 6m
M + 2m
95. A thin circular plate of mass M and radius R has its density
varying as r(r) = r0 r with r0 as constant and r is the
distance from its center. The moment of Inertia of the
circular plate about an axis perpendicular to the plate and
passing through its edge is I = a MR2. The value of the
coefficient a is:
[8 April 2019 I]
3
(a) 1 2
(b) 3 5
(c) 8 5
(d)
2
96. Let the moment of inertia of a hollow cylinder of length 30
cm (inner radius 10 cm and outer radius 20 cm), about its
axis be 1. The radius of a thin cylinder of the same mass
such that its moment of inertia about its axis is also I, is:
[12 Jan. 2019 I]
(a) 12 cm
(b) 16 cm
(c) 14 cm
(d) 18 cm
97. The moment of inertia of a solid sphere, about an axis
parallel to its diameter and at a distance of x from it, is ‘I(x)’.
Which one of the graphs represents the variation of I(x)
with x correctly?
[12 Jan. 2019 II]
I(x)
I(x)
(a)
(b)
x
O
x
O
I(x)
I(x)
(c)
(d)
O
x
O
x
98. An equilateral triangle ABC is cut from a thin solid sheet
of wood. (See figure) D, E and F are the mid-points of its
sides as shown and G is the centre of the triangle. The
moment of inertia of the triangle about an axis passing
through G and perpendicular to the plane of the triangle is
I0. If the smaller triangle DEF is removed from ABC, the
moment of inertia of the remaining figure about the same
axis is I. Then :
[11 Jan. 2019 I]
A
D
E
G
B
C
F
15
3
I0
(b) I = I0
16
4
9
I0
(c) I = I 0
(d) I =
16
4
99. a string is wound around a hollow cylinder of mass 5 kg
and radius 0.5 m. If the string is now pulled with a
horizontal force of 40 N, and the cylinder is rolling
without slipping on a horizontal surface (see figure), then
the angular acceleration of the cylinder will be (Neglect
the mass and thickness of the string) [11 Jan. 2019 II]
40 N
(a)
I=
(a) 20 rad/s2
(b) 16 rad/s2
2
(c) 12 rad/s
(d) 10 rad/s2
100. A circular disc D1 of mass M and radius R has two
identical discs D2 and D3 of the same mass M and radius
R attached rigidly at its opposite ends (see figure). The
moment of inertia of the system about the axis OO’,
passing through the centre of D1, as shown in the figure,
will be :
[11 Jan. 2019 II]
O'
D2
O
D3
D1
(a) MR2
(b) 3MR2
4
2
MR 2
MR 2
(d)
(c)
5
3
101. Two identical spherical balls of mass M and radius R
each are stuck on two ends of a rod of length 2R and
mass M (see figure). The moment of inertia of the
system about the axis passing perpendicularly through
the centre of the rod is:
[10 Jan. 2019 II]
P-88
Physics
(a)
(c)
137
MR 2
15
(b)
209
MR 2
15
(d)
17
MR 2
15
152
MR 2
15
m
are connected at the two ends of
2
a massless rigid rod of length l. The rod is suspended by
a thin wire of torsional constant k at the centre of mass
of the rod-mass system (see figure). Because of
torsional constant k, the restoring toruque is t= kq for
angular displacement q. If the rod is rotated by q0 and
released, the tension in it when it passes through its mean
position will be:
[9 Jan. 2019 I]
102. Two masses m and
105. From a uniform circular disc of radius R and mass 9 M, a
R
is removed as shown in the figure.
3
The moment of inertia of the remaining disc about an
axis perpendicular to the plane of the disc and passing,
through centre of disc is :
[2018]
small disc of radius
2R
3
R
40
37
MR2 (c) 10 MR2 (d)
MR2
9
9
106. A thin circular disk is in the xy plane as shown in the
figure. The ratio of its moment of inertia about z and z¢ axes
will be
[Online April 16, 2018]
(a) 4 MR2 (b)
z
z¢
O
3k q0 2
2k q0 2
k q0 2
k q0 2
(b)
(c)
(d)
l
l
l
2l
103. A rod of length 50 cm is pivoted at one end. It is raised
such that if makes an angle of 30° from the horizontal as
shown and released from rest. Its angular speed when
it passes through the horizontal (in rads–1) will be
(g = 10 ms–2)
[9 Jan. 2019 II]
x
(a)
(a) 1 : 2
(b) 1 : 4
(c) 1 : 3
(d) 1 : 5
107. A thin rod MN, free to rotate in the vertical plane about the
fixed end N, is held horizontal. When the end M is released
the speed of this end, when the rod makes an angle a with
the horizontal, will be proportional to: (see figure)
[Online April 15, 2018]
M
30°
(a)
30
7
(b)
30
20
30
(d)
3
2
104. Seven identical circular planar disks, each of mass M and
radius R are welded symmetrically as shown. The moment
of inertia of the arrangement about the axis normal to the
plane and passing through the point P is:
[2018]
(c)
P
O
(a)
19
MR 2
2
(b)
55
MR 2 (c)
2
73
181
MR 2 (d)
MR2
2
2
y
a
N
(a)
(b) cosa
cos a
(c) sin a
(d)
sin a
108. The moment of inertia of a uniform cylinder of length l and
radius R about its perpendicular bisector is I. What is the
ratio l/R such that the moment of inertia is minimum ?
[2017]
3
3
3
(a) 1
(b)
(c)
(d)
2
2
2
109. Moment of inertia of an equilateral triangular lamina ABC,
about the axis passing thr ough its centre O and
perpendicular to its plane is Io as shown in the figure. A
cavity DEF is cut out from the lamina, where D, E, F are
the mid points of the sides. Moment of inertia of the
remaining part of lamina about the same axis is :
[Online April 8, 2017]
(a)
7
Io
8
(b)
15
Io
16
(c)
3Io
4
(d)
31I o
32
C
E
F
O
A
D
B
P-89
System of Particles and Rotational Motion
R
is made in a thin uniform
4
disc having mass M and radius R, as shown in figure. The
moment of inertia of the remaining portion of the disc about
an axis passing through the point O and perpendicular to
the plane of the disc is :
[Online April 9, 2017]
219 MR 2
(a)
256
110. A circular hole of radius
R
237 MR 2
R/4
512
o'
O
19 MR 2
(c)
3R/4
512
197 MR 2
(d)
256
111. From a solid sphere of mass M and radius R a cube of
maximum possible volume is cut. Moment of inertia of cube
about an axis passing through its center and perpendicular
to one of its faces is :
[2015]
4MR 2
4MR 2
(a)
(b)
9 3p
3 3p
2
MR
MR 2
(c)
(d)
32 2p
16 2p
112. Consider a thin uniform square sheet made of a rigid
material. If its side is ‘a’ mass m and moment of inertia I
about one of its diagonals, then :[Online April 10, 2015]
(b)
(a)
I>
ma 2
12
(b)
ma 2
ma 2
<I<
24
12
ma 2
ma 2
(d) I =
24
12
113. A ring of mass M and radius R is rotating about its axis
with angular velocity w. Two identical bodies each of mass
m are now gently attached at the two ends of a diameter of
the ring. Because of this, the kinetic energy loss will be:
[Online April 25, 2013]
(c)
I=
Mm
m( M + 2m) 2 2
w2 R2
w R
(b)
( M + m)
M
Mm
( M + m) M 2 2
w R
w2 R2
(c)
(d)
( M + 2 m)
(M + 2m)
114. This question has Statement 1and Statement 2. Of the four
choices given after the Statements, choose the one that
best describes the two Statements.
Statement 1: When moment of inertia I of a body rotating
about an axis with angular speed w increases, its angular
momentum L is unchanged but the kinetic energy K
increases if there is no torque applied on it.
Statement 2: L = Iw, kinetic en ergy of rotation
1 2
= Iw
[Online May 12, 2012]
2
(a) Statement 1 is true, Statement 2 is true, Statement
2 is not the correct explanation of Statement 1.
(b) Statement 1 is false, Statement 2 is true.
(a)
(c) Statement 1 is true, Statement 2 is true, Statement
2 is correct explanation of the Statement 1.
(d) Statement 1 is true, Statement 2 is false.
115. A solid sphere having mass m and radius r rolls down an
inclined plane. Then its kinetic energy is
[Online May 7, 2012]
5
2
rotational and translational
(a)
7
7
2
5
(b)
rotational and translational
7
7
2
3
(c)
rotational and translational
5
5
1
1
(d)
rotational and translational
2
2
116. A circular hole of diameter R is cut from a disc of mass M
and radius R; the circumference of the cut passes through
the centre of the disc. The moment of inertia of the
remaining portion of the disc about an axis perpendicular
to the disc and passing through its centre is
[Online May 7, 2012]
(a)
æ 15 ö
2
çè ÷ø MR
32
(b)
æ 1ö
2
çè ÷ø MR
8
æ 3ö
æ 13 ö
2
2
(d) çè ÷ø MR
çè ÷ø MR
8
32
117. A mass m hangs with the help of a string wrapped around
a pulley on a frictionless bearing. The pulley has mass m
and radius R. Assuming pulley to be a perfect uniform
circular disc, the acceleration of the mass m, if the string
does not slip on the pulley, is:
[2011]
(c)
(a) g
(b)
2
g
3
(c)
g
3
(d)
3
g
2
118. A pulley of radius 2 m is rotated about its axis by a force
F = (20t – 5t2) newton (where t is measured in seconds)
applied tangentially. If the moment of inertia of the pulley
about its axis of rotation is 10 kg-m2 the number of rotations
made by the pulley before its direction of motion is
reversed, is:
[2011]
(a) more than 3 but less than 6
(b) more than 6 but less than 9
(c) more than 9
(d) less than 3
119. A thin uniform rod of length l and mass m is swinging
freely about a horizontal axis passing through its end. Its
maximum angular speed is w. Its centre of mass rises to
a maximum height of
[2009]
2
2
1 lw
1l w
(a)
(b)
6 g
2 g
2 2
1l w
1 l 2w2
(c)
(d)
6 g
3 g
120. Consider a uniform square plate of side ‘a’ and mass ‘M’.
The moment of inertia of this plate about an axis
perpendicular to its plane and passing through one of its
corners is
[2008]
P-90
Physics
(a)
5
Ma 2
6
(b)
1
Ma 2
12
7
2
Ma 2
Ma 2
(d)
12
3
121. For the given uniform square lamina ABCD, whose centre
is O,
[2007]
(c)
D
F
C
O
A
(a)
I AC = 2 I EF
B
E
(b)
2 I AC = I EF
(c) I AD = 3I EF
(d) I AC = I EF
122. Four point masses, each of value m, are placed at the
corners of a square ABCD of side l. The moment of inertia
of this system about an axis passing through A and parallel
to BD is
[2006]
(a)
2ml 2
(b)
3ml2
(c) 3ml 2
(d) ml 2
123. The moment of inertia of a uniform semicircular disc of
mass M and radius r about a line perpendicular to the
plane of the disc through the centre is
[2005]
1
2
Mr
(a)
(b)
Mr 2 `
4
5
(c)
1
Mr 2
2
(d)
(d)
IA dA
=
I B dB
where dA and dB are their densities.
125. A circular disc X of radius R is made from an iron plate of
thickness t, and another disc Y of radius 4R is made from an
t
. Then the relation between the
4
moment of inertia IX and II is
[2003]
iron plate of thickness
(a)
ΙY = 32 Ι X
(c) Ι Y = Ι X
(b)
(c)
æ M ö
çè
÷w
M + 4m ø 1
(d)
æ M ö
çè
÷ w1
M + 2mø
TOPIC 5 Rolling Motion
129.A uniform sphere of mass 500 g rolls without slipping
on a plane horizontal surface with its centre moving at a
speed of 5.00 cm/s. Its kinetic energy is:
[8 Jan. 2020 II]
(a) 8.75 × 10–4 J
(b) 8.75 × 10–3 J
(c) 6.25 × 10–4 J
(d) 1.13 × 10–3 J
130.
Mr 2
124. One solid sphere A and another hollow sphere B are of
same mass and same outer radii. Their moment of inertia
about their diameters are respectively IA and IB Such that
[2004]
(a) IA < IB
(b) IA > IB
(c) IA = IB
126. A particle performing uniform circular motion has angular
frequency is doubled & its kinetic energy halved, then the
new angular momentum is
[2003]
L
(a)
(b) 2 L
4
L
(c) 4 L
(d)
2
127. Moment of inertia of a circular wire of mass M and radius
R about its diameter is
[2002]
(a) MR2/2 (b) MR 2
(c) 2MR 2 (d) MR2/4
128. Initial angular velocity of a circular disc of mass M is
w 1. Then two small spheres of mass m are attached gently
to diametrically opposite points on the edge of the disc.
What is the final angular velocity of the disc? [2002]
æ M + mö
æ M + mö
(a) çè
(b) çè
÷w
÷w
M ø 1
m ø 1
Consider a uniform cubical box of side a on a rough floor
that is to be moved by applying minimum possible force F
at a point b above its centre of mass (see figure). If the
coefficient of friction is m = 0.4, the maximum possible value
b
for box not to topple before moving is
a
________.
[NA 7 Jan. 2020 II]
131.A solid sphere and solid cylinder of identical radii approach
an incline with the same linear velocity (see figure). Both
roll without slipping all throughout. The two climb
maximum heights hsph and hcyl on the incline. The ratio
of 100 ×
hsph
hcyl
is given by :
[8 Apr. 2019 II]
ΙY = 16 Ι X
(d) ΙY = 64 Ι X
(a)
2
5
(b) 1
(c)
14
15
(d)
4
5
P-91
System of Particles and Rotational Motion
132.The following bodies are made to roll up (without
slipping) the same inclined plane from a horizontal plane:
(i) a ring of radius R, (ii) a solid cylinder of radius
R
2
R
and (iii) a solid sphere of radius . If, in each case, the
4
speed of the center of mass at the bottom of the incline
is same, the ratio of the maximum heights they climb is:
[9 April 2019 I]
(a) 4 : 3 : 2
(b) 10 : 15 : 7
(c) 14 : 15 : 20
(d) 2 : 3 : 4
133. A homogeneous solid cylindrical roller of radius R and
mass M is pulled on a cricket pitch by a horizontal
force. Assuming rolling without slipping, angular
acceleration of the cylinder is:
[10 Jan. 2019 I]
(a)
3F
2mR
(b)
F
3m R
(c)
F
2mR
(d)
2F
3m R
134. A roller is made by joining together two cones at their
vertices O. It is kept on two rails AB and CD, which are
placed asymmetrically (see figure), with its axis
perpendicular to CD and its centre O at the centre of line
joining AB and Cd (see figure). It is given a light push so
that it starts rolling with its centre O moving parallel to
CD in the direction shown. As it moves, the roller will
tend to:
[2016]
B
D
O
Figure). If they roll on the incline without slipping such
sin qc
that their accelerations are the same, then the ratio
sin q s
is:
[Online April 9, 2014]
MC
A
M
S
B
qC
qS
C
(a)
8
7
D
15
14
(b)
8
15
(d)
7
14
137. A loop of radius r and mass m rotating with an angular
velocity w0 is placed on a rough horizontal surface.
The initial velocity of the centre of the hoop is zero.What
will be the velocity of the centre of the hoop when it ceases
to slip ?
[2013]
(c)
(a)
rw0
4
(b)
rw0
3
rw0
(d) rw0
2
138. A tennis ball (treated as hollow spherical shell) starting
from O rolls down a hill. At point A the ball becomes air
borne leaving at an angle of 30° with the horizontal. The
ball strikes the ground at B. What is the value of the
distance AB ?
(Moment of inertia of a spherical shell of mass m and radius
2
R about its diameter = mR 2 )
3
[Online April 22, 2013]
(c)
O
A
C
(a) go straight.
(b) turn left and right alternately.
(c) turn left.
(d) turn right.
135. A uniform solid cylindrical roller of mass ‘m’ is being
pulled on a horizontal surface with force F parallel to the
surface and applied at its centre. If the acceleration of the
cylinder is ‘a’ and it is rolling without slipping then the
value of ‘F’ is:
[Online April 10, 2015]
5
ma
(a) ma
(b)
3
3
ma
(c)
(d) 2 ma
2
136. A cylinder of mass Mc and sphere of mass Ms are placed
at points A and B of two inclines, respectively (See
2.0 m
30°
0.2 m
A
B
(a) 1.87 m
(b) 2.08 m
(c) 1.57 m
(d) 1.77 m
139. A thick-walled hollow sphere has outside radius R0. It rolls
down an incline without slipping and its speed at the bottom
is v0. Now the incline is waxed, so that it is practically
frictionless and the sphere is observed to slide down
(without any rolling). Its speed at the bottom is observed
to be 5v0/4. The radius of gyration of the hollow sphere
about an axis through its centre is [Online May 26, 2012]
(a) 3R0/2
(b) 3R0/4
(c) 9R0 /16
(d) 3R0
P-92
Physics
140. A solid sphere is rolling on a surface as shown in figure,
with a translational velocity v ms–1. If it is to climb the
inclined surface continuing to roll without slipping, then
minimum velocity for this to happen is
[Online May 12, 2012]
h
v
(a)
(b)
2gh
7
gh
5
7
10
gh
gh
(d)
2
7
141. A round uniform body of radius R, mass M and moment of
inertia I rolls down (without slipping) an inclined plane
making an angle q with the horizontal. Then its
acceleration is
[2007]
(c)
(a)
(c)
g sin q
2
1 - MR / I
g sin q
1 + MR 2 / I
(b)
(d)
g sin q
1 + I / MR 2
g sin q
1 - I / MR 2
P-93
System of Particles and Rotational Motion
1.
(3) Centre of mass of solid hemisphere of radius R lies at
a distance
2.
L
Læ
æ aL2 bL2 ö
bx 3 ö
xdm
=
ax
+
dx
=
+
ç
÷
ç
÷
ò
òç
ç 2
4 ÷ø
L2 ÷ø
è
0
0è
3R
above the centre of flat side of hemisphere.
8
3R 3 ´ 8
\ hcm =
=
= 3 cm
8
8
(23.00) Let s be the mass density of circular disc.
\
X CM
Original mass of the disc, m0 = pa 2 s
æ aL2 bL2 ö
+
çç
÷
2
4 ÷ø
è
=
bL
aL +
3
Þ X CM =
a2
s
Removed mass, m =
4
4.
3L æ 2a + b ö
ç
÷
4 è 3a + b ø
(b)
æ
a2 ö
æ 4p - 1ö
Remaining, mass, m ' = ç pa 2 - ÷ s = a 2 ç
s
è 4 ÷ø
4
è
ø
Y
a
2
a
2
For given Lamina
x y
m1 = 1, C1 = (1.5, 2.5)
m2 =3, C2 = (0.5, 1.5)
m x + m2 x2 1.5 + 1.5
=
= 0.75
X cm = 1 1
m1 + m2
4
X
1
New position of centre of mass
X CM
a2 a
2
m0 x0 - mx pa ´ 0 - 4 ´ 2
=
=
m0 - m
a2
pa 2 4
=
3.
a
-a 3 / 8
-a
-a
=
=
=1 ö 2 2(4p - 1) 8p - 2
23
æ
çè p - ÷ø a
4
\ x = 23
(b) Given,
æxö
Linear mass density, r( x ) = a + b ç ÷
èLø
ò xdm
X CM =
ò dm
L
ò dm = ò r( x)dx
0
Lé
æxö
= ò êa + b ç ÷
èLø
0 êë
2ù
bL
ú dx = aL +
3
úû
m1 y1 + m2 y2 2.5 + 4.5
=
= 1.75
m1 + m2
4
\ Coordinate of centre of mass of flag shaped lamina
(0.75, 1.75)
(a) Mass of sphere = volume of sphere x density of
sphere
Ycm =
2
5.
=
4 3
pR r
3
Mass of cavity M cavity =
4
p(1)3 r
3
Mass of remaining
4
4
M (Remaining) = pR3r – p(1)3 r
3
3
Centre of mass of remaining part,
M r + M 2 r2
X COM = 1 1
M1 + M 2
é4 3 ù
é4
ù
3
êë 3 pR rúû 0 + êë 3 p(1) (– r) úû [ R –1]
Þ –(2 – R ) =
4 3
4
pR r + p(1)3 (–r)
3
3
P-94
Physics
Þ
Þ
( R – 1)
3
( R –1)
= 2– R
=
( R –1)
( R –1)( R 2 + R + 1)
=2–R
b
12
So CM coordinates one
Þ (R2 + R + 1) (2 – R) = 1
6.
M a
´
4 4 =- a
M
12
M4
M ´0-
and yCM = -
(d)
a a 5a
- =
2 12 12
x0 =
and y0 =
Xcm =
m1 x1 + m2 x2 + m3 x3
m1 + m2 + m3
b b 5b
- =
2 12 12
2m (L,L)
10. (a)
Lö
æ
mç 2L, ÷
2ø
è
1 ´ 0 + 1.5 ´ 3 + 2.5 ´ 0 1.5 ´ 3
X cm =
=
= 0.9cm
1 + 1.5 + 2.5
5
Y
1 ´ 0 + 1.5 ´ 0 + 2.5 ´ 4 2.5 ´ 4
=
= 2cm
1 + 1.5 + 2.5
5
Hence, centre of mass of system is at point (0.9, 2)
Ycm =
7.
(c) x cm =
3L
2L m
m y + m2 y2 + m3 y3
Ycm = 1 1
m1 + m2 + m3
æ 5L ö
ç ,0 ÷
è 2 ø
X
x-coordinate of centre of mass is
5mL
2 = 13 L
Xcm =
4m
8
y-coordinate of centre of mass is
50 ´ 0 + 100 ´1 + 150 ´ 0.5
7
=
m
50 + 100 + 150
12
2mL + 2mL +
æ Lö
2m ´ L + m ´ ç ÷ + m ´ 0
è 2ø
5L
=
Ycm =
4m
8
11. (c) To produce maximum moment of force line of action
of force must be perpendicular to line AB.
A
y cm =
8.
50 ´ 0 + 100 ´ 0 + 150 ´
50 + 100 + 150
q
3
m
4
(a) Acceleration of centre of mass (acm) is given by
r
r
m a + m2 a2 + ........
r
\ acm = 1 1
m1 + m2 + ........
(2m)ajˆ + 3m ´ aiˆ + ma (-iˆ) + 4m ´ a(- ˆj )
2m + 3m + 4m + m
2aiˆ - 2ajˆ a ˆ ˆ
=
= (i - j )
10
5
(d) With respect to point q, the CM of the cut-off portion
=
9.
3
2 =
æa bö
ç , ÷ .
è 4 4ø
Using, xCM =
4m
q
2m
2 1
=
4 2
12. (c) According to principle of moments when a system is
stable or balance, the anti-clockwise moment is equal to
clockwise moment.
i.e., load × load arm = effort × effort arm
When 5 mg weight is placed, load arm shifts to left side,
hence left arm becomes shorter than right arm.
\ tan q =
13. (c) Centre of mass xcm
MX - mx
M -m
B
Þ
1 y y2
+ =
2 x x2
x
=
2
æ x ö1
(rx ) ç ÷ + ry y /2
è2ø2
r( x + y )
P-95
System of Particles and Rotational Motion
x
£1
L
With increase in the value of n, the centre of mass shift
towards the end x = L This is satisfied by only option (a).
A
Here
x
L
L
ò xdm
C
y
B (0,0)
xCM = 0
Bc y 1 + 3
=
= 1.37
AB x
2
14. (d) Let density of cone = r.
h
ò ypr
=
2
dyr
0
1 2
pR hr
3
h 2
=
ò0 r
=
dy h
R
C
For a cone, we know that
r y
y
=
\r= R
R n
n
L
xcm =
ò (ax +
0
L
0
n +1
L
é kx
ù
ê
nú
ëê (n + 1) L ûú 0
=
L(n + 1)
n+2
L
; n = 1,
2
2R
h
é y4 ù
3ê ú
3
ë 4 û0
=
ycm = 0 3
= h
3
4
h
h
15. (b) Centre of mass of the rod is given by:
3
ò 3 y dy
n
æxö
ò k çè L ÷ø dx
2L
3L
; n = 2, xCM =
;....
3
4
For n ® ¥xcm = L
Moment of inertia of a square plate about an axis through
its centre and perpendicular to its plane is.
19. (b) Let s be the mass per unit area of the disc.
Then the mass of the complete disc
= s(p(2R)2 )
r
h
0
L
xCM =
a
B
é x n+ 2 ù
kê
nú
êë (n + 2) L úû 0
For n = 0 , xCM =
A
y
0
n
æ xö
ò k çè L÷ø .xdx
L
...(i)
1 2
R h
3
ò l dx
0
ò ydm
ò dm
ydy
=
L
ò dm
\
Centre of mass, ycm =
0
=
L
L
ò x (l dx)
bx2
) dx
L
bx
ò (a + L )dx
0
2
aL bL2 L æ a + b ö
+
ç
÷
3 = è 2 3ø
= 2
bL
b
aL +
a+
2
2
a b
+
7L 2 3
=
Now
b
12
a+
2
On solving we get, b = 2a
16. (c)
17. (d)
æ xö
18. (a) The linear mass density l = k ç ÷
è Lø
O
The mass of the removed disc = s (pR 2 ) = psR 2
Let us consider the above situation to be a complete disc
of radius 2R on which a disc of radius R of negative mass
is superimposed. Let O be the origin. Then the above
figure can be redrawn keeping in mind the concept of
centre of mass as :
2
R
4ps R
2
O
–ps R
( 6p( 2R) ) ´ 0 + ( -6( pR )) R
=
2
xc.m
\ xc.m
2
4psR2 - psR2
\ xc.m =
n
R
-psR2 ´ R
3psR2
R
1
== aR Þ a =
3
3
20. (c) Initially,
P-96
Physics
m1 ( - x1 ) + m2 x2
Þ m1 x1 = m2 x2 ...(1)
m1 + m2
Let the particles is displaced through distanced away from
centre of mass
x1– d
x2– d ¢
d
d¢
m2
m1
O
0=
\0 =
24. (c) Here, rdr w2 r = rgdh
R
h
0
0
Þ w2 ò rdr = g ò dh
w
dh
m1 (d - x1 ) + m2 ( x2 - d ')
m1 + m2
dr
Þ 0 = m1d - m1 x1 + m2 x2 - m2 d '
m1
d
[From (1).]
m2
21. (a) The centre of mass of bodies B and C taken together
does not shift as no external force acts. The centre of mass
of the system continues its original path. It is only the
internal forces which comes into play while breaking.
l
22. (d)
B
A
y1
Þd'=
Þ
w2 R 2
= gh
2
(Given R = 5 cm)
w2 R 2 25w2
=
2g
2g
25. (c) Free body diagram in the frame of disc
\h =
kx
® mw2 (l0 + x )
¬¾¾ m ¾¾
F
y2
P
\ mw2 (l 0 + x ) = kx
2l
y
Þ x=
(0, 0)
C
To have translational motion without rotation, the force
uur
F has to be applied at centre of mass. i.e. the point ‘P’has
to be at the centre of mass
Taking point C at the origin position, positions of y, and y2
are r1 = 2l, r2 = l and ml = m and m2 = 2m
m y + m2 y2 m ´ 2 l + 2m ´ l 4l
=
=
y= 1 1
m1 + m2
3m
3
k – mw2
For k >> mw2
Þ
Fradial =
(Q r = l + x here)
kx = mlw2 + mxw2
2
N
a
\x=
mw acosq
q
P(a, b)
mg
mgsinq
For steady circular motion
2
mw a cos q = mg sin q
Þw=
g tan q
a
\w =
g ´ 8aC
= 2 2 gC
a
q
mv 2
= mr w2
r
\ kx = m(l + x)w2
dy
= tan q = 8Cx
dx
At P, tan q = 8Ca
w
x mw2
=
l0
k
26. (b) At elongated position (x),
23. (a) y = 4Cx 2 Þ
y
ml 0 w2
mlw2
k – mw2
T
x
0
l
2
27. (d) ò (- dT ) = ò (dm)w x
2
mw a
x
mgcosq
x
æm ö 2
– T = ò ç l dx ÷w x
è
ø
l
mw2 2
(l - x 2 )
l
It is a parabola between T and x.
or T =
P-97
System of Particles and Rotational Motion
28. (b) N sin q = mw2 (r/2)
...(i)
The rotational speed of the drum
v
g
10
<
<
R
R
1.25
The maximum rotational speed of the drum in revolutions
per minute
Þω<
sin q =
r/2 1
= Þ q = 30°
r
2
and N cos q = mg
60 10
< 27 .
2p 1.25
34. (b) Angular momentum, mvr = Iw
Moment of Inertia (I) of cubical block is given by
R
m.2
2
æ 2 æ ö2 ö÷
\ w=
çR
R
é R2 æ R ö 2 ù
I < m çç
∗ çç ÷÷÷ ÷÷÷
çç 6 èç 2 ø ÷÷
mê + ç ÷ ú
è
ø
êë 6 è 2 ø úû
ω(rpm) <
...(ii)
2
or tan q =
w r
2g
w2 r
or tan 30° =
2g
or
1
3
=
\ w2 =
w2 r
2g
12
3
10
=
=
= 5 rad / s .
8R 2 ´ 0.3 2
35. (b) Angular velocity is the angular displacement per
Dq
unit time i.e., w =
Dt
Here w1 = w2 and independent of f.
36. (c) Angular momentum, L = I w
Þw=
2g
.
r 3
29. (a) Using v2 = u2 + 2gy
v2 = 2gy
[\u = 0 at (0,0)]
[\v = wx]
Y
w
Þy
X
(0,0)
Axis
w2 x 2 (2 ´ 2p) 2 ´ (0.05) 2
Þ y=
=
; 2cm
2g
20
2
30. (c) mw R = Force µ
Þ w2 µ
1
R
R
n
(Force =
Þ wµ
n +1
Time period T =
1
l
t = Mg sin q.
2
Also t = la
l
\ l a = Mg sin q
2
1
l/ 2
n +1
R 2
m
m
l/ 2
2
æ l ö
I = m(0) 2 + m ç
´ 2 + m( 2l ) 2
è 2 ÷ø
n +1
R 2
=
w,a
Q
Q
é
Ml 2 ù
Ml 2
l
êQ I rod =
ú
.a = Mg sin q
3 û
3
2
ë
la
sin q
3g sin q
=g
Þ
\ a=
3
2
2l
33. (a) For just complete rotation
v < Rg at top point
m
mv2
)
R
2p
w
Time period, T µ
31. (b) decreasing speed
32. (c) Torque at angle q
l
2ml 2
+ 2ml 2 = 3ml 2
2
Angular momentum L = I w = 3ml 2 w
37. (20)
w
(M, L)
m
v
Before collision After collision
Using principal of conservation of angular momentum we
have
P-98
Physics
r
r
Li = L f Þ mvL = I w
Þ
æ ML2
ö
Þ mvL = ç
+ mL2 ÷ w
è 3
ø
1 æ 5 2 ö 9v 2
= 2mgl (1 - cos q )
ç ml ÷ø
2è3
25l 2
3
mv 2 = 2mgl (1 - cos q)
5´ 2
3
36
27
´
= 1 - cos q Þ 1 = cos q
10 2 ´ 10
50
23
or, cos q =
\q ; 63°.
50
40. (d) Vertical force = mg
Þ
æ 0.9 ´ 12
ö
Þ 0.1 ´ 80 ´ 1 = ç
+ 0.1 ´ 12 ÷ w
è 3
ø
4
æ 3 1ö
Þ8=ç + ÷wÞ8= w
è 10 10 ø
10
rad/sec.
Þ w = 20
38. (9.00)
Here M0 = 200 kg, m = 80 kg
Using conservation of angular momentum, Li = Lf
M0
2 l
Horizontal force = Centripetal force = mw sin q
2
l
Torque due to vertical force = mg sin q
2
l
2 l
Torque due to horizontal force = mw sin q cos q
2
2
m
FV
I1w1 = I 2 w 2
I1 = ( I M
I2 =
FH
æ M R2
ö
+ I m ) = ç 0 + mR 2 ÷
è 2
ø
l
w2 sin q
2
1
M 0 R 2 and w1 = 5 rpm
2
æ M R2
ö
5
\w2 = ç 0 + mR 2 ÷ ´
ç 2
÷ M R2
è
ø
0
2
5R 2
(80 + 100)
= 2 ´
= 9 rpm.
100
R
39. (a) Using conservation of angular momentum
æ
2 ml 2 ö
5
3v
mvl = ç ml 2 +
w Þ mvl = ml 2 w Þ w =
3 ÷ø
è
3
5l
or, w =
3 ´ 6 18
=
rad/s
5 ´1 5
w
M = 2 kg
mg
Net Torque = Angular momentum
l
l
l
ml 2 2
mg sin q - mw2 sin q cos q =
w sin q cos q
2
2
2
12
3 g
Þ cos q =
2 w 2l
41. (d) Net torque, tnet about B is zero at equilibrium
\ TA ´ 100 - mg ´ 50 - 2mg ´ 25 = 0
Þ TA ´ 100 = 100mg
Þ TA = 1 mg (Tension in the string at A)
q
TA
50 cm
25 cm
A
m = 1 kg
Now using energy conservation, after collision
1 2
l
I w = 2mg (1 - cos q) + mgl (1 - cos q)
2
2
TB
B
mg
2 mg
P-99
System of Particles and Rotational Motion
42. (a)
46. (d) Angular acceleration,
F
w - w0
25 ´ 2p - 0
=
= 10 p rad/s2
t
5
t = Ia
æ5
ö
æ5ö
mR 2 ÷ a » ç ÷ (5 ´10-3 )(10-4 )10p
Þ t = çè 4
ø
è4ø
a=
R
R–a N
Mg
x
f
a
= 2.0 × 10–5 Nm
47. (a)
48. (c) According to work-energy theorem
For step up, F ´ R ³ Mg ´ x
x = R 2 - ( R - a) 2 from figure
Fmin =
Mg
æ R-aö
´ R 2 - ( R - a )2 = Mg 1 - ç
÷
R
è R ø
2
43. (c)
About point O angular momentum
Linitial = Lfinal
Þ
mV
2
´
1 é 4mL2 mL2 ù
=ê
+
ú´ w
2 ë 12
4 û
2 × 10 × 10 = v2B – 52
Þ vB =15 m/s
Angular momentum about O,
LO = mvr
= 20 × 10–3 × 20
LO = 6 kg.m2/s
r F
F 3 ˆ
49. (a) Given, F1 = ( -ˆi) +
(- j)
2
2
r
r = 0iˆ + 6jˆ
1
Torque due to F1 force
æF
r
r r
F 3 ˆ ö
ˆ
t F1 = r1 ´ F1 = 6ˆj ´ ç ( -ˆi) +
( - j) ÷ = 3F(k)
ç2
÷
2
è
ø
Torque due to F2 force
r
ˆ ´ Fkˆ = 3Fiˆ + 2F( - ˆj)
t F2 = (2iˆ + 3j)
r
r
r
ˆ
t = t + t = 3Fiˆ + 2F(-ˆj) + 3F(k)
=
dx
= Vx
dt
dy
= v = bw cos(w t)
Þ vx = – aw1 sin (w1t), and
y
2
2
dt
dv y
dv x
= a = – aw2 cos (w t),
= ay = – bw22 sin (w2t)
x
1
1
dt
dt
At t = 0, x = x0 + a, y = y0
ax = – aw21, ay = 0
r r r
r r
Now, t = r ´ F = m(r´ a)
= [(x 0 + a) iˆ+ y 0 ˆj] ´ m( - a w12 ˆi) = + my0 a w12 kˆ
r
45. (d) We have given r = 2tiˆ - 3t 2 ˆj
r
r (at t = 2) = 4iˆ - 12 ˆj
r
r dr
= 2iˆ - 6tjˆ
Velocity, v =
dt
r
v (at t = 2) = 2iˆ - 12 ˆj
r
r r
L = mvr sin qnˆ = m(r ´ v )
= 2(4iˆ - 12 ˆj ) ´ (2iˆ - 12 ˆj ) = -48kˆ
1
1
mv 2B – mv 2A
2
2
2gh = v2B – v 2A
3 2V
7L
7 2L
44. (b) Given that, x = x0 + a cos w1t
y = y0 + b sin w2t
\w=
6V
mgh =
net
F1
F2
ˆ
= (3iˆ - 2jˆ + 3k)F
r r r
50. (a) Torque about the origin = t = r ´ F
= r F sin q Þ 2.5 = 1 × 5 sin q
1
sin q= 0.5 =
2
p
Þ q=
6
51. (d) Consider a strip of radius x and thickness dx,
Torque due to friction on this strip
Net torque = å Torque on ring
R
ò dt = ò
0
Þ t=
mF.2 pxdx
pR 2
2µF R 3
·
R2 3
2µFR
3
52. (a) Applying torque equation about point P.
t = I a = [2M0(2l)2 + 5M0l2]a
t=
P-100
Physics
Þ 5M0gl – 4 M0gl = [2M0(2l)2 + 5 M0l2]a
Þ M0gl = (13M0gl2)a
g
\a=
13l
53. (d) Given that, the rod is of uniform mass density and
AB = BC
In none of the cases, the perpendicular
æ R
ö
+ a÷
distance r^ is ç
è 2
ø
57. (a) Angular momentum,
L0 = mvr sin 90°
= 2 × 0.6 × 12 × 1 × 1
[As V = rw, Sin 90° = 1]
So, L0 = 14.4 kgm2/s
o 0.6m
0.8m
1m
O
58. (c) Torque working on the bob of mass m is, t = mg × l sin
q. (Direction parallel to plane of rotation of particle)
l
Let mass of one rod is m.
Balancing torque about hinge point.
mg (C1P) = mg (C2N)
æL
ö
æL
ö
mg ç sin q ÷ = mg ç cos q - L sin q ÷
è2
ø
è2
ø
3
mgL
Þ mgL sin q =
cos q
2
2
sin q 1
1
Þ
= or, tan q =
cos q 3
3
54. (a) Balancing torque w.r.t. point of suspension
æl
ö
mg x = Mg ç – x ÷
è2
ø
l
Þ mx = M – Mx
2
æ lö1
m = çM ÷ – M
è 2ø x
1
y= a –C
x
A
l/2-X
X
Mg
m
Straight line equation.
55. (a)
56. (a) We know that |L| = mvr^
a
V
O
R
2
V
a
a
A
R/ 2
l
m
mg
r
As t is perpendicular to L , direction of L changes but
magnitude remains same.
59. (c) Given : m = 0.160 kg
q = 60°
v = 10 m/s
®
®
Angular momentum L = r ´ m v
= H mv cos q
v2 sin 2 q
cos q
=
2g
B
é
v2 sin 2 q ù
êH =
ú
2g úû
êë
102 ´ sin 2 60°´ cos 60°
2 ´10
= 3.46 kg m2/s
60. (a)
61. (a) Angular momentum L = m (v × r)
æ dr ö
= 2 kg ç ´ r÷ = 2 kg(4t j ´ 5i - 2t 2 ˆj)
è dt ø
=
= 2 kg (–20 t k̂ ) = 2 kg × –20 ×2 m–2 s–1 k̂
y
D
q
C
a V
a
V
a
B
X
= –80 k̂
62. (b)
63. (d)
64. (c) Angular momentum, L = Iw Þ L = mr2w
As insect moves along a diameter, the effective mass and
hence moment of inertia (I) first decreases then increases
so from principle of conservation of angular momentum,
angular speed w first increases then decreases.
r
r r
65. (c ) L = m(r ´ v )
r
1
L = m éêv0 cos qt iˆ + (v0 sin qt - gt 2 ) ˆj ùú
ë
2
û
P-101
System of Particles and Rotational Motion
R
´ éë v0 cos q iˆ + (v0 sin q - gt ) ˆj ùû
é 1
ù
ˆ
= mv0 cos qt ê - gt ú k
ë 2 û
1
2
= - mgv0 t cos qkˆ
2
uur
uur d L
66. (d) We know Torque t c = c
dt
uur
where Lc = Angular momentum about the center of mass
of the body. Central forces act along the center of mass.
Therefore torque about center of mass is zero.
uur
dL
\t =
=0
Þ Lc = constt.
dt
67. (d) Applying conservation of angular momentum I¢w¢ = Iw
(mR2 + 2MR2)w¢ = mR2w
Þ (m + 2m)R2w¢ = mR2w
é m ù
Þ w' = wê
ë m + 2 M úû
ur ur uur
68. (c) Torque t = r ´ F = (iˆ - ˆj ) ´ ( - Fkˆ)
= F [- iˆ ´ kˆ + ˆj ´ kˆ] = F ( ˆj + iˆ) = F ( iˆ + ˆj)
q
From diagram,
r
R
R
= tan q =
or r = h
h
H
H
...(ii)
Mass of element, dm = r(pr 2 )dh
...(iii)
From eq. (i), (ii) and (iii),
Area of element, dA = 2prdl = 2pr
dh
cos q
2 Mh tan dh
Mass of element, dm =
R R 2 + H 2 cos q
(here, r = h tan q )
éSince kˆ ´ iˆ = ˆj and ˆj ´ kˆ = iˆù
ë
û
69. (b) Angular momentum will remain the same since
no external torque act in free space.
r r r
70. (d) We know that t = r ´ F
I = ò dI =
H
=
H
2
ò dm(r ) =
0
æ æR
ö
ò r çè p çè H × h÷ø
0
H
æR ö
2
ò r(pr )dh çè H × h÷ø
2
0
4ö
÷ dh
ø
F
Solving we get, I =
r
t
r
H
MR 2
2
dm
73. (b)
A
B
x
dx
Mass of the small element of the rod
r
r
r
Vector t is perpendicular to both r and F . We also
know that the dot product of two vectors which have an
angle of 90° between them is zero.
r r
r r
\ r × T = 0 and F × T = 0
71. (d) Angular momentum (L)
= (linear momentum) × (perpendicular distance of the line
of action of momentum from the axis of rotation)
As the particle moves with velocity V along line PC, the line
of motion passes through P.
\ L = mv × r
= mv × 0
=0
72. (d) Hollow ice-cream cone can be assume as several parts
of discs having different radius, so
I = ò dI = ò dm(r )
2
...(i)
dm = l × dx
Moment of inertia of small element,
xö
æ
dI = dm × x 2 = l 0 ç1 + ÷ × x 2 dx
è Lø
Moment of inertia of the complete rod can be obtained by
integration
L
æ
x3 ö
I = l 0 ò ç x 2 + ÷ dx
Lø
0è
L
é L3 L3 ù
x3 x4
= l0
+
= l0 ê + ú
3 4L 0
4û
ë3
7l 0 L3
12
Mass of the thin rod,
ÞI =
...(i)
P-102
Physics
Let I ADE = I BDF = I EFC = I 2
L
\ l0 =
\I =
I0
5I
= I0 Þ I 2 = 0
16
16
Hence, moment of inertia of DECB i.e., after removal part
ADE
\ 3 I 2 + I1 = I 0 Þ 3 I 2 +
2M
3L
7 æ 2M ö 3
7
2
ç
÷ L Þ I = ML
12 è 3L ø
18
74. (d) By angular momentum conservation, Lc = L f
wI + 3 I ´ 0 = 4 I w ' Þ w ' =
w
4
æ 5 I ö æ I ö 11I 0 NI 0
= 2 I 2 + I1 = 2 ç 0 ÷ + ç 0 ÷ =
=
16
è 16 ø è 16 ø 16
Therefore value of N = 11.
1
2
76. (20) As we know moment of inertia disc, I disc = MR
2
w1
3I
w
R1=R
æ MR 2 MR 2
MR 2
´ w + 0 = çç
+
2
8
è 2
Initial K.E., K i =
2
1
I w2
æ wö
= ´ (4 I ) ´ ç ÷ =
è 4ø
2
8
1
1
3
DKE = I w 2 - I w 2 = I w 2
2
8
8
3 2
Iw
3
DKE 8
=
=
= .
\ Fractional loss in K.E.
1 2 4
KEli
Iw
2
75. (11) Let mass of triangular lamina = m, and length of side
= l, then moment of inertia of lamina about an axis passing
through centroid G perpendicular to the plane.
I 0 µ ml 2
R2(M)
2
I0
ml 2
æ m öæ l ö
So, I1 µ ç ÷ç ÷ µ
or I1 =
16
4
2
16
è øè ø
A
F
1 2 1 æ MR 2 ö 2 MR 2 w2
I w = çç
÷w =
2
2 è 2 ÷ø
4
1 æ MR 2 MR 2 ö 16 2 MR 2 w2
+
Final K.E., K f = çç
÷ w =
2è 2
8 ÷ø 25
5
Percentage loss in kinetic energy % loss
MR 2 w2 MR 2 w2
4
5
=
´ 100 = 20% = P%
MR 2 w2
4
Hence, value of P = 20.
77. (c) Let p be the density of the discs and t is the thickness
of discs.
Moment of inertia of disc is given by
4
B
Let moment of inertia of DEF = I1 about G
I 2 æ R2 ö
16
=ç ÷ Þ
= a4 Þ a = 2
I1 è R1 ø
1
78. (b) Moment of inertia of rectangular sheet about an axis
passing through O,
IO =
E
M 2
M
(a + b 2 ) = [(80) 2 + (60) 2 ]
12
12
y
O'
O
80 cm
10
0
l/2
G
m/4
ö
4
÷÷ w f Þ w f = w
5
ø
I=
G
B
R1(M)
MR 2 [r(pR 2 )t ]R 2
=
2
2
4
(As r and t are same)
I µR
A
m/4
w
M
Using angular momentum conservation
I1w1 + I 2w2 = ( I1 + I 2 ) ´ w f
1 2
Iw
2
1
( KE ) f = (3I + I )w '2
2
D
w2
I2
R2=R/2
I1
I
( KE )i =
I 0 = kml 2
M
cm
L
3l L
xö
æ
M = ò l dx = ò l 0 ç1 + ÷ dx = 0
è Lø
2
0
0
C
60 cm
x
P-103
System of Particles and Rotational Motion
From the parallel axis theorem, moment of inertia about O ',
IO ' = IO + M (50)2
81. (b) Initial angular momentum = I1w1 + I 2 w 2
Let w be angular speed of the combined system.
M
(802 + 602 )
IO
1
12
=
=
IO ' M
4
(802 + 602 ) + M (50) 2
12
79. (c) Let there be a cylinder of mass m length L and radius
R. Now, take elementary disc of radius R and thickness dx
at a distance of x from axis OO' then moment of inertia
about OO' of this element.
O
R
Final angular momentum = I1w + I 2 w
According to conservation of angular momentum
( I1 + I 2 )w = I1w1 + I 2 w 2
Þw=
I1w1 + I 2 w 2 0.1 ´ 10 + 0.2 ´ 5 20
=
=
0.1 + 0.2
3
I1 + I 2
Final rotational kinetic energy
Kf =
1
1
1
æ 20 ö
I1w 2 + I 2 w 2 = (0.1 + 0.2) ´ ç ÷
è 3ø
2
2
2
2
20
J
3
82. (a) Moment of inertia,
Þ Kf =
dx
dI =
O'
2
dmR
+ dmx 2
4
Þ I = ò dI = ò
Given : I =
ÞI =
dmR 2
+
4
n =- L / 2
ò
n=L / 2
M
dx ´ x 2
L
MR 2 ML2
+
4
12
M V ML2
MV ML2
´
+
ÞI=
+
4 pL 12
4pL 12
dI
mV
M ´ 2L
=+
=0
2
dL
12
4pL
ÞV =
and AO =
L
3
=
R
2
80. (25)
Moment of inertia of the system about axis XE.
X
F
rF
\
a
a
rG
G
I = I E + I F + IG
Þ I = m(rE )2 + m(rF )2 + m(rG )2
2
2
é 2 æ d ö2
æ d ö ù
I 0 = 3I1 = 3 ê m ç ÷ + m ç
÷ ú
êë 5 è 2 ø
è 3 ø úû
Þ I0 =
13
Md 2
10
é 2 æ d ö2
ù 2 æ d ö2
And I A = 2 ê M ç ÷ + Md 2 ú + M ç ÷
êë 5 è 2 ø
úû 5 è 2 ø
23
Md 2
10
13
Md 2
IO 10
13
\
=
=
23
I A 23
Md 2
10
Þ IA =
60°
E
d
3
Moment of inertia about ‘O’
2 3
2
pL Þ pR 2 L = pL3
3
3
a
2
2 æd ö
2
I1 = m ç ÷ + m ( AO )
5 è2ø
5
25
æaö
Þ I = m ´ 02 + m ç ÷ + ma 2 = ma 2 = ma 2
2
4
20
è ø
\ N = 25.
83. (15) Here, length of bar, l = 1 m
angle, q = 30°
DPE = DKE or mgh =
1 2
Iw
2
P-104
Physics
l
Þ (mg) sin 30° =
2
1 æ ml 2 ö 2
ç
÷w
2 çè 3 ÷ø
l 1 1 æ ml 2 ö 2
Þ mg ´ = ç
÷w
2 2 2 çè 3 ÷ø
Þ w = 15 rad/s
84. (a) Using principal of conservation of energy
1
1
(m1 – m2 ) gh = (m1 + m2 )v 2 + I w2
2
2
1
2 1
2
Þ (m1 – m2 ) gh = (m1 + m2 )(wR ) + I w
2
2
(Q v = wR )
Þ
(m1 – m2 ) gh =
Þ w=
w2 é
(m1 + m2 ) R 2 + I ù
û
2 ë
2(m1 – m2 ) gh
2
(m1 + m2 ) R + I
85. (a) When the bob covered a distance ‘h’
Using mgh =
1 2 1 2
mv + I w
2
2
1
1 mr 2
= m(w r) 2 + ´
´ w2 (Q v = wr no slipping )
2
2
2
3
2 2
Þ mgh = mw r
4
1 4 gh
r
3
3r
86. (c) Moment inertia of the rod passing through a point
Þw=
4 gh
2
=
l
away from the centre of the rod
4
I = Ig + ml2
Þ I=
æ I 2 ö 7 MI 2
MI 2
+ M ´ç ÷ =
ç 16 ÷
12
48
è ø
2
Using I = MK =
7MI 2
(K = radius of gyration)
48
7
Þ K=
I
48
87. (a) Given,
mass per unit area of circular disc, s = A + Br
Area of the ring = 2 prdr
Mass of the ring, dm = s2prdr
Moment of inertia,
I = ò dmr 2 = ò s 2prdr.r 2
a
é Aa 4 Ba5 ù
Þ I = 2 p ò ( A + Br )r 3 dr = 2p ê
+
ú
5 úû
êë 4
0
é A Ba ù
Þ I = 2pa 4 ê +
ú
ë4 5 û
b
2
88. (c) I = ò (dm)r
a
b
æ s0
ö 2 2 ps0 3 b
| r |a
= ò ç r ´ 2pr dr ÷ r =
è
ø
3
a
2ps0 3
(b - a 3 )
3
Mass of the disc,
=
b
m=
ò
a
s0
´ 2pr dr = 2ps (b – a)
0
r
Radius of gyration,
k=
I
m
(2ps0 / 3)(b3 - a3 )
=
2ps0 (b - a )
=
a 2 + b2 + ab
3
89. (d) As no external torque is acting so angular momentum should be conserved
(I1 + I2) w =I1w1 + I2w2 [wc = common angular velocity
of the system, when discs are in contact]
Iw
I1w1 + 1 1
4 æ 5 ´ 2 öw
wc =
I1 çè 4 3 ÷ø 1
I1 +
2
5w1
wc =
6
1
1
1
E f - Ei = ( I1 + I2 ) wc2 - I1w12 - I2 w22
2
2
2
5w1
Put I2 = I1/2 and wc =
5w1/6
6
We get :
I w2
E f – Ei = – 1 1
24
90. (b) As from the question density (s) = kr 2
R
2
Mass of disc M = ò (kr )2prdr = 2 pk
0
Þk=
R 4 pkR 4
=
4
2
2M
....(i)
pR 4
\ Moment of inertia about the axis of the disc.
l = ò dl = ò ( dm ) r 2 = ò sdAr 2
= ò (kr 2 )(2prdr)r 2
æ 2M ö
p´ ç
´ R6
4÷
pkR
2
è
p
ø
R
5
= ò 2pk r dr =
=
= MR 2
3
3
3
R
6
0
[putting value of k from eqn ....(i)]
P-105
System of Particles and Rotational Motion
91. (b)
Using parallel axis theorem
5
æ 1 1 ö 16pr0 R
\ I = IC + MR 2 = 2pr0 R 5 ç + ÷ =
15
è3 5ø
8 é2
8
ù
= ê pr0 R 3 ú R 2 = MR 2
5 ë3
5
û
14
æ 7M ö
ö
21 æ 7
I1 = ç
= ç ´ 4 ÷ MR2 = mR 2
÷ (2 R)
8
2
16
8
è
ø
è
ø
2æM ö
I2 = ç ÷ r 2
5è 8 ø
Þ I2 =
2æ M
ç
5è 8
öæ R
÷ çç
øè 4
2
ö MR 2
÷=
÷
80
ø
M
Mass of smaller triangle = 4
l
2
Moment of inertia of removed triangle
2
Mæaö
= 4ç ÷
è2ø
2
M æaö
ç
÷
I
2
\ removed = 4 . è ø2
M
Ioriginal
(a)
I0
I removed =
16
14 3 ù
é4 3
ê 3 pr r = 8 3 pR ´ r ú
ê
ú
êëÞ r = R /2
úû
Length of smaller triangle =
I1 14 ´ 80
=
= 140
I2
8
92. (d)
1 2
I w = kQ 2
2
æ 2k ö
or w = çç
÷÷ Q
è I ø
dw
2 K æ dQ ö æ 2k ö
or a =
=
÷w
ç
÷ =ç
p
I è dt ø çè I ÷ø
æ 2k öæ 2k ö
=ç
q 2k q
֍
ç I ÷ç
÷÷ =
I
I
è
øè
ø
93. (b) w = at = 20t
1 2
Given, I w = 1200
2
1
2
or ´ 1.5 ´ (20t ) = 1200
2
or t = 2 s
94. (c) Iiwi = If wf
æ ML2
or çç 12
è
2
æ ML2
ö
æ Lö ö
ç
2
m
w
=
+
÷ 0
ç ÷ ÷÷ w f
÷
ç 12
è2ø ø
ø
è
æ M w0 ö
\w f = ç
÷
è M + 6m ø
95. (c) Taking a circular ring of radius r and thickness dr as a
mass element, so total mass,
R
M = ò r0 r ´ 2prdr =
0
R
2pr0 R 3
3
IC = ò r0 r ´ 2prdr ´ r 2 =
0
96. (b)
97. (d) According to parallel axes theorem
2
I = mR 2 + mx 2
5
Hence graph (d) correctly depicts I vs x.
98. (a) Let mass of the larger triangle = M
Side of larger triangle = l
Moment of inertia of larger triangle = ma2
2pr0 R
5
5
I
15I0
So, I = I0 - 0 =
16
16
99. (b)
40
O
f
P
From newton’s second law
40 + f = m (Ra)
Taking torque about 0 we get
40 × R – f × R = Ia
40 × R – f × R = mR2 a
40 – f = mR a
Solving equation (i) and (ii)
40
a=
=16rad / s 2
mR
a =Ra
.....(i)
...(ii)
100. (b) Moment of inertia of disc D1 about OO¢ = I1 =
M.O.I of D2 about OO¢
1 æ MR 2 ö
MR 2
2
2
= I2 = 2 ç 2 ÷ + MR = 4 + MR
è
ø
M.O.I of D3 about OO¢
1 æ MR 2 ö
MR 2
2
+ MR 2
÷ + MR =
= I3 = ç
2è 2 ø
4
MR 2
2
P-106
Physics
104. (d) Using parallel axes theorem, moment of inertia about ‘O’
Io = Icm + md2
so, resultant M.O.I about OO¢ is I = I1 + I2 + I3
æ MR 2
ö
MR 2
+ 2ç
+ MR 2 ÷
2
è 4
ø
ÞI =
2
2
MR MR
2
+
+ 2MR 2 = 3 MR
2
2
101. (a) For Ball
=
using parallel axes theorem, for ball moment of inertaia,
2
22
2
2
2
Iball = MR + M ( 2R ) = MR
5
5
22
For two balls Iballs= 2×
MR2 =and,
5
M ( 2R ) MR
=
12
3
Isystem = Iballs + Irod
2
7MR 2
55MR 2
=
+ 6(M ´ (2R)2 ) =
2
2
2R
Again, moment of inertia about
point P, Ip = Io + md2
2R
2R
2R
O
2R
2R
55MR 2
181
+ 7M(3R) 2 =
MR 2
2
2
105. (a) Let s be the mass per unit area.
=
R/ 3
O'
2
Irod =
=
O
44
MR 2 137
MR 2 +
=
MR 2
5
3
15
102. (c) As we know, w =
w=
2R/3
é
êëQ I
3k
ml 2
k
I
The total mass of the disc
= s × p R2 = 9M
The mass of the circular disc cut
1 2ù
rod = ml ú
3
û
Tension when it passes through the mean position,
= mw2 q02
2
l
3k
l
= m 2 q02 = kq0
3
3
ml
l
103. (d)
= 50 cm
2
Initial position
o
Final position
By the low of conservation of energy,
P.E. of rod = Rotational K.E.
l
1
Sin q = Iw2
2
2
l 1 1 ml 2
1 ml 2 2
l
w
ω Þ mg ´ =
Þ mg Sin 30° =
2 3
2
2 2 2 3
2
For complete length of rod,
30
w = 3g 2 ( 2l ) =
rods –1
2
1
æ Rö
´M ´ç ÷
è 3ø
2
\ M.I. (I2) of the cut out portion about an axis
passing through O and perpendicular to the plane of
disc
2
2
é1
æ Rö
æ 2R ö ù
= ê ´M ´ç ÷ + M ´ç ÷ ú
è 3ø
è 3 ø ú
êë 2
û
[Using perpendicular axis theorem]
\ The total M.I. of the system about an axis passing
through O and perpendicular to the plane of the disc
is
I = I1 + I2
2
2
é1
1
æ Rö
æ 2R ö ù
2
= 9MR - ê ´ M ´ ç ÷ + M ´ ç ÷ ú
è 3ø
è 3 ø ú
2
êë 2
û
=
30
mg
2
p R2
æ Rö
=M
= s´pç ÷ = s ´
è 3ø
9
Let us consider the above system as a complete disc
of mass 9M and a negative mass M super imposed
on it.
Moment of inertia (I1 ) of the complete disc =
1
9MR 2 about an axis passing through O and
2
perpendicular to the plane of the disc.
M.I. of the cut out portion about an axis passing
through O' and perpendicular to the plane of disc
=
9MR 2 9MR 2
(9 - 1)MR 2
= 4 MR 2
=
2
18
2
P-107
System of Particles and Rotational Motion
106. (c) As we know, moment of inertia of a disc about an axis
passing through C.G. and perpendicular to its plane,
2
mR
2
Moment of inertia of a disc about a tangential axis
perpendicular to its own plane,
Iz =
3
Iz' = mR 2
2
q
q
r
mR 2 3mR 2
Iz Iz' =
=1 3
\
2
2
107. (a) When the rod makes an angle a
l
Displacement of centre of mass = cos a
2
l
l
mg cos a = I w 2
2
2
l
ml 2 2
mg cos a =
w (Q M.I. of thin uniform rod
2
6
about an axis passing through its centre of mass and
perpendicular to the rod I =
Þ
w=
Speed of end = w ´ l = 3g cos al
i.e., Speed of end, w µ cos a
108. (c) As we know, moment of inertia of a solid cylinder
about an axis which is perpendicular bisector
I=
l
m é V l2 ù
= ê + ú
4 ë pl 3 û
V
2l
=
2
3
pl
Þ
ÞV =
2
3
dl m é -V 2l ù
=
+
=0
dl 4 êë pl 2 3 úû
2 pl 3
3
l
3
3
l
2pl
Þ 2 = or, =
2
2
R
3
R
109. (b) According to theorem of perpendicular axes, moment
of inertia of triangle (ABC)
I0 = kml2
..... (i)
BC = 1
Moment of inertia of a cavity DEF
pR 2 l =
IDEF = K
=
mæ lö
ç ÷
4 è 2ø
k
ml 2
16
I0
16
Moment of inertia of remaining part
IDEF =
I0 15I 0
=
16
16
110. (b) Moment of Inertia of complete disc about 'O' point
MR 2
Itotal =
2
Radius of removed disc = R/4
\ Mass of removed disc = M/16
[As M µ R2]
M.I of removed disc about its own axis (O')
Iremain = I0 -
2
1 Mæ Rö
MR 2
çè ÷ø =
2 16 4
512
M.I of removed disc about O
Iremoved disc = Icm + mx2
=
2
MR 2 M æ 3R ö
19 MR 2
+ ç ÷ =
512
512 16 è 4 ø
M.I of remaining disc
=
2
237
Iremaining = MR - 19 MR 2 =
MR 2
512
2
512
2
R
111. (a) Here a =
3
4 3
pR
M
Now,
=3 3
M¢
a
ml 2
)
12
3 g cos a
l
mR 2 ml 2
+
4
12
é
m
l2 ù
I = ê R2 + ú
4ë
3û
From equation (i),
4 3
pR
3
= 3
=
p.
3
2
æ 2 ö
R÷
ç
è 3 ø
M¢=
a
2M
3p
Moment of inertia of the cube about the given axis,
I=
M ¢a 2
6
2
2M æ 2 ö
´ç
R÷
2
3p è 3 ø = 4 MR
=
9 3p
6
112. (d) For a thin uniform square sheet
I1 = I2 = I3 =
ma 2
12
2
I1
I2
I3
P-108
Physics
113. (c) Kinetic energy (rotational) KR =
M.I. of complete disc can also be written as
ITotal = Iremoved hole + Iremaining disc
1 2
Iw
2
1
Mv 2
2
M.I.(initial) Iring = MR2; winitial = w
M.I.(new) I¢(system) = MR 2 + 2mR 2
Kinetic energy (translational) KT =
(v = Rw)
Mw
M + 2m
Solving we get loss in K.E.
Mm
w2 R 2
=
(M + 2m)
114. (b) As L = Iw so L increases with increase in w.
Kinetic energy(rotational) depends on an angular velocity
‘w’ and moment of inertia of the body I.
w¢(system) =
i.e., K .E. ( rotational) =
115. (b) K .Erotational =
=
mg
1
MR2
2
R
1
2
Þ T.R = mR a
2
Also, acceleration, a = Ra
1
1
\ T = mRa = ma
2
2
Substituting the value of T is equation (1) we get mg -
...(i)
Circular hole of
diameter R (radius = R/2)
Disc mass = M
radius = R
Mass of circular hole (removed)
M
As M = pR 2t \ M µ R 2
=
4
M.I. of removed hole about its own axis
(
)
2
1 æ M ö æ Rö
1
MR2
ç ÷ç ÷ =
2 è 4 ø è 2ø
32
M.I. of removed hole about O¢
Iremoved hole = Icm + mx2
MR 2 M
+
32
4
æ Rö
çè ÷ø
2
1
2
ma = ma Þ a = g
2
3
118. (a) Given,
Force, F = (20t – 5t2)
Radius, r = 2m
Torque, T = r f = Ia
Þ 2(20t – 5t2) = 10a
\a = 4t – t2
w
Þ w = 2t 2 q
ò
0
MR 2 MR 2 3MR 2
=
+
=
32
16
32
(4t - t ) dt
2
t3
(as w = 0 at t = 0, 6s)
3
6 æ
t3 ö
d q = ò ç 2t 2 - ÷ dt
3ø
0 è
Þ q = 36 rad Þ 2 p n = 36 Þ n =
36
<6
2p
O
C. M
h
2
t
dw
= 4t - t 2 Þ ò d w = ò
Þ
dt
0
0
119. (c)
=
=
R
m
116. (d) M.I. of complete disc about its centre O.
O
m
1 2
Iw
2
1 2
mv
2
K .Erotational
2
=
\ K .E
5
translational
Hence option (b) is correct
R/2
O¢
MR 2 3MR 2 æ 13 ö
= ç ÷ MR2
è 32 ø
2
32
117. (b) For translational motion,
mg – T = ma
.....(1)
For rotational motion,
T.R = I a
=
T
K .Etranslational =
ITotal =
...(ii)
1
3MR 2
MR 2 =
+ I remaining disc
2
32
Þ Iremaining disc
1 2
Iw
2
2 2ö
æ
çèQ ISolid sphere = mr ÷ø
5
12 2 2
wr d
25
3MR 2
+ Iremaining disc
32
From eq. (i) and (ii),
ITotal =
C. M
Reference
level for P.E.
B
A
P-109
System of Particles and Rotational Motion
The moment of inertia of the rod about O is
1 2
ml . The
3
1 2
I w where I is
2
the moment of inertia of the rod about O. When the rod is in
position B, its angular velocity is zero. In this case, the
energy of the rod is mgh where h is the maximum height to
which the centre of mass (C.M) rises
Gain in potential energy = Loss in kinetic energy
1 2 1 æ 1 2ö 2
\ mgh = I w = 2 çè 3 ml ÷ø w
2
kinetic energy of the rod at position A =
Þ h=
l 2 w2
6g
120. (d) Inn' =
1
Ma 2
M (a 2 + a 2 ) =
12
6
n
Again, by the same theorem I = IAC + IBD = 2 IAC
(\ IAC = IBD by symmetry of the figure)
I
2
From (i) and (ii), we get, IEF = IAC.
...(ii)
\ I AC =
122. (c)
l
D
n
A
2
l/
C
O
B
n'
Inn' = M.I due to the point mass at B +
M.I due to the point mass at D +
M.I due to the point mass at C.
m
A
D
æ l ö
Inn' = m ç
è 2 ÷ø
2
æ l ö
+m ç
è 2 ÷ø
2
+m
O
( 2l)
2
2
B
C
n
m
1
DB
2a
a
=
=
2
2
2
By parallel axes the orem, moment of inertia of plate about
an axis through one of its corners.
Also, DO =
2
Ma 2 Ma 2
æ a ö
Imm ' = I nn ' + M ç
=
+
÷
è 2ø
6
2
2
2
Ma + 3Ma
2
= Ma 2
6
3
121. (d) By the theorem of perpendicular axes,
I = IEF + IGH
Here, I is the moment of inertia of square lamina about an
axis through O and perpendicular to its plane.
\ IEF = IGH (By Symmetry of Figure)
=
Z
Y
F
D
æ l ö
+ m( 2l) 2
Þ I nn ' = 2 ´ m ç
è 2 ÷ø
= ml2 + 2ml2 = 3ml2
123. (c) The disc may be assumed as combination of two semi
circular parts. Therefore, circular disc will have twice
the mass of semicircular disc.
1
(2m)r2 = Mr2
2
Let I be the moment of inertia of the uniform semicircular
disc
Moment of inertia of disc =
Mr 2
2
124. (a) The moment of inertia of solid sphere A about its
Þ 2 I = 2Mr 2 Þ I =
2
MR 2 .
5
The moment of inertia of a hollow sphere B about its
2
2
diameter I B = MR .
3
diameter I A =
\ I A < IB
C
125. (d) We know that density (d ) =
X
O
mass( M )
volume(V )
\ M = d ´ V = d ´ ( pR 2 ´ t ) .
The moment of inertia of a disc is given by I =
A
\ I EF =
I
2
E
B
\ Ix =
...(i)
1
1
MxRx2 = (d ´ pR 2 ´ t ) R 2
2
2
1
MR 2
2
P-110
=
Physics
pd
t ´ R4
2
Iy =
129. (a) K.E of the sphere = translational K.E + rotational K.E
1
1
= mv 2 + I w2
2
2
Where, I = moment of inertia,
w = Angular, velocity of rotation
m = mass of the sphere
v = linear velocity of centre of mass of sphere
( )
1
1é
æ 1ö ù
2
M R2 =
p 4R 2 ç ÷ d ú ´ ( 4 R )
è 4ø û
2 y y 2 êë
1
t ´ R4
I X t X RX4
=
=
=
4
t
64
IY
4
tY RY
´ (4 R )
4
1 2
126. (a) Rotational kinetic energy = I w ,
2
\
Angular momentum, L = Iw Þ I =
\ K .E. =
Q Moment of inertia of sphere I =
L
w
1L
1
´ w 2 = Lw
2w
2
2K.E
w
When w is doubled and K.E is haled.
New angular momentum,
\ K .E =
1 2 1 2
mv + ´ mR 2 ´ w2
2
2 5
Þ K .E =
1 2 1 2
ævö
mv + ´ mR 2 ´ ç ÷
2
2 5
è Rø
L' =
1æ2
öæ v ö
Þ KE = ç mR2 + mR 2 ÷ç ÷
2è5
øè R ø
2K.E
L' = 2
2w
Þ KE =
L
4
127. (a) M. I of a circular wire about an axis nn' passing through
the centre of the circle and perpendicular to the plane of the
Z
circle = MR2
Y
v ö
æ
çQw = ÷
Rø
è
2
1
7 v2
7 1 25
mR 2 ´ ´
= ´ ´
2
5 R 2 10 2 104
35
´ 10 –4 joule
4
Þ KE = 8.75 × 10–4 joule
130. (50) For the box to be slide
F = mmg = 0.4 mg
For no toppling
n
X
n'
As shown in the figure, X-axis and Y-axis lie along diameter
of the ring . Using perpendicular axis theorem
IX + IY = IZ
Here, IX and IY are the moment of inertia about the diameter.
Þ 2 IX = MR2 [Q IX = IY (by symmetry) and IZ = MR2]
1
MR 2
2
128. (c) Moment of inertia of circular disc
1
I1 = MR 2
2
When two small sphere are attached on the edge of the
disc, the moment of inertia becomes
1
I 2 = MR 2 + 2mR 2
2
When two small spheres of mass m are attached gently,
the external torque, about the axis of rotation, is zero and
therefore the angular momentum about the axis of rotation
is constant.
I
\ I1w1 = I 2 w 2 Þ w 2 = 1 w1
I2
\ IX =
2
2
Þ KE =
Þ \ L' =
\ w2 = 1
2
mR 2
5
1
MR 2
2
MR 2 + 2mR 2
´ w1 =
M
w1
M + 4m
a
æa
ö
F ç + b ÷ £ mg
2
è2
ø
a
æa
ö
Þ 0.4 mg ç + b ÷ £ mg
2
è2
ø
Þ 0.2 a + 0.4 b £ 0.5 a
b 3
£
a 4
i.e. b £ 0.75 a but this is not possible.
As the maximum value of b can be equal to 0.5a.
Þ
100b
= 50
a
131. (c) For sphere,
Þ
1 2
1
mv + Iw2 = mgh
2
2
2
7v 2
1 2 1 æ2
2öv
=
mR
mgh
or
or h =
mv +
ç
÷ 2
10 g
2
2 è5
øR
For cylinder
1 2 1 æ mR 2 ö
mv + ç
÷ = mgh '
2
2 çè 2 ÷ø
P-111
System of Particles and Rotational Motion
3
F = mRa
2
2F
a=
3mR
134. (c) As shown in the diagram, the normal reaction of AB
on roller will shift towards O.
This will lead to tending of the system of cones to turn
left.
D
B
3v 2
4g
or
h' =
\
h 7v 2 / 10 g 14
=
=
h ' 3v 2 / 4 g 15
1 2
1
2
132.(Bonus) mgh = mvcm + I cm w
2
2
1 2
1
æv ö
= mvcm
+ I cm ç cm ÷
2
2
è R ø
I
1æ
= ç m + cm
2è
R2
2
ö 2
÷ vcm
ø
O
mR 2 ö 2
1æ
For ring : mgh = 2 çç m + 2 ÷÷ vcm
R ø
è
\h =
2
vcm
g
A
135. (c) From figure,
ma = F – f
....(i)
a
.
1æ
mR
For solid cylinder, mgh = 2 çç m +
2R2
è
2
Mass = m
ö 2
÷ vcm
÷
ø
F
O
f
3v 2
\ h = cm
4g
And, torque t = Ia
1æ
2 mR 2
m
=
+
ç
For sphere, mgh 2 ç
5 R2
è
\h =
C
ö 2
÷ vcm
÷
ø
2
7vcm
10 g
3 7
Ratio of heights 1: : Þ 20 :15 :14
4 10
mR 2
a = fR
2
aù
é
mR 2 a
= fR êQ a = ú
R
ë
û
2 R
ma
= f
...(ii)
2
Put this value in equation (i),
ma
3ma
or F =
2
2
136. (d) As we know,
ma = F –
133. (d)
Acceleration, a =
For cylinder, a c =
F – fr = ma
mR
a
2
for pure rolling
a = aR
from (1) (2) and (3)
mRa
F–
= maR
2
frR = Ia =
Mc +
...(i)
2
...(ii)
mg sin q
I
m+
r2
M c .g. sin qc
1 McR
2 R2
2
2
g sin qc
3
For sphere,
or,
ac =
as =
Ms g sin qs = Ms g sin qs
I
2 MR 2
Ms +
M s + s2
5 R2
r
...(iii)
=
M c .g. sin qc
Mc +
McR 2
2R 2
P-112
or,
i.e.,
\
Physics
2
1
2 1
2 V0
mV
+
mk
=
0
2
2
R02
5
g sin qs
7
given, ac = as
as =
2
5
g sin qc = g sin qs
3
7
5
g
sin qc 7
15
=
=
2
sin qs
14
g
3
r
137. (c)
o
o
From conservation of angular momentum about any fix
point on the surface,
mr2w0 = 2mr2w
w0 r
[Q v = rw ]
2
138. (b) Velocity of the tennis ball on the surface of the earth
or ground
Þ w = w0/2 Þ v =
2gh
v=
1+
shell =
k2
R2
( where k = radius of gyration of spherical
=
1 æ5 ö
m v0
2 çè 4 ÷ø
2
...(ii)
1 2 é K2 ù
mv ê1 +
ú
2 0 êë R02 úû
P.E.
=
1 25
P.E.
´ ´ mV02
2 16
139. (b) When body rolls dawn on inclined plane with
velocity V0 at bottom then body has both rotational
and translational kinetic energy.
Therefore, by law of conservation of energy,
P.E. = K.Etrans + K.Erotational
1
1
mV02 + I w 2
2
2
=
K2
9
25
K2
= 1+ 2 Þ 2 =
16
16
R0
R0
3
R0 .
4
140. (d) Minimum velocity for a body rolling without slipping
or, K =
2 gh
K2
R2
For solid sphere,
2
ö
÷÷ sin(2 ´ 30°)
ø
= 2.08 m
g
only
Dividing (i) by (ii) we get
1+
v2 sin 2q
g
V ù
ú
R0 û
\ P.E. = K.Etrans
2
R)
3
æ
2gh
çç
1+ k2 / R2
=è
=
é
2
êQ I = mk , w =
ë
When body is sliding down then body has
translatory motion.
v=
Horizontal range AB =
…(i)
\
v=
2 gh
1+
K
2
K2
R
2
=
=
2
5
10
gh
7
R2
141. (b) Acceleration of the body rolling down an inclined
plane is given by.
g sin q
a=
I
1+
MR 2
7
P-113
Gravitation
Gravitation
TOPIC 1 Kepler's Laws of Planetary
Motion
1.
4.
If the angular momentum of a planet of mass m, moving
around the Sun in a circular orbit is L, about the center of
the Sun, its areal velocity is:
[9 Jan. 2019 I]
L
4L
L
2L
(b)
(c)
(d)
m
m
2m
m
Figure shows elliptical path abcd of a planet around the
(a)
2.
1
sun S such that the area of triangle csa is the area of the
4
ellipse. (See figure) With db as the semimajor axis, and ca
as the semiminor axis. If t1 is the time taken for planet to go
over path abc and t2 for path taken over cda then:
[Online April 9, 2016]
TOPIC 2 Newton's Universal Law of
Gravitation
5.
c
d
(a) 500 days
(b) 320 days
(c) 260 days
(d) 220 days
The time period of a satellite of earth is 5 hours. If the
separation between the earth and the satellite is increased
to 4 times the previous value, the new time period will
become
[2003]
(a) 10 hours
(b) 80 hours
(c) 40 hours
(d) 20 hours
A straight rod of length L extends from x = a to x = L + a.
The gravitational force it exerts on point mass ‘m’ at x = 0,
if the mass per unit length of the rod is A + Bx2, is given
by:
[12 Jan. 2019 I]
(a)
é æ 1
1ö
ù
Gm ê A ç
- ÷ - BL ú
ë èa+L aø
û
(b)
é æ1
1 ö
ù
Gm ê A ç ÷ - BL ú
a
a
+
L
ø
ë è
û
(c)
é æ 1
1ö
ù
- ÷ + BL ú
Gm ê A ç
ë èa+L aø
û
(d)
é æ1
1 ö
ù
Gm ê A ç ÷ + BL ú
a
a
+
L
è
ø
ë
û
b
S
a
3.
(a) t1 = 4t2
(b) t1 = 2t2
(c) t1 = 3t2
(d) t1 = t2
India’s Mangalyan was sent to the Mars by launching it
into a transfer orbit EOM around the sun. It leaves the
earth at E and meets Mars at M. If the semi-major axis of
Earth’s orbit is ae = 1.5 × 1011 m, that of Mars orbit am =
2.28 × 1011 m, taken Kepler’s laws give the estimate of time
for Mangalyan to reach Mars from Earth to be close to:
[Online April 9, 2014]
Mars orbit
O
M
am
ae
Sun
E
6.
Take the mean distance of the moon and the sun from the
earth to be 0.4 × 106 km and 150 × 106 km respectively.
Their masses are 8 × 1022 kg and 2 × 1030 kg respectively.
The radius of the earth is 6400 km. Let DF1 be the difference
in the forces exerted by the moon at the nearest and farthest
points on the earth and DF2 be the difference in the force
exerted by the sun at the nearest and farthest points on
the earth. Then, the number closest to
Earth’s orbit
(a) 2
(b) 6
DF1
is:
DF2
[Online April 15, 2018]
(c) 10–2
(d) 0.6
P-114
7.
Physics
Four particles, each of mass M and equidistant from each
other, move along a circle of radius R under the action of
their mutual gravitational attraction. The speed of each
particle is:
[2014]
GM
R
(a)
(
GM
R
)
(
(a)
)
GM
1 GM
1+ 2 2
1+ 2 2
(d)
R
2 R
From a sphere of mass M and radius R, a smaller sphere of
R
radius
is carved out such that the cavity made in the
2
original sphere is between its centre and the periphery
(See figure). For the configuration in the figure where the
distance between the centre of the original sphere and the
removed sphere is 3R, the gravitational force between the
two sphere is:
[Online April 11, 2014]
(c)
8.
2 2
(b)
12. The acceleration due to gravity on the earth’s surface at
the poles is g and angular velocity of the earth about the
axis passing through the pole is w. An object is weighed at
the equator and at a height h above the poles by using a
spring balance. If the weights are found to be same, then h
is : (h<<R, where R is the radius of the earth)
13.
R 2w 2
2g
(b)
R 2w 2
[5 Sep. 2020 (II)]
g
R 2w 2
R 2w 2
(c)
(d)
4g
8g
The height 'h' at which the weight of a body will be the
same as that at the same depth 'h' from the surface of the
earth is (Radius of the earth is R and effect of the rotation
of the earth is neglected) :
[2 Sep. 2020 (II)]
(a)
5
R-R
2
(b)
R
2
5R - R
3R - R
(d)
2
2
14. A box weighs 196 N on a spring balance at the north pole.
Its weight recorded on the same balance if it is shifted to
the equator is close to (Take g = 10 ms –2 at the north pole
and the radius of the earth = 6400 km): [7 Jan. 2020 II]
(a) 195.66 N
(b) 194.32 N
(c) 194.66 N
(d) 195.32 N
15. The ratio of the weights of a body on the Earth’s surface to
that on the surface of a planet is 9:4. The mass of the
(c)
3R
(a)
9.
10.
41 GM 2
(c)
(d)
450 R 2
225 R 2
450 R 2
Two particles of equal mass ‘m’ go around a circle of radius
R under the action of their mutual gravitational attraction.
The speed of each particle with respect to their centre of
mass is
[2011 RS]
Gm
Gm
Gm
Gm
(a)
(b)
(c)
(d)
4R
3R
R
2R
Two spherical bodies of mass M and 5M & radii R & 2R
respectively are released in free space with initial separation
between their centres equal to 12 R. If they attract each
other due to gravitational force only, then the distance
covered by the smaller body just before collision is [2003]
(a) 2.5 R
(b) 4.5 R
(c) 7.5 R
(d) 1.5 R
3600 R 2
(b)
GM 2
59 GM 2
41 GM 2
TOPIC 3 Acceleration due to Gravity
11.
The value of acceleration due to gravity is g1 at a height
R
(R = radius of the earth) from the surface of the
2
earth. It is again equal to g1 and a depth d below the sur-
h=
ædö
face of the earth. The ratio ç ÷ equals : [5 Sep. 2020 (I)]
è Rø
(a)
4
9
(b)
5
9
(c)
1
3
(d)
7
9
1
th of that of the Earth. If ‘R’ is the radius of the
9
Earth, what is the radius of the planet ? (Take the planets
to have the same mass density).
[12 April 2019 II]
planet is
R
R
R
R
(b)
(c)
(d)
3
4
9
2
16. The value of acceleration due to gravity at Earth’s surface
is 9.8 ms– 2. The altitude above its surface at which the
acceleration due to gravity decreases to 4.9 ms– 2, is close
to : (Radius of earth = 6.4 × 106 m)
[10 April 2019 I]
6
6
(a) 2.6×10 m
(b) 6.4×10 m
(c) 9.0×106 m
(d) 1.6×106 m
17. Suppose that the angular velocity of rotation of earth is
increased. Then, as a consequence.
[Online April 16, 2018]
(a) There will be no change in weight anywhere on the
earth
(b) Weight of the object, everywhere on the earth, wild
decrease
(c) Weight of the object, everywhere on the earth, will
increase
(d) Except at poles, weight of the object on the earth will
decrease
(a)
P-115
Gravitation
18. The variation of acceleration due to gravity g with distance
d from centre of the earth is best represented by (R =
Earth's radius):
[2017, Online May 7, 2012]
g
g
(a)
(b)
d
R
O
R
g
g
(c)
(d)
d
d
O
O
R
19. The mass density of a spherical body is given by r (r) =
k
for r < R and r (r) = 0 for r > R,
r
where r is the distance from the centre.
The correct graph that describes qualitatively the acceleration, a, of a test particle as a function of r is :
[Online April 9, 2017]
a
a
(a)
(b)
r
R
R
(c)
(d)
r
r
R
If the Earth has no rotational motion, the weight of a person
on the equator is W. Determine the speed with which the
earth would have to rotate about its axis so that the person
R
20.
r
a
a
3
at the equator will weight W . Radius of the Earth is
4
2
6400 km and g =10 m/s .
[Online April 8, 2017]
(a) 1.1×10–3 rad/s
(c)
0.63 × 10–3 rad/s
(b) 0.83×10–3 rad/s
(d) 0.28×10–3 rad/s
21. The change in the value of acceleration of earth towards
sun, when the moon comes from the position of solar
eclipse to the position on the other side of earth in line
with sun is:
= 7.36 × 1022 kg, radius of the moon’s
(mass of the moon
orbit = 3.8 × 108 m).
(a)
6.73 × 10–5 m/s2
(c)
6.73 × 10–2 m/s2
[Online April 22, 2013]
(b)
6.73 × 10–3 m/s2
(d)
6.73 × 10–4 m/s2
g
(where g = the acceleration due to gravity on
9
the surface of the earth) in terms of R, the radius of the
earth, is
[2009]
R
(a)
(b) R / 2
(c)
(d) 2 R
2R
2
24. The change in the value of ‘g’ at a height ‘h’ above the
surface of the earth is the same as at a depth ‘d’ below the
surface of earth. When both ‘d’ and ‘h’ are much smaller
than the radius of earth, then which one of the following is
correct?
[2005]
h
3h
(a) d =
(b) d =
2
2
(c) d = h
(d) d =2 h
25. Average density of the earth
[2005]
(a) is a complex function of g
(b) does not depend on g
(c) is inversely proportional to g
(d) is directly proportional to g
becomes
d
O
22. Assuming the earth to be a sphere of uniform density, the
acceleration due to gravity inside the earth at a distance of
r from the centre is proportional to[Online May 12, 2012]
(a) r
(b) r–1
(c) r2
(d) r–2
23. The height at which the acceleration due to gravity
Gravitational Field and
Potential Energy
TOPIC 4
26. Two planets have masses M and 16 M and their radii are a
and 2a, respectively. The separation between the centres
of the planets is 10a. A body of mass m is fired from the
surface of the larger planet towards the smaller planet along
the line joining their centres. For the body to be able to
reach the surface of smaller planet, the minimum firing
speed needed is :
[6 Sep. 2020 (II)]
(a) 2
GM
a
(b) 4
GM
a
3 5GM
GM 2
(d)
2
a
ma
27. On the x-axis and at a distance x from the origin, the
gravitational field due to a mass distribution is given by
(c)
Ax
2
( x + a 2 )3/2
in the x-direction. The magnitude of
gravitational potential on the x-axis at a distance x, taking
its value to be zero at infinity, is :
[4 Sep. 2020 (I)]
(a)
A
( x2 + a2 )
1
(b)
2
(c) A( x 2 + a 2 )
1
2
A
( x2 + a2 )
3
2
2
2
(d) A( x + a )
3
2
P-116
28.
Physics
The mass density of a planet of radius R varies with the
æ
r2 ö
distance r from its centre as r(r ) = r0 çç1 - 2 ÷÷ . Then the
è R ø
gravitational field is maximum at :
[3 Sep. 2020 (II)]
(a) r = 3 R
4
1
R
Gravitational field E
(b) 1.16
GM
a
(c) 1.21
GM
a
(d) 1.41
GM
a
K
. Identify the
r2
correct relation between the radius R of the particle’s orbit
and its period T:
[8 April 2019 II]
2
3
(a) T/R is a constant
(b) T /R is a constant
field produced by a mass density r (r ) =
(d) r =
4
(c) T/R2 is a constant
(d) TR is a constant
34. A body of mass m is moving in a circular orbit of radius R
about a planet of mass M. At some instant, it splits into
two equal masses. The first mass moves in a circular orbit
of radius
3
radius
1
1
1
2
3
4
5
radius R
2
1
1
1
(b)
(c)
(d)
3
6
2
3
30. An asteroid is moving directly towards the centre of the
earth. When at a distance of 10 R (R is the radius of the
earth) from the earths centre, it has a speed of 12 km/s.
Neglecting the effect of earths atmosphere, what will be
the speed of the asteroid when it hits the surface of the
earth (escape velocity from the earth is 11.2 km/ s)? Give
your answer to the nearest integer in kilometer/s _____.
[NA 8 Jan. 2020 II]
31. A solid sphere of mass ‘M’ and radius ‘a’ is surrounded by
a uniform concentric spherical shell of thickness 2a and
mass 2M. The gravitational field at distance ‘3a’ from the
centre will be:
[9 April 2019 I]
2
(b)
GM
2
(c)
GMm
GMm
GMm
GMm
(b) +
(c) (d)
6
R
6
R
2R
2R
35. From a solid sphere of mass M and radius R, a spherical
portion of radius R/2 is removed, as shown in the figure.
Taking gravitational potential V = 0 at r = ¥, the potential at
the centre of the cavity thus formed is :
(G = gravitational constant)
[2015]
(a)
(a)
2GM
R
, and the other mass, in a circular orbit of
2
3R
. The difference between the final and initial
2
total energies is:
[Online April 15, 2018]
2
2
0
(a)
GM
a
33. A test particle is moving in circular orbit in the gravitational
(b) r = R
5
R
3
9
29. Consider two solid spheres of radii R1 = 1m, R2=2m and
masses M1 and M2, respectively. The gravitational field
m1
due to sphere 1 and 2 are shown. The value of m is:
2
[8 Jan. 2020 I]
(c) r =
(a) 1.35
GM
2
(d)
2GM
9a
9a
3a
3a 2
32. Four identical particles of mass M are located at the corners
of a square of side ‘a’. What should be their speed if each
of them revolves under the influence of others’
gravitational field in a circular orbit circumscribing the
square ?
[8 April 2019 I]
-
-2GM
-2GM
-GM
-GM
(b)
(c)
(d)
3R
R
2R
R
Which of the following most closely depicts the correct
variation of the gravitational potential V(r) due to a large
planet of radius R and uniform mass density ? (figures
are not drawn to scale)
[Online April 11, 2015]
(a)
36.
V(r)
V(r)
(a)
V(r)
(c)
(b)
r
O
O
r
(d)
r
O
V(r)
O
r
P-117
Gravitation
37. The gravitational field in a region is given by
(
)
®
(a)
2Gm
r
potential energy of a particle of mass 1 kg when it is taken
from the origin to a point (7 m, – 3 m) is:
[Online April 19, 2014]
(c)
2Gm æ
1 ö
1÷
r çè
2ø
g = 5N / kgiˆ + 12N / kgjˆ . The change in the gravitational
(a) 71 J
m1
38.
(b) 13 58J (c) – 71 J
m2
v1 v2
(d) 1 J
Two hypothetical planets of masses m1 and m2 are at rest
when they are infinite distance apart. Because of the
gravitational force they move towards each other along
the line joining their centres. What is their speed when
their separation is ‘d’?
[Online April 12, 2014]
(Speed of m1 is v1 and that of m2 is v2)
(a) v1 = v2
(b)
v1 = m 2
2G
2G
v 2 = m1
d ( m1 + m 2 )
d ( m1 + m 2 )
(c)
v1 = m1
2G
2G
v 2 = m2
d ( m1 + m 2 )
d ( m1 + m 2 )
(d)
2G
2G
v1 = m 2
v2 = m 2
m1
m2
39. The gravitational field, due to the 'left over part' of a uniform
sphere (from which a part as shown, has been 'removed
out'), at a very far off point, P, located as shown, would be
(nearly) :
[Online April 9, 2013]
Mass of complete
sphere = M
Removed
Part
R
P
R
(b)
Gm
r
(d)
2Gm
r
42. Two bodies of masses m and 4 m are placed at a distance r.
The gravitational potential at a point on the line joining
them where the gravitational field is zero is:
[2011]
(a)
d
2 -1
-
4Gm
r
(b)
-
6Gm
r
(c)
-
9Gm
r
(d) zero
43. This question contains Statement-1 and Statement-2. Of
the four choices given after the statements, choose the
one that best describes the two statements.
[2008]
Statement-1 : For a mass M kept at the centre of a cube of
side ‘a’, the flux of gravitational field passing through its
sides 4 p GM. and
Statement-2: If the direction of a field due to a point source
is radial and its dependence on the distance ‘r’ from the
source is given as
1
, its flux through a closed surface
r2
depends only on the strength of the source enclosed by
the surface and not on the size or shape of the surface.
(a) Statement -1 is false, Statement-2 is true
(b) Statement -1 is true, Statement-2 is true; Statement -2
is a correct explanation for Statement-1
(c) Statement -1 is true, Statement-2 is true; Statement 2 is not a correct explanation for Statement-1
(d) Statement -1 is true, Statement-2 is false
44. A particle of mass 10 g is kept on the surface of a uniform
sphere of mass 100 kg and radius 10 cm. Find the work to
be done against the gravitational force between them to
take the particle far away from the sphere
(you may take G = 6.67× 10 -11 Nm 2 / kg 2 )
[2005]
x
(a)
5 GM
6 x2
(b)
8 GM
9 x2
(c)
7 GM
8 x2
(d)
6 GM
7 x2
40. The mass of a spaceship is 1000 kg. It is to be launched
from the earth's surface out into free space. The value of g
and R (radius of earth) are 10 m/s2 and 6400 km respectively.
The required energy for this work will be
[2012]
11
8
(a) 6.4 × 10 Joules
(b) 6.4 × 10 Joules
(c) 6.4 × 109 Joules
(d) 6.4 × 1010 Joules
41. A point particle is held on the axis of a ring of mass m and
radius r at a distance r from its centre C. When released, it
reaches C under the gravitational attraction of the ring. Its
speed at C will be
[Online May 26, 2012]
(a) 3.33 × 10 -10 J
(b) 13.34 × 10 -10 J
(c) 6.67 × 10 -10 J
(d) 6.67 × 10 -9 J
45. If ‘g’ is the acceleration due to gravity on the earth’s
surface, the gain in the potential energy of an object of
mass ‘m’ raised from the surface of the earth to a height
equal to the radius ‘R' of the earth is
[2004]
(a)
1
mgR (b)
4
1
mgR (c) 2 mgR
2
(d) mgR
46. Energy required to move a body of mass m from an orbit of
radius 2R to 3R is
[2002]
(a) GMm/12R2
(b) GMm/3R2
(c) GMm/8R
(d) GMm/6R.
P-118
Physics
Motion of Satellites, Escape
TOPIC 5
Speed and Orbital Velocity
47.
48.
49.
A satellite is in an elliptical orbit around a planet P. It is
observed that the velocity of the satellite when it is farthest
from the planet is 6 times less than that when it is closest
to the planet. The ratio of distances between the satellite
and the planet at closest and farthest points is :
[NA 6 Sep. 2020 (I)]
(a) 1 : 6
(b) 1 : 3
(c) 1 : 2
(d) 3 : 4
A body is moving in a low circular orbit about a planet of
mass M and radius R. The radius of the orbit can be taken
to be R itself. Then the ratio of the speed of this body in
the orbit to the escape velocity from the planet is :
1
(a)
(b) 2
[4 Sep. 2020 (II)]
2
(c) 1
(d) 2
A satellite is moving in a low nearly circular orbit around
the earth. Its radius is roughly equal to that of the earth’s
radius Re. By firing rockets attached to it, its speed is
instantaneously increased in the direction of its motion so
3
times larger. Due to this the farthest
2
distance from the centre of the earth that the satellite
reaches is R. Value of R is :
[3 Sep. 2020 (I)]
(a) 4Re
(b) 2.5Re
(c) 3Re
(d) 2Re
a circular orbit at this height is E2. The value of h for
which E1 and E2 are equal, is:
[9 Jan. 2019 II]
3
(a) 1.6 × 10 km
(b) 3.2 × 103 km
3
(c) 6.4 × 10 km
(d) 28 × 104 km
53. Planet A has mass M and radius R. Planet B has half the
mass and half the radius of Planet A. If the escape velocities
from the Planets A and B are vA and vB, respectively, then
vA n
= .
vB 4 The value of n is :
(a) 4
(b) 1
(c) 2
K
over
r
a large distance 'r' from its centre. In that region, a small
star is in a circular orbit of radius R. Then the period of
revolution, T depends on R as :
[2 Sep. 2020 (I)]
The mass density of a spherical galaxy varies as
(a)
T 2 µ R (b) T 2 µ R3 (c)
T2 µ
1
R3
(d) T µ R
51. A body A of mass m is moving in a circular orbit of radius
R about a planet. Another body B of mass
m
collides with
2
r
æ vö
A with a velocity which is half çè ÷ø the instantaneous
2
r
velocity v or A. The collision is completely inelastic.
Then, the combined body:
[9 Jan. 2020 I]
(a) continues to move in a circular orbit
(b) Escapes from the Planet’s Gravitational field
(c) Falls vertically downwards towards the planet
(d) starts moving in an elliptical orbit around the planet
52. The energy required to take a satellite to a height 'h'
above Earth surface (radius of Eareth = 6.4 × 10 3 km) is
E1 and kinetic energy required for the satellite to be in
(d) 3
54. A satellite of mass m is launched vertically upwards
with an initial speed u from the surface of the earth.
After it reaches height R (R = radius of the earth), it
m
so that subsequently the
10
satellite moves in a circular orbit. The kinetic energy of
the rocket is (G is the gravitational constant; M is the
mass of the earth):
[7 Jan. 2020 I]
ejects a rocket of mass
(a)
m æ 2 113 GM ö
çu +
÷
20 è
200 R ø
that it become
50.
[9 Jan. 2020 II]
3m æ
5GM ö
u+
(c)
ç
8 è
6 R ÷ø
2
æ 2 119 GM ö
(b) 5m çè u ÷
200 R ø
mæ
2GM ö
(d) 20 çè u - 3 R ÷ø
2
55. A spaceship orbits around a planet at a height of 20 km
from its surface. Assuming that only gravitational field of
the planet acts on the spaceship, what will be the number
of complete revolutions made by the spaceship in 24 hours
around the planet ? [Given : Mass of Planet = 8×10 22 kg,
Radius of planet = 2×10 6 m, Gravitational constant
G = 6.67×10–11Nm2/kg2]
[10 April 2019 II]
(a) 9
(b) 17
(c) 13
(d) 11
56. A rocket has to be launched from earth in such a way that
it never returns. If E is the minimum energy delivered by
the rocket launcher, what should be the minimum energy
that the launcher should have if the same rocket is to be
launched from the surface of the moon? Assume that the
density of the earth and the moon are equal and that the
earth’s volume is 64 times the volume of the moon.
[8 April 2019 II]
(a)
E
64
(b)
E
32
(c)
E
4
(d)
E
16
57. A satellite of mass M is in a circular orbit of radius R about
the centre of the earth. A meteorite of the same mass, falling
towards the earth collides with the satellite completely in
elastically. The speeds of the satellite and the meteorite are
the same, Just before the collision. The subsequent motion
of the combined body will be
[12 Jan. 2019 I]
P-119
Gravitation
(a) such that it escape to infinity
(b) In an elliptical orbit
(c) in the same circular orbit of radius R
(d) in a circular orbit of a different radius
58. Two satellites, A and B, have masses m and 2m respectively.
A is in a circular orbit of radius R, and B is in a circular orbit
of radius 2R around the earth. The ratio of their kinetic
energies, TA/TB, is :
[12 Jan. 2019 II]
(a)
1
2
(b) 1
1
2
59. A satellite is revolving in a circular orbit at a height h from
the earth surface, such that h << R where R is the radius of
the earth. Assuming that the effect of earth’s atmosphere
can be neglected the minimum increase in the speed
required so that the satellite could escape from the
gravitational field of earth is:
[11 Jan. 2019 I]
(c) 2
(d)
(a)
2gR
(b)
gR
(c)
gR
2
(d)
gR
(
)
2 -1
60. A satellite is moving with a constant speed v in circular
orbit around the earth. An object of mass ‘m’ is ejected
from the satellite such that it just escapes from the
gravitational pull of the earth. At the time of ejection,
the kinetic energy of the object is:
[10 Jan. 2019 I]
2
2
(a) 2 m v
(b) m v
(c)
1
m v2
2
(d)
3
m v2
2
61. Two stars of masses 3 × 1031 kg each, and at distance
2 × 1011 m rotate in a plane about their common centre
of mass O. A meteorite passes through O moving
perpen-dicular to the star’s rotation plane. In order to
escape from the gravitational field of this double star,
the minimum speed that meteorite should have at O is:
(Take Gravitational constant G = 66 × 10–11 Nm2 kg–2)
[10 Jan. 2019 II]
4
(a) 2.4 × 10 m/s
(b) 1.4 × 105 m/s
(c) 3.8 × 104 m/s
(d) 2.8 × 105 m/s
62. A satellite is revolving in a circular orbit at a height 'h' from
the earth's surface (radius of earth R; h < < R). The minimum
increase in its orbital velocity required, so that the satellite
could escape from the earth's gravitational field, is close
to : (Neglect the effect of atmosphere.)
[2016]
(a)
gR / 2
(b)
gR
(c)
2gR
(d)
gR
(
)
2-1
63. An astronaut of mass m is working on a satellite orbiting
the earth at a distance h from the earth's surface. The radius
of the earth is R, while its mass is M. The gravitational pull
FG on the astronaut is :
[Online April 10, 2016]
(a) Zero since astronaut feels weightless
GMm
(b)
(R + h)
(c)
FG =
2
< FG <
GMm
R2
GMm
(R + h) 2
(d) 0 < FG <
GMm
R2
64. A very long (length L) cylindrical galaxy is made of
uniformly distributed mass and has radius R(R < < L). A
star outside the galaxy is orbiting the galaxy in a plane
perpendicular to the galaxy and passing through its centre.
If the time period of star is T and its distance from the
galaxy’s axis is r, then :
[Online April 10, 2015]
(a) T µ r
(b)
(c) T µ r2
(d) T2 µ r3
Tµ r
65. What is the minimum energy required to launch a satellite
of mass m from the surface of a planet of mass M and
radius R in a circular orbit at an altitude of 2R? [2013]
5GmM
(b)
6R
2GmM
(c)
3R
GmM
GmM
(d)
2R
2R
66. A planet in a distant solar system is 10 times more massive
than the earth and its radius is 10 times smaller. Given that
the escape velocity from the earth is 11 km s–1, the escape
velocity from the surface of the planet would be [2008]
(a) 1.1 km s–1
(b) 11 km s–1
(c) 110 km s–1
(d) 0.11 km s–1
67. Suppose the gravitational force varies inversely as the nth
power of distance. Then the time period of a planet in circular
orbit of radius ‘R’ around the sun will be proportional to
(a)
(a)
R
n
æ n +1ö
çè
÷ø
(b)
R
æ n -1ö
çè
÷
2 ø
æ n- 2ö
ç
÷
[2004]
(c) R 2
(d) Rè 2 ø
68. The time period of an earth satellite in circular orbit is
independent of
[2004]
(a) both the mass and radius of the orbit
(b) radius of its orbit
(c) the mass of the satellite
(d) neither the mass of the satellite nor the radius of its
orbit.
P-120
69.
A satellite of mass m revolves around the earth of radius R
at a height x from its surface. If g is the acceleration due to
gravity on the surface of the earth, the orbital speed of the
satellite is
[2004]
(a)
70.
Physics
gR 2
R+ x
(b)
gR
R-x
(c) gx
æ gR2 ö 1/ 2
(d) ç
÷
è R + xø
The escape velocity for a body projected vertically
upwards from the surface of earth is 11 km/s. If the body is
projected at an angle of 45°with the vertical, the escape
velocity will be
[2003]
(a) 11 2 km / s
(b) 22 km/s
(c) 11 km/s
(d)
11
2
km / s
71. The kinetic energy needed to project a body of mass m from
the earth surface (radius R) to infinity is
[2002]
(a) mgR/2 (b) 2mgR (c) mgR
(d) mgR/4.
72. If suddenly the gravitational force of attraction between
Earth and a satellite revolving around it becomes zero,
then the satellite will
[2002]
(a) continue to move in its orbit with same velocity
(b) move tangentially to the original orbit in the same
velocity
(c) become stationary in its orbit
(d) move towards the earth
73. The escape velocity of a body depends upon mass as
[2002]
(a) m0
(b) m1
(c) m2
(d) m3
P-121
Gravitation
1.
(c) Areal velocity;
dA =
a + L Gm
(A + Bx 2 )dx
2
a
dA
dt
Þ F=ò
dA 1 2 dq 1 2
= r
= r w
dt 2 dt 2
Also, L = mvr = mr2w
dA 1 L
=
dt 2 m
(c) Let area of ellipse abcd = x
x x
Area of SabcS = + (i .e., ar of abca + SacS)
2 4
(Area of half ellipse + Area of triangle)
3x
=
4
6.
\
é æ1
1 ö
ù
= Gm ê A ç ÷ø + BL ú
è
a
a
L
+
ë
û
(a) As we know, Gravitational force of attraction,
GMm
F=
R2
GM e m
GM e M s
F1 =
and F2 =
2
r1
r22
DF1 =
c
d
x
4
t1
t2
r13
7.
(d)
2 F cos 45° + F ¢ =
Where F =
GM 2
( 2 R) 2
Gmdm
x
2
Mv 2
(From figure)
R
and F ¢ =
GM 2
4R2
F
F'
M
3
æR ö 2
é 4R ù 2
= 5´ ê ú
T2 = T1 ç 2 ÷
ë R û
è R1 ø
= 5 × 23 = 40 hours
(d) Given l = (A + Bx2),
Taking small element dm of length dx at a distance x
from x = 0
dx
x=0
dF =
r23
3
(b)
(c) According to Kepler’s law of periods T2 µ R3
2
3
æ T2 ö
æ R2 ö
\ ç
÷ = çR ÷
è T1 ø
è 1ø
m dF
so, dm = l dx
dm = (A + Bx2)dx
GM e M s
Dr2
Using Dr1 = Dr2 = 2 Rearth; m = 8 × 1022 kg;
Ms = 2 × 1030 kg
r1 = 0.4 × 106 km and r2 = 150 × 106 km
F
Þ
5.
Dr1 and DF2 =
DF1 æ 8 ´ 1022 ö æ 150 ´ 106 ö
=ç
֍
÷ ´1 @ 2
DF2 è 2 ´ 1030 ø è 0.4 ´ 106 ø
t1
= 3 or, t = 3t
1
2
t2
3
2GM e m
æ m ö æ r23 ö æ Dr1 ö
DF1 mDr1 r23
= 3
=ç
÷
֍ ֍
DF2
r1 Ms Dr2 è Ms ø è r13 ø è Dr2 ø
b
S
a
3x
=
Area of SadcS = x 4
Area of SabcS 3x / 4 =
=
Area of SadcS x / 4
3.
4.
a+L
é A
ù
= Gm ê - + Bx ú
ë x
ûa
1 2
r dq
2
Þ
2.
x
M
R
o
M
M
2 ´ GM 2
Þ
Þ
2( R 2)2
+
GM 2
4 R2
=
Mv 2
R
GM 2 é 1 1 ù
+
= Mv 2
ê
ú
R ë4
2û
\ v=
GM æ 2 + 4 ö 1 GM
(1 + 2 2 )
ç
÷=
R çè 4 2 ÷ø 2
R
P-122
8.
Physics
(a) Volume of removed sphere
1
S = ut + at 2
2
1
\ x5 M = a5 M t 2
....(ii)
2
For mass M
u = 0, s = xM, t = t, a = aM
1 2
\ s = ut + at
2
1
2
Þ xM = aM t
… (iii)
2
Dividing (ii) by (iii)
1
a5 M t 2 a
x5 M
1
2
=
= 5M =
1
xM
aM
5 [From (i)]
aM t 2
2
\ 5x5M = xM
....(iv)
From the figure it is clear that
x5M + xM = 9R
....(v)
Where O is the point where the two spheres collide.
From (iv) and (v)
3
4 æRö
4
æ 1ö
p ç ÷ = pR3 ç ÷
3 è2ø
3
è 8ø
Volume of the sphere (remaining)
Vremo =
4 3 4 3æ1ö
4
æ7ö
pR - pR ç ÷ = pR 3 ç ÷
3
3
3
è8ø
è8ø
Therefore mass of sphere carved and remaining sphere
1
7
are at respectively M and M.
8
8
Therefore, gravitational force between these two sphere,
7M 1
GM m G 8 ´ 8 M
7 GM 2
F=
=
=
r2
64 ´ 9 R 2
(3R )2
Vremain =
41 GM 2
3600 R 2
(a) As two masses revolve about the common centre of
mass O.
\ Mutual gravitational attraction = centripetal force
;
9.
Gm 2
( 2R)
Þ
Þ
2
R
= mw 2 R
Gm
4 R3
w=
m
=w
O
2
m
\
Gm
4 R3
If the velocity of the two particles with respect to the centre
of gravity is v then
v = wR
v=
10.
xM
+ xM = 9 R
5
\ 6xM = 45R
Gm
3
11.
xM =
(b) According to question, gh = g d = g1
h = R/2
d
Gm
4R
´R =
4R
(c) We know that
Force = mass × acceleration.
(R-d)
9R
R
45
R = 7.5R
6
xM
gh =
x5M
2R
12R
The gravitational force acting on both the masses is the
same.
F1 = F2
ma1 = ma2
9M
5M
Þ
=
=5
95M
M
9M
1
Þ
=
95M
5
Let t be the time taken for the two masses to collide and
x5M, xM be the distance travelled by the mass 5M and M
respectively.
For mass 5M
u = 0,
GM
Rö
æ
çè R + ÷ø
2
GM
æ 3R ö
çè ÷ø
2
2
=
2
and g d =
GM ( R - d )
R
3
Þ
GM ( R - d )
R3
4 (R - d )
=
9
R
Þ 4 R = 9 R - 9d Þ 5 R = 9 d
d 5
=
R 9
12. (b) Value of g at equator, g A = g × - Rw2
Value of g at height h above the pole,
\
æ 2h ö
gB = g × ç1 - ÷
Rø
è
As object is weighed equally at the equator and poles, it
means g is same at these places.
P-123
Gravitation
g A = gB
æ
h ö
g h = g ç1 +
è R e ÷ø
æ 2h ö
Þ g - Rw2 = g ç1 - ÷
Rø
è
R2 w2
2 gh
Þ
h
=
Þ Rw =
2g
R
(c) The acceleration due to gravity at a height h is given
by
æ
h ö
= ç1 2 è R e ÷ø
1
2
13.
g=
GM
( R + h)2
Here, G = gravitation constant
M = mass of earth
The acceleration due to gravity at depth h is
g' =
GM æ
ç1 R2 è
hö
÷
Rø
h = Re
\
- R ± R2 + 4(1) R2
2
- R + 5R ( 5 - 1)
=
R
2
2
14. (d) Weight at pole, w = mg = 196 N
Þ m = 19.6 kg
Weight at equator, w’ = mg’ = m(g – w2R)
=
é
ù
2p ö
æ
3
= 19.6 ê10 – ç
÷ø ´ 6400 ´ 10 ú N
è
24 ´ 3600
êë
úû
2p ö
æ
çèQ w = ÷ø
T
= 19.6 [10 – 0.034] = 195.33 N
2
We mge 9
ge 9
15. (d) W = mg = 4 or g = 4
p
p
p
or
GM / R 2
G (M / 9) /
\ Rp =
R 2p
=
9
4
R
2
16. (a) Given
Acceleration due to gravity at a height h from earth’s
surface is
-2
[as h <<< Re]
)
2 -1
If d < R, g =
GM
GM æ
hö
= 2 ç1 - ÷
( R + h)2
R è Rø
\ R3 = ( R + h)2 ( R - h) = ( R 2 + h2 + 2hR )( R - h)
Þ R 3 = R3 + h2 R + 2hR 2 - R 2 h - h3 - 2h 2 R
Þ h3 + h 2 (2 R - R ) - R 2 h = 0
Þ h3 + h2 R - R 2 h = 0
Þ h 2 + hR - R 2 = 0
æ
h ö
Þ 4.9 = 9.8 ç 1 +
è R e ÷ø
h = 6400 × 0.414 km = 2.6 × 106 m
17. (d) With rotation of earth or latitude, acceleration due to
gravity vary as g' = g – w2R cos2 f
Where f is latitude, there will be no change in gravity
at poles as f = 90°
At all other points as w increases g' will decreases
hence, weight, W = mg decreases.
18. (b) Variation of acceleration due to gravity, g with
distance 'd ' from centre of the earth
Given, g = g'
Þh=
(
-2
If d = R, gs =
If d > R, g =
19.
Gm
R2
Gm
.d
R2
Gm
2
i.e., g µ d (straight line)
i.e., g µ
1
d2
mass ö
(b) Given that, mass density æç
of a spherical
è volume ÷ø
k
body r(r) =
r
M k
= for inside r £ R
V r
kv
M=
..... (i)
r
Inside the surface of sphere Intensity
GMr
F
I= 3
Q I=
m
R
mg
GMr
=g
ginside =
or I =
3
m
R
=
d
G kv
. .r = constant
R3 r
From eq. (i),
Similarly, gout = GM
r2
Hence, option (2) is correct graph.
20. (c) We know, g' = g – w2R cos2 q
3g
= g - w2R
4
3
Given, g ' = g
4
g
w 2R =
4
g
10
w=
=
4R
4 ´ 6400 ´ 103
1
=
= 0.6 ´ 10 -3 rad/s
2 ´ 8 ´ 100
P-124
21.
22.
Physics
(a)
(a) Acceleration due to gravity at depth d from the
surface of the earth or at a distance r from the centre ‘O’ of
4
the earth g¢ = prGr
3
Hence g ' µ r
d
g¢
r
r = ( R – d)
O
23.
(d) On earth’s surface g =
GM
R2
At height above earth’s surface
gh =
25.
26.
GM
\
gn
R2
=
g ( R + h)2
Þ
g /9 é R ù
=ê
g
ë R + h úû
=
2
... (i)
GMm G (16M )m
GMm G (16 M ) m
+ KE = 8a
2a
2a
8a
é 1 16 1 16 ù
KE = GMm ê +
ë 8a 2a 2a 8a úû
-
1 2
é 45 ù
mv = GMm ê ú Þ v =
2
ë 8a û
90GM
8a
3 5GM
a
2
27. (a) Given : Gravitational field,
EG =
R
1
=
R+h 3
\
h = 2R
(d) Value of g with altitude is,
Vx
ò
é 2h ù
gh = g ê1 - ú ;
Rû
ë
Value of g at depth d below earth’s surface,
é dù
gd = g ê1 - ú
ë Rû
Equating gh and gd, we get d = 2h
(d) Value of g on earth’s surface,
GM Gr ´ V
g= 2 =
R
R2
4
G ´ r ´ pR 3
3
Þ g=
R2
4
g = rpG. R where r ® average density
3
æ 3g ö
r= ç
è 4pGR ÷ø
Þ r is directly proportional to g.
(d)
16M
A
x
(10a - x )2
1
4
=
Þ 4 x = 10a - x Þ x = 2a
x (10a - x)
Using conservation of energy, we have
Þv=
a
G (16 M )
Þ
Þ
( R + h )2
M
2
é1 + 64 - 4 - 16 ù
Þ KE = GMm ê
úû
8a
ë
Þ
24.
GM
x
g
R
Let A be the point where gravitation field of both planets
cancel each other i.e. zero.
2a
(10a – x)
10a
V¥
Ax
2
( x + a 2 )3/ 2
, V¥ = 0
x
r r
dV = - ò EG × d x
¥
x
Þ Vx - V¥ = - ò
¥
\ Vx =
Ax
2
( x + a 2 )3/ 2
A
2
2 1/ 2
(x + a )
-0=
dx
A
2
( x + a 2 )1/ 2
28. (d)
r
dx
x
Mass of small element of planet of radius x and thickness dx.
æ
x2 ö
dm = r ´ 4px 2 dx = r0 ç1 - 2 ÷ ´ 4px 2 dx
è R ø
Mass of the planet
r
æ
x4 ö
M = 4pr0 çç x 2 - 2 ÷÷ dx
R ø
0è
ò
Þ M = 4pr0
r3
r5
- 2
3 5R
P-125
Gravitation
Gravitational field,
E=
GM
r
2
=
Þ F =
ær
r ö
´ 4pr0 ç - 2 ÷
r
è 3 5R ø
G
3
5
Y
2
æ r r3 ö
Þ E = 4pGr0 ç - 2 ÷
ç 3 5R ÷
è
ø
dE
E is maximum when
=0
dr
A
æ 1 3r 2 ö
dE
Þ
= 4pGr0 ç - 2 ÷ = 0
ç 3 5R ÷
dr
è
ø
B
\
Gm
r2
Gm1
r12
and E2 =
Gm2
r22
From the diagram given in question,
E1 2
=
E2 3 (r1 = 1m, R2 = 2m given)
2
\
2
E1 æ r2 ö æ m1 ö
2 æ 2ö æ m ö
=ç ÷ ç ÷ Þ =ç ÷ ç 1÷
3 è 1 ø è m2 ø
E2 è r1 ø è m2 ø
30. (16.00)
Using law of conservation of energy
Total energy at height 10 R = total energy at earth
GM E m 1
GM E m 1
+ mV02 = + mV 2
–
10 R
2
R
2
GMm ù
é
êëQ Gravitational potential energy = – r úû
GM E æ
1 ö V02 V 2
=
çè 1– ÷ø +
R
10
2
2
9
Þ V 2 = V02 + gR
5
9
Þ V = V02 + gR » 16 km / s
5
[Q V0 = 12 km/s given]
GM G (2 M )
GM
Eg =
+
2
2 =
(3a )
(3a )
3a 2
Þ
AC a 2
a
=
=
2
2
2
Resultant force on the body
32. (b) AC = a 2
B=
2a 2
D
r
a
45°
a
C
X
Mv 2
GM 2 æ
1ö
= 2 ç 2+ ÷
2ø
æ a ö
a è
çè
÷ø
2
Q r=
GM 2 ˆ GM 2 ˆ GM 2
i + 2 j+
(cos 45°iˆ + sin 45° ˆj )
a2
a
(a 2)2
1 ö
æ
çè1 +
÷
2 2ø
Þ v2 =
GM
a
Þ v=
GM æ
1 ö
GM
1+
= 1.16
ç
÷
è
ø
a
a
2 2
33. (a) F =
æ m1 ö
1
Þ çè m ÷ø = 6
2
31. (c)
GM 2
Mv 2
= Resultant force towards centre
r
5
R
3
29. (b) Gravitation field at the surface
\ E1 =
a2
( 2) +
O
Þr=
E=
GM 2
r(dV )m
GMm
=òa
r
r2
R
k 4 pr 2 dr
= mG ò 2
r2
0r
R
æ 1ö
= - 4pkGm ç ÷
è rø0
4pkGm
R
Using Newton’s second law, we have
=-
mv02 4pkGm
=
R
R
or v0 = C (const.)
2pR 2pR
Time period, T = v = C
0
or =
T
= constant.
R
34. (c) Initial gravitational potential energy, Ei = –
GMm
2R
Final gravitational potential energy,
GMm / 2 GMm / 2
GMm GMm
–
–
Ef = –
= –
æRö
æ 3R ö
2R
6R
2ç ÷
2ç
÷
2
2
è ø
è
ø
4GMm
2GMm
== –
6R
3R
\ Difference between initial and final energy,
GMm æ 2 1 ö
GMm
Ef – Ei =
ç– + ÷ = –
R è 3 2ø
6R
P-126
35.
Physics
(d) Due to complete solid sphere, potential at point P
-GM é 2 æ R ö
ê3R - ç ÷
Vsphere =
è2ø
2R 3 êë
GM
-GM æ 11R 2 ö
=
ç
÷ = -11
3 ç 4 ÷
8R
2R è
ø
Solid
sphere
By conservation of linear momentum
2ù
v1
m
m
= - 2 Þ v2 = – 1 v1
v2
m1
m2
Putting value of v2 in equation (1), we get
ú
úû
m1v1 + m2 v 2 = 0 or
2
æ m v ö
2Gm1m 2
m1v12 + m 2 çç - 1 1 ÷÷ =
m
d
2 ø
è
2
2 2
m1m 2 v1 + m1 v1 2Gm1m 2
=
m2
d
v1 =
P
Cavity
2Gm 22
2G
= m2
d (m1 + m 2 )
d (m1 + m 2 )
Similarly v 2 = - m1
Due to cavity part potential at point P
GM
3 8
3GM
Vcavity = =2 R
8R
2
So potential at the centre of cavity
= Vsphere - Vcavity
39. (c)
37.
I =-
38.
1
1
Gm 1m 2
m 1 v12 + m 2 v 22 2
2
d
From conservation of energy,
Initial energy = Final energy
Final energy =
\0=
Gm 1 m 2
1
1
m 1 v12 + m 2 v 2 2 2
2
d
1
1
Gm 1 m 2
m 1 v12 + m 1 v 22 =
...(1)
2
2
d
M 7
= M
8 8
Therefore gravitational field due to the left over part of the
sphere
GM ' 7 GM
= 2 =
8 x2
x
(d) The work done to launch the spaceship
M' = M -
dv
dr
= - éë35 + ( -36 ) ùû = 1 J / kg
i.e., change in gravitational potential 1 J/kg.
Hence change in gravitational potential energy 1 J
(b) We choose reference point, infinity, where total
energy of the system is zero.
So, initial energy of the system = 0
or
M
8
Mass of the left over part of the sphere
Þm=
)
y
éx
ù
v = - ê ò I x dx + ò I y dy ú
êë 0
úû
0
= – éë I x .x + I y .y ùû
= – éë5 ( 7 - 0 ) + 12 ( -3 - 0 ) ùû
R
(from figure)
2
M
m
=
4 3 4 æ R ö3
pR
pç ÷
3
3 è2ø
11GM æ 3 GM ö -GM
-ç÷=
8R
R
è 8 R ø
GM
2
2
(c) As, V = – 3 (3R – r )
2R
Graph (c) most closely depicts the correct variation of v(r).
(d) Gravitational field, I = 5iˆ + 12ˆj N/kg
(
Let mass of smaller sphere (which has to be removed) is m
Radius =
=-
36.
2G
d ( m1 + m 2 )
40.
¥ ur uur
¥
GMm
W = - ò F × dr = - ò 2 dr
r
R
R
W =+
GMm
R
… (i)
The force of attraction of the earth on the spaceship, when
it was on the earth's surface
GMm
F=
R2
GM
GMm
g= 2
Þ mg =
… (ii)
2 Þ
R
R
Substituting the value of g in (i) we get
gR 2m
R
Þ W = mgR
Þ W = 1000 × 10 × 6400 × 103
= 6.4 × 1010 Joule
W=
P-127
Gravitation
41.
(c) Let 'M' be the mass of the particle
Now, Einitial = Efinal
GMm
i.e.
2r
+0=
GM m 1
+ MV 2
r
2
1
1 ù
2 GMm é
or, 2 MV = r ê1 ú
2û
ë
1 2 Gm é
1 ù
Þ 2 V = r ê1 ú
2û
ë
or, V =
42.
2Gm æ
1 ö
1ç
÷
r è
2ø
(c) Let P be the point where gravitational field is zero.
Gm
4Gm
\ x 2 = (r - x )2
1
2
Þ =
Þ r – x = 2x
x r-x
P
m
4m
x
r
r
Þ x=
3
46. (d) Gravitational potential energy of mass m in an orbit
of radius R
GMm
R
Energy required = potential energy at 3R – potential energy
a 2R
u= –
=
-GMm æ -GMm ö
-ç
è 2R ÷ø
3R
=
-GMm GMm
+
3R
2R
-2GMm + 3GMm GMm
=
6R
6R
47. (a) By angular momentum conservation
=
rmin vmax = mrmax vmin
vmin
rmin
rmax
planet
Gravitational potential at P,
V =-
43.
44.
vmax
Gm 4Gm
9Gm
=r
2r
r
3
3
(b) Gravitational field, E = –
GM
r2
uuur r
2
Flux, f = ò E g × dS =| E × 4p r |= -4p GM
where, M = mass enclosed in the closed surface
r
1
This relationship is valid when | E g | µ 2 .
r
GMm
(c) Initial P.E. Ui = –
R
When the particle is far away from the sphere, the P.E. of
the system is zero.
\ Uf = 0
é -GMm ù
W = DU = U f - U i = 0 - ê
ë R úû
Given, vmin =
rmin vmin 1
=
=
rmax vmax 6
48. (a) Orbital speed of the body when it revolves very close
to the surface of planet
\
GM
...(i)
R
Here, G = gravitational constant
Escape speed from the surface of planet
V0 =
2GM
R
Dividing (i) by (ii), we have
Ve =
6.67 ´10 -11 ´ 100 10
´
= 6.67 × 10–10 J
0.1
1000
(b) On earth’s surface potential energy,
W=
45.
GmM
R
At a height R from the earth's surface, P.E. of system =
U=
-
GmM
2R
-GmM GmM
;
+
2R
R
GmM
Þ DU =
2R
GM
GM
Now 2 = g ; \
= gR
R
R
vmax
6
V0
=
Ve
49. (c)
GM
1
R
=
2GM
2
R
hv =
3
V
2
Re
\ DU =
1
\ DU = mgR
2
Rmax
Vmin.
...(ii)
P-128
Physics
GM
Re
Orbital velocity, V0 =
–
From energy conversation,
æ 1
GMe m h
1 ö
Þ E1 =
´
Þ E1 = GMe m ç
–
( Re + h) Re
è R e R e + h ÷ø
Gravitational attraction
2
-
GMm 1 æ 3 ö
GMm 1
2
+ mç V÷ =
+ mVmin
Re
Rmax 2
2 è 2 ø
...(1)
FG = mac =
From angular momentum conversation
3
VRe = Vmin Rmax
2
50.
...(2)
mv 2 =
Solving equation (1) and (2) we get,
Rmax = 3Re
(a) According to question, mass density of a spherical
galaxy varies as
k
.
r
ÞM =
ò
0
k
4pr 2 dr
r
M
0
FG =
Þ
G
R
m
4pkR02
= 2pkR2
2
R02
=
mw02 R
MB =
2GM A
RA
MA
R
, RB = A
2
2
\ VB =
(= FC )
4pkR 2
2 = w2 R Þ w =
0
0
R2
2pKG
R
2p ö
æ
çQ w =
÷
T ø
è
2p
2p R
2pR
2pR
=
=
Þ T2 =
w0
KG
KG
2pKG
Q 2p, K and G are constants
\T =
\ T 2 µ R.
GMe m
mv2
=
2
2( Re + h )
E1 = E2
E2 =
where MA and RA be the mass and radius of the planet
A.
According to given problem
R0
GMm
GM e m
(Re + h)
53. (a) Escape velocity of the planet A is VA =
Þ M = 4pk ò r dr
or, M =
mv 2
GM em
=
( R e + h) ( Re + h )2
Clearly, h = 1 Þ h = R e = 3200km
Re 2
2
Mass, M = ò rdV
r = R0
GMe m
GMe m
+ E1 = –
Re
( Re + h )
r
r
51. (d) From law of conservation of momentum, pi = p f
m1u1 + m2u2 = MVf
mv ö
æ
çè mv +
÷
4 ø 5v
Þ vf =
=
3m
6
2
Clearly, vf < vi \ Path will be elliptical
52. (b) K.E. of satellite is zero at earth surface and at height
h from energy conservation
Usurface + E1 = Uh
MA
2 \
RA
2
2G
VA
=
VB
Þ n=4
M
M
R
u
54. (b) R
Þ R
m
2GM A
RA
n
= =1
2GM A 2 4
RA 2
v
1
– GMm 1 2 –GMm
mu 2 +
= mv +
2
R
2
2R
Þ
1
– GMm
m( v 2 – u 2 ) =
2
2R
Þ V = V = u2 –
v0 =
GM
2R
GM
R
\ vrad =
...(i)
m´v
= 10 v
æ mö
çè ÷ø
10
Ejecting a rocket of mass
\
m
10
GM
m
GM
9m
´
= ´ vt Þ Vt2 = 81
10
2 R 10
2R
P-129
Gravitation
Kinetic energy of rocket,
1M 2
VT + Vr2
KE rocket =
2 10
1 m æ 2 GM
GM ö
)100 + 81
= ´ ´ ç (u –
÷
2 10 è
R
R ø
(
=
)
m
GM 81 GM ö
æ
´ 100 ç u 2 –
+
÷
è
20
200 R ø
R
vT
M
vR
9M
10
r
r3
= 2p
GM
GM
(202)3 ´ 1012
6.67 ´ 10 -11 ´ 8 ´ 1022
vc =
=
'
For moon, vc =
Given,
sec
8
prGR 2
3
8
2
prGRm
3
4 3
4 3
R
pR = 64 ´ pRm
or Rm =
3
3
4
'
\ ve =
2
v
8
æ Rö
prG ç ÷ = c
è 4ø
3
4
1 2
mve
v2
v
E
= 2
= 'e2 = e = 16
E' 1
æ ve ö
mv 'e2 vc
çè 4 ÷ø
2
GM
r
Kinetic energy of satellite A,
1
2
2
TA = m A VA
Kinetic energy of satellite B,
1
2
2
TB = m BVB
GM
m´
TA
R
Þ T =
GM = 1
B
2m ´
2R
v0 = g(R + h) » gR
2GrV
R
2GS ´ 4pR3
=
R
1
GM
´
R
2
59. (d) For a satellite orbiting close to the earth, orbital
velocity is given by
T = 7812.2 s
T ; 2.17 hr Þ 11 revolutions.
56. (d) Escape velocity,
2GM
=
R
2
58. (b) Orbital, velocity, v =
Substituting the values, we get
T = 2p
2
v
r
æ vö
æ vö
Þ v= ç ÷ +ç ÷ =
è 2ø
è 2ø
2
=
M
10
2R
119 GM ö
æ
= 5m ç u 2 –
÷
è
200 R ø
55. (d) Time period of revolution of satellite,
2 pr
T=
v
GM
v=
r
\ T = 2p r
E
16
r
57. (b) mviˆ + mvjˆ = 2mv
v
r v
Þ v = ˆi + ˆj
2
2
or E’ =
Escape velocity (ve) is
ve = 2g(R + h) » 2gR
[Q h <<R]
Dv = ve - v0 = ( 2 - 1) gR
60. (b) At height r from center of earth, orbital velocity
GM
r
By principle of energy conservation
æ GMm ö
KE of ‘m’ + çè –
÷ =0+0
r ø
(Q At infinity, PE = KE = 0)
v=
2
GMm æ GM ö
=ç
m = mv2
or KE of ‘m’ =
r
r ÷ø
è
61. (d) Let M is mass of star m is mass of meteroite
By energy convervation between 0 and ¥.
–
GMm –GMm 1
+
+ m Vese2 = 0 + 0
r
r
2
\v=
4GM
=
r
; 2.8 ´105 m / s
4 ´ 6.67 ´ 10 –11 ´ 3 ´ 1031
1011
P-130
Physics
62. (b) For h << R, the orbital velocity is
gR
Escape velocity = 2gR
\ The minimum increase in its orbital velocity
= 2gR – gR = gR ( 2 – 1)
63.
(c) According to universal law of Gravitation,
GMm
Gravitational force F =
(R + h)2
Astronaut
h
67. (c) Gravitational force, F = KR–n
This force provides the centripetal force MRw2 to the
planet at height h above earth’s surface.
\ F = KR–n = MRw2
Þ w2 = KR–(n+1)
Þ
w=
2p
µR
T
- ( n +1)
KR 2
- (n +1)
2
+ ( n +1)
µR 2
\T
68. (c) Time period of satellite is given by
( R + h )3
GM
Where R + h = radius of orbit of satellite
M = mass of earth.
Time period is independent of mass of satellite.
69. (d) Gravitational force provides the necessary centripetal
force.
\ Centripetal force on a satellite = Gravitational force
T = 2p
R
Earth
64.
(a)
F=
mv 2 2GM
2GM
=
m
m or,
r
Lr
Lr
2
65.
2GMm é
æ 2p ö
2p ù
mr ç ÷ =
êQv = r w and w = T ú
T
Lr
ë
û
è ø
Þ T µr
(a) As we know,
-GMm
Gravitational potential energy =
r
and orbital velocity, v0 = GM / R + h
1
GMm 1 GM GMm
Ef = mv02 = m
2
3R
2 3R
3R
GMm æ 1 ö - GMm
=
ç - 1÷ =
3R è 2 ø
6R
-GMm
Ei =
+K
R
Ei = Ef
Therefore minimum required energy, K =
66.
(c) Escape velocity on earth,
ve =
(ve )e
=
Rp
2GM e
Re
=
Mp
Me
´
\
mv 2
n!
æ GM ö R 2
= mç
è R 2 ÷ø ( R + x )2 r !( n - r ) !
( R + x)
\
mv 2
R2
= mg
( R + x)
( R + x) 2
1/2
æ gR 2 ö
gR 2
\v =
÷÷
Þ v = çç
R+ x
èR+xø
5GMm
6R
2GMe
= 11 km s–1
Re
(ve ) p
mv 2
GmM
GM
=
also g = 2
2
( R + x) ( R + x)
R
2
2GM p
\
\
Re
=
Rp
10M e
Re
=
´
= 10
Me
R e /10
\ (ve ) p = 10 ´ (ve )e = 10 ´ 11 = 110 km / s
70. (c) ve = 2 gR
Clearly escape velocity does not depend on the angle at
which the body is projected.
1
2
71. (c) K.E = m ve
2
Here ve = escape velocity is independent of mass of the
body
Escape velocity, ve = 2gR
Substituting value of ve in above equation we get
1
K.E = m ´ 2 gR = mgR
2
72. (b) Due to inertia of motion it will move tangentially to
the original orbit with the same velocity.
73. (a) Escape velocity, ve = 2 gR =
2GM
R
Þ ve µ m0
Where M, R are the mass and radius of the planet
respectively. Clearly, escape velocity is independent of
mass of the body
8
P-131
Mechanical Properties of Solids
Mechanical Properties
of Solids
5.
TOPIC 1 Hooke's Law & Young's Modulus
1.
If the potential energy between two molecules is given by
A
B
U = - 6 + 12 , then at equilibrium, separation between
r
r
molecules, and the potential energy are: [Sep. 06, 2020 (I)]
1
ö6
B
A2
(a) æç ÷ , è 2 Aø
2B
1
A2
æ 2B ö 6
(c) ç ÷ , è Aø
4B
2.
3.
4.
1
Bö 6
(b) æç ÷ , 0
è Aø
[10 April 2019 II]
6.
1
A2
æ 2B ö 6
(d) ç ÷ , è Aø
2B
A body of mass m = 10 kg is attached to one end of a wire of
length 0.3 m. The maximum angular speed (in rad s–1) with
which it can be rotated about its other end in space station is
(Breaking stress of wire = 4.8 × 107 Nm–2 and area of crosssection of the wire = 10–2 cm2) is _______ .
[9 Jan 2020 I]
A uniform cylindrical rod of length L and radius r, is made
from a material whose Young’s modulus of Elasticity
equals Y. When this rod is heated by temperature T and
simultaneously subjected to a net longitudinal
compressional force F, its length remains unchanged. The
coefficient of volume expansion, of the material of the
rod, is (nearly) equal to :
[12 April 2019 II]
(a) 9F/(pr 2YT)
(b) 6F/(pr 2YT
(c) 3F/(pr 2YT)
(d) F/(3pr 2YT)
In an environment, brass and steel wires of length 1 m
each with areas of cross section 1 mm2 are used. The
wires are connected in series and one end of the combined
wire is connected to a rigid support and other end is
subjected to elongation. The stress required to produce a
net elongation of 0.2 mm is,
[Given, the Young’s modulus for steel and brass are,
respectively, 120×109N/m2 and 60×109N/m2]
[10 April 2019 II]
(a) 1.2×106 N/m 2
(b) 4.0×106 N/m 2
(c) 1.8×106 N/m2
(d) 0.2×106 N/m2
The elastic limit of brass is 379 MPa. What should be the
minimum diameter of a brass rod if it is to support a 400
N load without exceeding its elastic limit?
7.
8.
9.
(a) 1.00 mm
(b) 1.16 mm
(c) 0.90 mm
(d) 1.36 mm
A steel wire having a radius of 2.0 mm, carrying a load of
4kg, is hanging from a ceiling. Given that g = 3.1 À ms–2,
what will be the tensile stress that would be developed in
the wire?
[9 April 2019 I]
(a) 6.2 × 106 Nm–2
(b) 5.2 × 106 Nm–2
(c) 3.1 × 106 Nm–2
(d) 4.8 × 106 Nm–2
A steel wire having a radius of 2.0 mm, carrying a load of
4kg, is hanging from a ceiling. Given that g = 3.1 À ms–2,
what will be the tensile stress that would be developed in
the wire?
[8 April 2019 I]
6
–2
6
(a) 6.2 × 10 Nm
(b) 5.2 × 10 Nm–2
6
–2
(c) 3.1 × 10 Nm
(d) 4.8 × 106 Nm–2
Young’s moduli of two wires A and B are in the ratio 7 : 4.
Wire A is 2 m long and has radius R. Wire B is 1.5 m long
and has radius 2 mm. If the two wires stretch by the same
length for a given load, then the value of R is close to :
[8 April 2019 II]
(a) 1.5 mm (b) 1.9 mm (c) 1.7 mm
(d) 1.3 mm
As shown in the figure, forces of 105N each are applied in
opposite directions, on the upper and lower faces of a
cube of side 10cm, shifting the upper face parallel to itself
by 0.5cm. If the side of another cube of the same material
is, 20cm, then under similar conditions as above, the
displacement will be:
[Online April 15, 2018]
F
F
(a) 1.00cm
(c) 0.37cm
(b) 0.25cm
(d) 0.75cm
P-132
10.
11.
12.
Physics
A thin 1 m long rod has a radius of 5 mm. A force of 50 pkN
is applied at one end to determine its Young's modulus.
Assume that the force is exactly known. If the least count
in the measurement of all lengths is 0.01 mm, which of the
following statements is false ? [Online April 10, 2016]
(a) The maximum value of Y that can be determined is
2 × 1014N/m2.
DY
(b)
gets minimum contribution from the uncertainty
Y
in the length
DY
(c)
gets its maximum contribution from the
Y
uncertainty in strain
(d) The figure of merit is the largest for the length of the
rod.
A uniformly tapering conical wire is made from a material
of Young's modulus Y and has a normal, unextended length
L. The radii, at the upper and lower ends of this conical
wire, have values R and 3R, respectively. The upper end of
the wire is fixed to a rigid support and a mass M is
suspended from its lower end. The equilibrium extended
length, of this wire, would equal : [Online April 9, 2016]
(a)
æ 2 Mg ö
L ç1 +
è 9 pYR 2 ÷ø
(b)
æ 1 Mg ö
L ç1 +
è 9 pYR 2 ÷ø
(c)
æ 1 Mg ö
L ç1 +
è 3 pYR 2 ÷ø
(d)
æ 2 Mg ö
L ç1 +
è 3 pYR 2 ÷ø
The pressure that has to be applied to the ends of a steel
wire of length 10 cm to keep its length constant when its
temperature is raised by 100ºC is:
(For steel Young’s modulus is 2 ´ 1011 Nm -2 and
coefficient of thermal expansion is 1.1 ´ 10-5 K -1 ) [2014]
(a)
(b)
2.2 ´ 10 9 Pa
(d) 2.2 ´ 106 Pa
Two blocks of masses m and M are connected by means
of a metal wire of cross-sectional area A passing over a
frictionless fixed pulley as shown in the figure. The system
is then released. If M = 2 m, then the stress produced in
the wire is :
[Online April 25, 2013]
(c)
13.
2.2 ´ 108 Pa
2.2 ´ 10 7 Pa
14. A copper wire of length 1.0 m and a steel wire of length
0.5 m having equal cross-sectional areas are joined end to
end. The composite wire is stretched by a certain load
which stretches the copper wire by 1 mm. If the Young’s
modulii of copper and steel are respectively 1.0 × 1011 Nm–
2 and 2.0 × 1011 Nm–2, the total extension of the composite
wire is :
[Online April 23, 2013]
(a) 1.75 mm (b) 2.0 mm (c) 1.50 mm (d) 1.25 mm
15. A uniform wire (Young’s modulus 2 × 1011 Nm–2) is
subjected to longitudinal tensile stress of 5 × 107 Nm–2. If
the overall volume change in the wire is 0.02%, the
fractional decrease in the radius of the wire is close to :
[Online April 22, 2013]
(a) 1.0 × 10–4
(b) 1.5 × 10–4
(c) 0.25 × 10–4
(d) 5 × 10–4
16. If the ratio of lengths, radii and Young's moduli of steel
and brass wires in the figure are a, b and c respectively,
then the corresponding ratio of increase in their lengths is :
[Online April 9, 2013]
Steel
M
Brass
2M
3a
2ac
2a 2 c
(c)
(d)
2
2b c
b2
b
2ab2
17. A steel wire can sustain 100 kg weight without breaking. If
the wire is cut into two equal parts, each part can sustain
a weight of
[Online May 19, 2012]
(a) 50 kg
(b) 400 kg (c) 100 kg (d) 200 kg
18. A structural steel rod has a radius of 10 mm and length of
1.0 m. A 100 kN force stretches it along its length. Young’s
modulus of structural steel is 2 × 1011 Nm–2. The percentage
strain is about
[Online May 7, 2012]
(a) 0.16% (b) 0.32% (c) 0.08% (d) 0.24%
19. The load versus elongation graphs for four wires of same
length and made of the same material are shown in the
figure. The thinnest wire is represented by the line
[Online May 7, 2012]
Load
D
(a)
3c
(b)
C
B
A
T
O
m
T
M
(a)
2mg
3A
(b)
4mg
3A
(c)
mg
A
(d)
3mg
4A
Elongation
(a) OA
(b) OC
(c) OD
(d) OB
20. Two wires are made of the same material and have the
same volume. However wire 1 has cross-sectional area A
and wire 2 has cross-sectional area 3A. If the length of wire
1 increases by Dx on applying force F, how much force is
needed to stretch wire 2 by the same amount?
[2009]
(a) 4 F
(b) 6 F
(c) 9 F
(d) F
P-133
Mechanical Properties of Solids
21.
a
A wire elongates by l mm when a load W is hanged from it. If
the wire goes over a pulley and two weights W each are hung
at the two ends, the elongation of the wire will be (in mm)
[2006]
(a) l
(b) 2l
(c) zero
(d) l/2
TOPIC 2
Bulk and Rigidity Modulus and
Work Done in Stretching a Wire
22. Two steel wires having same length are suspended from a
ceiling under the same load. If the ratio of their energy
stored per unit volume is 1 : 4, the ratio of their diameters
is:
[9 Jan 2020 II]
(a)
b
2 :1
26.
(b) 1 : 2
(c) 2 : 1
(d) 1 : 2
23. A boy’s catapult is made of rubber cord which is 42 cm
long, with 6 mm diameter of cross-section and of negligible
mass. The boy keeps a stone weighing 0.02 kg on it and
stretches the cord by 20 cm by applying a constant force.
When released, the stone flies off with a velocity of 20 ms–
1. Neglect the change in the area of cross-section of the
cord while stretched. The Young’s modulus of rubber is
closest to :
[8 April 2019 I]
(a) 106 N/m–2
(b) 104 N/m–2
(c) 108 N/m–2
(d) 103 N/m–2
24. A solid sphere of radius r made of a soft material of bulk
modulus K is surrounded by a liquid in a cylindrical
container. A massless piston of area a floats on the surface
of the liquid, covering entire cross-section of cylindrical
container. When a mass m is placed on the surface of the
piston to compress the liquid, the fractional decrement in
æ dr ö
the radius of the sphere çè ÷ø ,is :
[2018]
r
27.
Ka
Ka
mg
mg
(b)
(c)
(d)
mg
3mg
3Ka
Ka
A bottle has an opening of radius a and length b. A cork of
length b and radius (a + Da) where (Da < < a) is compressed
to fit into the opening completely (see figure). If the bulk
modulus of cork is B and frictional coefficient between the
bottle and cork is m then the force needed to push the cork
into the bottle is :
[Online April 10, 2016]
29.
(a)
25.
28.
(a) (pmB b) a
(b) (2pmBb) Da
(c) (pmB b) Da
(d) (4 pmB b) Da
Steel ruptures when a shear of 3.5 × 108 N m–2 is applied.
The force needed to punch a 1 cm diameter hole in a steel
sheet 0.3 cm thick is nearly:
[Online April 12, 2014]
(a) 1.4 × 104 N
(b) 2.7 × 104 N
(c) 3.3 × 104 N
(d) 1.1 × 104 N
The bulk moduli of ethanol, mercury and water are given
as 0.9, 25 and 2.2 respectively in units of 109 Nm–2. For a
given value of pressure, the fractional compression in
DV
volume is
. Which of the following statements about
V
DV
for these three liquids is correct ?[Online April 11, 2014]
V
(a) Ethanol > Water > Mercury
(b) Water > Ethanol > Mercury
(c) Mercury > Ethanol > Water
(d) Ethanol > Mercury > Water
In materials like aluminium and copper, the correct order of
magnitude of various elastic modului is:
[Online April 9, 2014]
(a) Young’s modulus < shear modulus < bulk modulus.
(b) Bulk modulus < shear modulus < Young’s modulus
(c) Shear modulus < Young’s modulus < bulk modulus.
(d) Bulk modulus < Young’s modulus < shear modulus.
If ‘S’ is stress and ‘Y’ is young’s modulus of material of a
wire, the energy stored in the wire per unit volume is [2005]
2
2Y
S
S
(b) 2S 2Y (c)
(d)
2Y
S2
2Y
30. A wire fixed at the upper end stretches by length l by
applying a force F. The work done in stretching is [2004]
(a)
(a) 2Fl
(b) Fl
(c)
F
2l
(d)
Fl
2
P-134
1.
Physics
(c) Given : U =
-A
r
6
+
B
Young modulus, Y =
12
r
For equilibrium,
dU
= -( A( -6r -7 )) + B ( -12r -13 ) = 0
dr
6 A 12 B
6A
1
Þ 0 = 7 - 13 Þ
= 6
12 B r
r
r
Let s be the stress
F=
æ 2B ö
\ Separation between molecules, r = ç ÷
è Aø
Potential energy,
sL1 sL2
Total elongation Dlnet = Y + Y
1
2
1/ 6
é1 1ù
Dlnet = s ê + ú
ë Y1 Y2 û
æ YY ö
s = Dl ç 1 2 ÷
è Y1 + Y2 ø
- A2 A2 - A2
+
=
2B 4B
4B
(4) Given : Wire length, l = 0.3 m
N
æ 120 ´ 60 ö
= 0.2 ´ 10-3 ´ ç
´ 109 = 8 ´ 106 2
è 180 ÷ø
m
Mass of the body, m = 10 kg
5.
(b) Stress =
Breaking stress, s = 4.8 × 107 Nm–2
Þ d2 =
Area of cross-section, a = 10–2 cm2
Maximum angular speed w = ?
T = Mlw2
s=
6.
T ml w
=
A
A
2
(
)
48 ´ 107 A
ml w 2
7
2
£ 48 ´ 10 Þ w £
A
ml
( 48 ´ 10 )(10 ) = 16 Þ w
7
Þ w2 £
3.
-6
max
10 ´ 3
= 4 rad/s
7.
(c) Dtemp = Dforce
r = 3a =
3F
pr 2YT
8.
(Bonus)
9.
Brass
Steel
F
400 ´ 4
F
mg
4 ´ 3.1p
=
=
= 3.1 ´ 106 N/m2
A p (r ) 2 p ´ (2 ´ 10 -3 ) 2
(c) D1 = D2
or
.
F 400 ´ 4
=
= 379 ´106 N/m 2
2
A
pd
379 ´ 106 p
d = 1.15 mm
(c) Given,
Radius of wire, r = 2 mm
Mass of the load m = 4 kg
F
mg
Stress = A =
p(r ) 2
4 ´ 3.1p
=
= 3.1 ´ 106 N/m2
p ´ (2 ´ 10-3 )2
(c) Given,
Radius of wire, r = 2 mm
Mass of the load m = 4 kg
Stress =
FL
F
FL
= 2
\ a=
or La (DT) =
AYT
AY
pr YT
Coefficient of volume expression
4.
[Q L1 = L2 = 1m]
1/ 6
æ
A
B
æ 2B ö ö
U çr = ç ÷ ÷ = + 2 2
è
ø
A
B
A
2
/
4B / A
è
ø
=
2.
Stress
æ Dl ö
çè ÷ø
L
Fl1
pr12 y1
=
Fl2
pr22 y2
or
2
1.5
= 2
R ´7 2 ´4
2
\ R = 1.75 mm
(b) For same material the ratio of stress to strain is same
For first cube
Stress1 =
105
force1
=
area1 (0.12 )
P-135
Mechanical Properties of Solids
change in length1 0.5 ´ 10-2
Strain1 =
=
original length1
0.1
For second block,
force2
105
=
area 2 (0.22 )
strain2 =
change in length 2
x
=
original length 2
0.2
stress1 stress2
=
strain1 strain 2
105
(0.1)2
or,
0.5 ´ 10 -2
0.1
=
(0.2)2
x
0.2
F Dl
/
A l
l
< 0.01 mm
Χl
11.
l
= 2 × 1014 N / m2.
D
pr l
(c) Consider a small element dx of radius r,
r<
The equilibrium extended length of wire = L + DL
æ 1 Mg ÷ö
< L çç1 ∗
÷
èç 3 p YR 2 ÷ø
3p R 2 Y
MgL
stress
strain
stress = Y ´ strain
Stress in steel wire = Applied pressure
Pressure = stress = Y × strain
DL
= α DT
L
(As length is constant)
= 2 × 1011 × 1.1 × 10–5 × 100 = 2.2 × 108 Pa
Strain =
Fl
pr 2Χl
Given, radius r = 5mm, force F = 50pk N,
\ Y=
é 2R
ù2
pê
x ∗ Rú y
êë L
úû
12. (a) Young's modulus Y =
(a) Young's modulus Y =
Y<
Mg dx
é
ù
ê
ú
ê
ú
Mg ê
1
L ú
MgL
ΧL <
≥ ú<
ê,
L
p y ê é 2Rx
2R ú 3p R 2 y
ù
ê ê
ú
∗ Rú
ê êë L
ú
úû 0
ë
û
< L∗
Solving we get, x = 0.25 cm
10.
ò dL < ò
Taking limit from 0 to L
x is the displacement for second block.
10.5
1
0
stress2 =
For same material,
At equilibrium change in length of the wire
F
2
æ 2mM ö
13. (b) Tension in the wire, T = ç
÷g
èm+Mø
Force / Tension
2mM
=
g
Stress =
Area
A(m + M)
=
2R
x∗R
L
2(m ´ 2m)g
(M = 2 m given)
A(m + 2m)
=
R
14. (d)
4m 2
4mg
g=
3mA
3A
Yc ´ ( DLc / Lc ) = Ys ´ (DLs / Ls )
æ 1´ 10-3 ö
11 æ DL ö
Þ 1´1011 ´ ç
÷ = 2 ´10 ´ ç s ÷
ç 1 ÷
è 0.5 ø
è
ø
x
r
dx
0.5 ´10-3
= 0.25 mm
2
Therefore, total extension of the composite wire
\ DLs =
L
= DLc + DLs
= 1 mm + 0.25 m = 1.25 m
3R
Mg
P-136
15.
Physics
19. (a) From the graph, it is clear that for the same value of
load, elongation is maximum for wire OA. Hence OA is the
thinnest wire among the four wires.
(c) Given , y = 2 ´ 1011 Nm -2
æ Fö
Stress ç ÷ = 5 ´ 107 Nm -2
è Aø
l
DV = 0.02% = 2 × 10–4 m3
20. (c)
A
Y
Dr
=?
r
g=
Wire (1)
æ Dl ö
g
stress
Þ strain ç ÷ =
strain
è l 0 ø stress
… (i)
DV = 2prl 0 Dr - pr 2 Dl
and solving we get
Dr
= 0.25 ´10-4
r
(c) According to questions,
Dl b =
prs2 .ys
2Mgl b
prb2 .y b
Y=
[Q Fs = (M + 2M)g]
[Q Fb = 2Mg]
Þ
Stress
Strain
105
pr 2Y
Therefore % strain =
=
For wire 2 ,
...(ii)
21. (a) Case (i)
T
T
T
W
W
Stress
F
Strain =
=
Y
AY
=
...(i)
F l
F'
l
´
=
´
Þ F¢ = 9F
A Dx 3 A 3Dx
(c) Breaking force a area of cross section of wire
Load hold by wire is independent of length of the wire.
(a) Given: F = 100 kN = 105 N
Y = 2 × 1011 Nm–2
l0 = 1.0 m
radius r = 10 mm = 10– 2 m
From formula, Y =
F/A
D x/l
F '/ 3 A
Dx /( l / 3)
From (i) and (ii) we get,
3Mgl s
18.
l
3
Y=
Dl s
pr 2 .y
3a
= s s = 2
\ Dl b 2Mg.l b 2b C
prb2 .y b
17.
For wire 1
Length, L1 = l
Area, A1 = A
For wire 2
Area, A2 = 3A
As the wires are made of same material, so they will have
same young’s modulus.
For wire 1,
Fl
Fl
As, y = ADl Þ Dl = Ay
Dl s =
l/3
Wire (2)
Length, L2 =
ls
r
y
Dls
= a, s = b, s = c,
=?
lb
rb
yb
Dl b
3mgl s
Y
… (ii)
From eqns (i) and (ii) putting the value of Dl , l 0 and DV
16.
3A
At equilibrium, T = W
105
1
-4
11 = 628
3.14 ´ 10 ´ 2 ´ 10
1
´100 = 0.16%
628
Young’s modules, Y =
Elongation, l =
W L
´
A Y
W/A
.....(1)
l/L
W
P-137
Mechanical Properties of Solids
Case (ii) At equilibrium T = W
\ Young’s moduls, Y =
Fractional change in volume
W/A
l/2
L/2
Using eq. (i) & (ii)
W L
W/A
´
Þ l=
A Y
l/ L
Þ Elongation is the same.
Þ
Y=
\
22. (a) If force F acts along the length L of the wire of crosssection A, then energy stored in unit volume of wire
is given by
1
Energy density = stress × strain
2
F
X ö
1 F
F æ
= ´ ´
çèQ stress = and strain =
÷
A
AY ø
2 A AY
3dr mg
=
r
Ka
dr mg
=
(fractional decrement in radius)
r 3Ka
Stress <
N
N
Normal force
< <
A
(2
p
a)b
Area
Stress = B×strain
N
2paΧa ≥ b
<B
(2pa)b
pa 2 b
ÞN<B
(2pa)2 Χab 2
pa 2 b
Force needed to push the cork.
D
26. (c)
If u1 and u2 are the densities of two wires, then
4
d1
d
u1 æ d 2 ö
14
= ( 4) Þ 1 = 2 :1
=ç ÷ Þ
d2
d2
u2 è d1 ø
F
23. (a) When a catapult is stretched up to length l, then the
stored energy in it = Dk. E Þ
1 æ YA ö
1
. ç ÷ ( DI )2 = mv 2
2 è Lø
2
Þy=
0.02 ´ 400 ´ 0.42 ´ 4
p ´ 36 ´ 10 –6 ´ 0.04
D (Dl ) 2
Shearing strain is created along the side surface of the
punched disk. Note that the forces exerted on the disk are
exerted along the circumference of the disk, and the total
force exerted on its center only.
Let us assume that the shearing stress along the side
surface of the disk is uniform, then
F=
= 2.3 × 106 N/m2
(c)
Bulk modulus, K =
K=
Þ
dFmax =
ò
s max dA = s max
surface
ò
dA
surface
æDö
= ò s max .A = smax .2p ç ÷ h
è2ø
volumetric stress
volumetric strain
8 æ1
-2 ö
-2
= 3.5 ´10 ´ ç ´ 10 ÷ ´ 0.3 ´10 ´ 2p
è2
ø
= 3.297 ´ 104 ; 3.3 ´ 10 4 N
mg
æ dV ö
aç
÷
èV ø
1
Bulk modulus
As bulk modulus is least for ethanol (0.9) and maximum for
mercury (25) among ehtanol, mercury and water. Hence
27. (a) Compressibility =
dV mg
=
V
Ka
volume of sphere, V =
ò
surface
So, order is 106.
24.
h
mv 2 L
m = 0.02 kg
v = 20 ms–1
L = 0.42 m
A = (p d2)/(4)
d = 6 × 10–3 m
Dl = 0.2 m
y=
...(ii)
f < m N < m 4pbΧaB = (4pmBb)Da
1 F2
1 F 2 ´ 16 1 F 2 ´ 16
=
=
2 A2Y 2 (pd 2 )2 Y 2 pd 4Y
=
25. (d)
dV 3dr
=
V
r
....(i)
4 3
pR
3
DV
V
Ethanol > Water > Mercury
compression in volume
P-138
28. (c)
Physics
Poisson’s ratio, s =
lateral strain ( b )
longitudinal strain ( a )
For material like copper, s = 0.33
And, Y = 3k (1 – 2 s)
9 1 3
Also,
= +
Y k h
1
´ stress ´ strain
2
We know that,
Y=
stress
strain
1
stress 1 S 2
´ stress ´
= ×
2
Y
2 Y
30. (d) Let A and L be the area and length of the wire.
Work done by constant force in displacing the wire by a
distance l.
= change in potential energy
E=
Y = 2h (1+ s)
Hence, h < Y < k
29. (a) Energy stored in the wire per unit volume,
E=
stress
Y
On substituting the expression of strain in equation (i) we
get
Þ strain =
...(i)
1
= × stress × strain × volume
2
=
1 F l
Fl
´ ´ ´ A´ L =
2 A L
2
9
P-139
Mechanical Properties of Fluids
Mechanical Properties
of Fluids
Pressure, Density, Pascal's Law
TOPIC 1 and Archimedes' Principle
1.
2.
3.
A hollow spherical shell at outer radius R floats just
submerged under the water surface. The inner radius of
the shell is r. If the specific gravity of the shell material is
27
w.r.t water, the value of r is :
[5 Sep. 2020 (I)]
8
8
4
2
1
R
R
R
R
(a)
(b)
(c)
(d)
9
9
3
3
An air bubble of radius 1 cm in water has an upward
acceleration 9.8 cm s–2. The density of water is 1 gm
cm–3 and water offers negligible drag force on the bubble. The
mass of the bubble is (g = 980 cm/s2)
[4 Sep. 2020 (I)]
(a) 4.51 gm (b) 3.15 gm (c) 4.15 gm
(d) 1.52 gm
Two identical cylindrical vessels are kept on the ground
and each contain the same liquid of density d. The area of
the base of both vessels is S but the height of liquid in one
vessel is x1 and in the other, x2. When both cylinders are
connected through a pipe of negligible volume very close
to the bottom, the liquid flows from one vessel to the other
until it comes to equilibrium at a new height. The change in
energy of the system in the process is : [4 Sep. 2020 (II)]
(a) gdS ( x22 + x12 )
(b) gdS ( x2 + x1 )2
3
1
gdS ( x2 - x1 )2
gdS ( x2 - x1 )2
(d)
4
4
A leak proof cylinder of length 1 m, made of a metal which
has very low coefficient of expansion is floating vertically
in water at 0°C such that its height above the water surface
is 20 cm. When the temperature of water is increased to
4°C, the height of the cylinder above the water surface
becomes 21 cm. The density of water at T = 4°C, relative to
the density at T = 0°C is close to:
[8 Jan 2020 (I)]
(a) 1.26
(b) 1.04
(c) 1.01
(d) 1.03
Consider a solid sphere of radius R and mass density
(c)
4.
5.
æ
r2 ö
r(r) = r0 ç 1 - 2 ÷ , 0 < r £ R. The minimum density of a
è R ø
liquid in which it will float is:
[8 Jan 2020 (I)]
r0
r0
2r0
2r0
(a)
(b)
(c)
(d)
3
5
5
3
6.
M
5m
N
5m
O
Two liquids of densities r1 and r2(r2 = 2r1) are filled up
behind a square wall of side 10 m as shown in figure. Each
liquid has a height of 5 m. The ratio of the forces due to these
liquids exerted on upper part MN to that at the lower part NO
is (Assume that the liquids are not mixing): [8 Jan 2020 (II)]
(a) 1/3
(b) 2/3
(c) 1/2
(d) 1/4
7. A cubical block of side 0.5 m floats on water with 30% of
its volume under water. What is the maximum weight that
can be put on the block without fully submerging it under
water? [Take, density of water = 103 kg/m3]
[10 April 2019 (II)]
(a) 46.3 kg (b) 87.5 kg (c) 65.4 kg
(d) 30.1 kg
8. A submarine experiences a pressure of 5.05×106 Pa at depth
of d1 in a sea. When it goes further to a depth of d2, it
experiences a pressure of 8.08×106 Pa. Then d1– d1 is
approximately (density of water=103 kg/m3 and acceleration
due to gravity = 10 ms–2):
[10 April 2019 (II)]
(a) 300 m
(b) 400 m (c) 600 m
(d) 500 m
4
9. A wooden block floating in a bucket of water has of its
5
volume submerged. When certain amount of an oil poured
into the bucket, it is found that the block is just under the
oil surface with half of its volume under water and half in
oil. The density of oil relative to that of water is:
[9 April 2019 (II)]
(a) 0.5
(b) 0.8
(c) 0.6
(c) 0.7
10. A load of mass M kg is suspended from a steel wire of
length 2 m and radius 1.0 mm in Searle’s apparatus
experiment. The increase in length produced in the wire is
4.0 mm. Now the load is fully immersed in a liquid of relative
density 2. The relative density of the material of load is 8.
The new value of increase in length of the steel wire is :
[12 Jan. 2019 (II)]
(a) 3.0 mm (b) 4.0 mm (c) 5.0 mm
(d) Zero
P-140
Physics
11. A soap bubble, blown by a mechanical pump at the mouth
of a tube, increases in volume, with time, at a constant rate.
The graph that correctly depicts the time dependence of
pressure inside the bubble is given by: [12 Jan. 2019 (II)]
P
P
(a)
(b)
1
t
log(t)
P
P
(c)
(d)
13
t
t
12. A liquid of density r is coming out of a hose pipe of radius
a with horizontal speed v and hits a mesh. 50% of the liquid
passes through the mesh unaffected. 25% looses all of its
momentum and 25% comes back with the same speed. The
resultant pressure on the mesh will be: [11 Jan. 2019 (I)]
1 2
3 2
1 2
ρv
ρv (c)
ρv
(b)
(d) ρv 2
4
4
2
13. A thin uniform tube is bent into a circle of radius r in the
vertical plane. Equal volumes of two immiscible liquids, whose
densities are r1 and r1 (r1 > r2) fill half the circle. The
angle q between the radius vector passing through the common
interface and the vertical is
[Online April 15, 2018]
-1 é p æ r - r ö ù
(a) q = tan ê ç 1 2 ÷ ú
ë 2 è r1 + r2 ø û
-1 p æ r1 - r2 ö
(b) q = tan
ç
÷
2 è r1 + r2 ø
(a)
ær ö
(c) q = tan -1 p ç 1 ÷
è r2 ø
(d) None of above
14. There is a circular tube in a vertical plane. Two liquids
which do not mix and of densities d1 and d2 are filled in the
tube. Each liquid subtends 90º angle at centre. Radius
joining their interface makes an angle a with vertical. Ratio
d1
is:
d2
[2014]
a
d1
d2
1 + sin a
1 + cos a
(b)
1 - sin a
1 - cos a
1 + tan a
1 + sin a
(c)
(d)
1 - tan a
1 - cos a
15. A uniform cylinder of length L and mass M having crosssectional area A is suspended, with its length vertical, from
a fixed point by a massless spring such that it is half
submerged in a liquid of density s at equilibrium position.
The extension x0 of the spring when it is in equilibrium
is:
[2013]
(a)
(a)
Mg
k
(b)
Mg æ
LAs ö
ç1 –
÷
k è
M ø
(c)
Mg æ
LAs ö
ç1 –
÷
k è
2M ø
(d)
Mg æ
LAs ö
ç1 +
÷
k è
M ø
16. A ball is made of a material of density r where roil < r < rwater
with roil and rwater represe-nting the densities of oil and
water, respectively. The oil and water are immiscible. If the
above ball is in equilibrium in a mixture of this oil and water,
which of the following pictures represents its equilibrium
position ?
[2010]
water
(a)
oil
oil
(b)
water
water
(c)
oil
(d)
oil
water
17. Two identical charged spheres are suspended by strings
of equal lengths. The strings make an angle of 30°
with each other. When suspended in a liquid of density
0.8g cm–3, the angle remains the same. If density of the
material of the sphere is 1.6 g cm–3 , the dielectric constant
of the liquid is
[2010]
(a) 4
(b) 3
(c) 2
(d) 1
18. A jar is filled with two non-mixing liquids 1 and 2 having
densities r1 and, r2 respectively. A solid ball, made of a
material of density r3 , is dropped in the jar. It comes to
equilibrium in the position shown in the figure.Which of
the following is true for r1, r2and r3?
[2008]
P-141
Mechanical Properties of Fluids
r1
r3
(a) r3 < r1 < r2
(c) r1 < r2 < r3
(b) r1 > r3 > r2
(d) r1 < r3 < r2
Fluid Flow, Reynold's Number
TOPIC 2 and Bernoulli's Principle
19. A fluid is flowing through a horizontal pipe of varying crosssection, with speed v ms–1 at a point where the pressure is
P
P Pascal. At another point where pressure is
Pascal its
2
–1
speed is V ms . If the density of the fluid is r kg m–3 and the
flow is streamline, then V is equal to : [6 Sep. 2020 (II)]
P
2P 2
(a)
(b)
+v
+v
r
r
P
P 2
(c)
(d)
+ v2
+v
2r
r
20. Water flows in a horizontal tube (see figure). The pressure
of water changes by 700 Nm–2 between A and B where the
area of cross section are 40 cm2 and 20 cm2, respectively.
Find the rate of flow of water through the tube.
(density of water = 1000 kgm–3)
[9 Jan. 2020 (I)]
A
B
(a) 3020 cm3/s
(b) 2720 cm3/s
(c) 2420 cm3/s
(d) 1810 cm3/s
21. An ideal fluid flows (laminar flow) through a pipe of nonuniform diameter. The maximum and minimum diameters of
the pipes are 6.4 cm and 4.8 cm, respectively. The ratio of
the minimum and the maximum velocities of fluid in this
pipe is:
[7 Jan. 2020 (II)]
81
9
3
3
(a)
(b)
(c)
(d) 256
16
4
2
22. Water from a tap emerges vertically downwards with an
initial speed of 1.0 ms–1. The cross-sectional area of the
tap is 10–4 m 2. Assume that the pressure is constant
throughout the stream of water and that the flow is streamlined.
The cross-sectional area of the stream, 0.15 m below the tap
would be : [Take g = 10 ms–2)
[10 April 2019 (II)]
(a) 2×10–5 m2
(b) 5×10–5 m2
(c) 5×10–4 m2
(d) 1×10–5 m2
23. Water from a pipe is coming at a rate of 100 liters per
minute. If the radius of the pipe is 5 cm, the Reynolds
number for the flow is of the order of : (density of water =
1000 kg/m3, coefficient of viscosity of water = 1 mPa s)
[8 April 2019 I]
(a) 10 3
(b) 10 4
(c) 10 2
(d)
10 6
24. Water flows into a large tank with flat bottom at the rate
of 10–4 m3 s–(1) Water is also leaking out of a hole of area
1 cm2 at its bottom. If the height of the water in the tank
remains steady, then this height is:
[10 Jan. 2019 I]
(a) 5.1 cm
(b) 7 cm (c) 4 cm
(d)
9 cm
25. The top of a water tank is open to air and its water lavel
is maintained. It is giving out 0.74m3 water per minute
through a circular opening of 2 cm radius in its wall. The
depth of the centre of the opening from the level of water
in the tank is close to:
[9 Jan. 2019 (II)]
(a) 6.0 m
(b) 4.8 m (c) 9.6 m
(d) 2.9 m
26. When an air bubble of radius r rises from the bottom to the
5r
. Taking the
4
atmospheric pressure to be equal to 10m height of water
column, the depth of the lake would approximately be
(ignore the surface tension and the effect of temperature):
[Online April 15, 2018]
(a) 10.5m
(b) 8.7m (c) 11.2m
(d) 9.5m
27. Two tubes of radii r1 and r2, and lengths l1 and l2, respectively, are connected in series and a liquid flows through
each of them in streamline conditions. P1 and P2 are pressure
surface of a lake, its radius becomes
differences across the two tubes. If P2 is 4P1 and l2 is l1 ,
4
then the radius r2 will be equal to :
[Online April 9, 2017]
r1
2
28. Consider a water jar of radius R that has water filled up to
height H and is kept on a stand of height h (see figure).
Through a hole of radius r (r << R) at its bottom, the water
leaks out and the stream of water coming down towards
the ground has a shape like a funnel as shown in the figure.
If the radius of the cross–section of water stream when it
hits the ground is x. Then :
[Online April 9, 2016]
(a) r1
(b) 2r1
(c) 4r1
(d)
R
H
2r
h
2x
1
ö4
æ H
(a) x = r ç
è H + h ÷ø
2
æ H ö
(b) x = r çè
÷
H + hø
1
æ H ö
æ H ö2
(c) x = r ç
(d) x = r ç
è H + h ÷ø
è H + h ÷ø
29. If it takes 5 minutes to fill a 15 litre bucket from a water tap
of diameter
2
cm then the Reynolds number for the
p
flow is (density of water = 103 kg/m3) and viscosity of
water = 10–3 Pa.s) close to :
[Online April 10, 2015]
(a) 1100
(b) 11,000 (c) 550
(d) 5500
P-142
Physics
30. An open glass tube is immersed in mercury in such a way
that a length of 8 cm extends above the mercury level. The
open end of the tube is then closed and sealed and the tube
is raised vertically up by additional 46 cm. What will be
length of the air column above mercury in the tube now?
(Atmospheric pressure = 76 cm of Hg)
[2014]
(a) 16 cm
(b) 22 cm (c) 38 cm
(d) 6 cm
31. In the diagram shown, the difference in the two tubes of
the manometer is 5 cm, the cross section of the tube at A
and B is 6 mm2 and 10 mm2 respectively. The rate at which
water flows through the tube is (g = 10 ms–2)
[Online April 19, 2014]
5 cm
(a) Depends on H
(b) 1 : 1
(c) 2 : 2
(d) 1 : 2
36. Water is flowing through a horizontal tube having crosssectional areas of its two ends being A and A¢ such that the
ratio A/A¢ is 5. If the pressure difference of water between
the two ends is 3 × 105 N m–2, the velocity of water with
which it enters the tube will be (neglect gravity effects)
[Online May 12, 2012]
(a) 5 m s–1
(b) 10 m s–1
(c) 25 m s–1
(d) 50 10 m s–1
37. A square hole of side length l is made at a depth of h and
a circular hole of radius r is made at a depth of 4h from the
surface of water in a water tank kept on a horizontal surface.
If l << h, r << h and the rate of water flow from the holes is
the same, then r is equal to
[May 7, 2012]
B
A
h
4h
(a) 7.5 cc/s (b) 8.0 cc/s (c) 10.0 cc/s (d) 12.5 cc/s
32. A cylindrical vessel of cross-section A contains water to a
height h. There is a hole in the bottom of radius ‘a’. The
time in which it will be emptied is: [Online April 12, 2014]
2A h
2A h
(a)
(b)
2 g
pa
pa 2 g
2 2A h
A
h
(d)
2
2 g
g
pa
2pa
33. Water is flowing at a speed of 1.5 ms–1 through horizontal
tube of cross-sectional area 10–2 m2 and you are trying to
stop the flow by your palm. Assuming that the water stops
immediately after hitting the palm, the minimum force
that you must exert should be
(density of water = 103 kgm–3) [Online April 9, 2014]
(a) 15 N
(b) 22.5 N (c) 33.7 N
(d) 45 N
34. Air of density 1.2 kg m–3 is blowing across the horizontal
wings of an aeroplane in such a way that its speeds above
and below the wings are 150 ms –1 and 100 ms –1 ,
respectively. The pressure difference between the upper
and lower sides of the wings, is : [Online April 22, 2013]
(a) 60 Nm–2
(b) 180 Nm–2
(c) 7500 Nm–2
(d) 12500 Nm–2
35. In a cylindrical water tank, there are two small holes A and
B on the wall at a depth of h1, from the surface of water and at
a height of h2 from the bottom of water tank. Surface of water
is at heigh H from the bottom of water tank. Water coming out
from both holes strikes the ground at the same point S. Find
the ratio of h1 and h2
[Online May 26, 2012]
(c)
h1
A
H
B
h2
S
A
v1
B
v2
l
l
l
l
(b)
(c)
(d)
2p
3p
3p
2p
38. Water is flowing continuously from a tap having an internal
diameter 8 × 10–3 m. The water velocity as it leaves the tap
is 0.4 ms–1. The diameter of the water stream at a distance 2
× 10–1 m below the tap is close to:
[2011]
(a) 7.5 × 10–3 m
(b) 9.6 × 10–3 m
(c) 3.6 × 10–3 m
(d) 5.0 × 10–3 m
39. A cylinder of height 20 m is completely filled with water.
The velocity of efflux of water (in ms-1) through a small hole
on the side wall of the cylinder near its bottom is [2002]
(a) 10
(b) 20
(c) 25.5
(d) 5
(a)
TOPIC 3 Viscosity and Terminal Velocity
40. In an experiment to verify Stokes law, a small spherical ball of
radius r and density r falls under gravity through a distance h
in air before entering a tank of water. If the terminal velocity of
the ball inside water is same as its velocity just before entering
the water surface, then the value of h is proportional to :
(ignore viscosity of air)
[5 Sep. 2020 (II)]
4
3
(a) r
(b) r
(c) r (d)
r2
41. A solid sphere, of radius R acquires a terminal velocity
v1 when falling (due to gravity) through a viscous fluid
having a coefficient of viscosity h. The sphere is broken
into 27 identical solid spheres. If each of these spheres
acquires a terminal velocity, v2, when falling through the
same fluid, the ratio (v1/v2) equals: [12 April 2019 (II)]
(a) 9
(b) 1/27 (c) 1/9
(d) 27
P-143
Mechanical Properties of Fluids
42. Which of the following option correctly describes the
variation of the speed v and acceleration 'a' of a point mass
falling vertically in a viscous medium that applies a force
F = –kv, where 'k' is a constant, on the body? (Graphs are
schematic and not drawn to scale)
[Online April 9, 2016]
a
(a)
v
t
v
(b)
a
47. The terminal velocity of a small sphere of radius a in a
viscous liquid is proportional to [Online May 26, 2012]
(a) a 2
(b) a 3
(c) a
(d) a–1
–3
48. If a ball of steel (density r = 7.8 g cm ) attains a terminal
velocity of 10 cm s–1 when falling in water (Coefficient
of viscosity hwater = 8.5 × 10–4 Pa.s), then, its terminal
velocity in glycerine (r = 1.2 g cm–3, h = 13.2 Pa.s) would
be, nearly
[2011 RS]
(a) 6.25 × 10–4 cm s–1
(b) 6.45 × 10–4 cm s–1
(c) 1.5 × 10–5 cm s–1
(d) 1.6 × 10–5 cm s–1
49. A spherical solid ball of volume V is made of a material
of density r1. It is falling through a liquid of density
r1 (r2< r1). Assume that the liquid applies a viscous force
on the ball that is proportional to the square of its speed
v, i.e., Fviscous = –kv2 (k > 0). The terminal speed of the ball
is
[2008]
(a)
t
v
Vg (r1 – r2 )
k
(b)
Vg r1
k
Vgr1
Vg (r1 – r2 )
(d)
k
k
50. If the terminal speed of a sphere of gold (density
= 19.5 kg/m3) is 0.2 m/s in a viscous liquid (density = 1.5
kg/m 3), find the terminal speed of a sphere of silver
(density = 10.5 kg/m3) of the same size in the same liquid
[2006]
(a) 0.4 m/s
(b) 0.133 m/s
(c) 0.1 m/s
(d) 0.2 m/s
51. Spherical balls of radius ‘R’ are falling in a viscous fluid
of viscosity ‘h’ with a velocity ‘v’. The retarding viscous
force acting on the spherical ball is
[2004]
(a) inversely proportional to both radius ‘R’ and velocity ‘v’
(b) directly proportional to both radius ‘R’ and velocity ‘v’
(c) directly proportional to ‘R’ but inversely proportional
to ‘v’
(d) inversely proportional to ‘R’ but directly proportional
to velocity ‘v’
(c)
(c)
a
t
v
(d)
a
t
43. The velocity of water in a river is 18 km/hr near the surface.
If the river is 5 m deep, find the shearing stress between the
horizontal layers of water. The co-efficient of viscosity of
water = 10–2 poise.
[Online April 19, 2014]
(a) 10–1 N/m2
(b) 10–2 N/m2
(c) 10–3 N/m2
(d) 10–4 N/m2
44. The average mass of rain drops is 3.0 × 10–5 kg and their
avarage terminal velocity is 9 m/s. Calculate the energy
transferred by rain to each square metre of the surface at a
place which receives 100 cm of rain in a year.
[Online April 11, 2014]
(a) 3.5 × 105 J
(b) 4.05 × 104 J
(c) 3.0 × 105 J
(d) 9.0 × 104 J
45. A tank with a small hole at the bottom has been filled with
water and kerosene (specific gravity 0.8). The height of
water is 3m and that of kerosene 2m. When the hole is
opened the velocity of fluid coming out from it is nearly:
(take g = 10 ms–2 and density of water = 103 kg m–3)
[Online April 11, 2014]
(a) 10.7 ms–1
(b) 9.6 ms–1
(c) 8.5 ms–1
(d) 7.6 ms–1
46. In an experiment, a small steel ball falls through a liquid at
a constant speed of 10 cm/s. If the steel ball is pulled upward
with a force equal to twice its effective weight, how fast
will it move upward ?
[Online April 25, 2013]
(a) 5 cm/s (b) Zero (c) 10 cm/s
(d) 20 cm/s
TOPIC 4
Surface Tension, Surface
Energy and Capillarity
52. When a long glass capillary tube of radius 0.015 cm is
dipped in a liquid, the liquid rises to a height of 15 cm
within it. If the contact angle between the liquid and glass
to close to 0°, the surface tension of the liquid, in
milliNewton m–1, is [r(liquid) = 900 kgm–3, g = 10 ms–2]
(Give answer in closest integer) __________.
[NA 3 Sep. 2020 (I)]
53. Pressure inside two soap bubbles are 1.01 and 1.02
atmosphere, respectively. The ratio of their volumes is :
[3 Sep. 2020 (I)]
(a) 4 : 1
(b) 0.8 : 1 (c) 8 : 1
(d) 2 : 1
54. A capillary tube made of glass of radius 0.15 mm is dipped
vertically in a beaker filled with methylene iodide (surface
tension = 0.05 Nm–1, density = 667 kg m–3) which rises
P-144
Physics
to height h in the tube. It is observed that the two tangents
drawn from liquid-glass interfaces (from opp. sides of
the capillary) make an angle of 60° with one another. Then
h is close to (g = 10 ms–2).
[2 Sep. 2020 (II)]
(a) 0.049 m
(b) 0.087 m
(c) 0.137 m
(d) 0.172 m
55. A small spherical droplet of density d is floating exactly
half immersed in a liquid of density r and surface tension
T. The radius of the droplet is (take note that the surface
tension applies an upward force on the droplet):
[9 Jan. 2020 (II)]
(a) r =
2T
3(d + r) g
(c) r =
T
(d + r) g
(b) r =
(d) r =
T
(d - r) g
3T
(2d - r) g
56. The ratio of surface tensions of mercury and water is
given to be 7.5 while the ratio of their densities is 13.6.
Their contact angles, with glass, are close to 135o and 0o,
respectively. It is observed that mercury gets depressed
by an amount h in a capillary tube of radius r1, while water
rises by the same amount h in a capillary tube of radius
r2. The ratio, (r1/r2), is then close to :
[10 April 2019 (I)]
(a) 4/5
(b) 2/5
(c) 3/5
(d) 2/3
57. If ‘M’ is the mass of water that rises in a capillary tube of
radius ‘r’, then mass of water which will rise in a capillary
tube of radius ‘2r’ is :
[9 April 2019 I]
M
(a) M
(b)
(c) 4 M
(d)
2M
2
58. If two glass plates have water between them and are
separated by very small distance (see figure), it is very
difficult to pull them apart. It is because the water in
between forms cylindrical surface on the side that gives
rise to lower pressure in the water in comparison to
atmosphere. If the radius of the cylindrical surface is R
and surface tension of water is T then the pressure in water
between the plates is lower by : [Online April 10, 2015]
Cylindrical surface
of water
2T
4T
T
T
(a)
(b)
(c)
(d)
R
R
4R
R
59. On heating water, bubbles being formed at the bottom of
the vessel detach and rise. Take the bubbles to be spheres
of radius R and making a circular contact of radius r with
the bottom of the vessel. If r << R and the surface tension
of water is T, value of r just before bubbles detach is:
(density of water is rw)
[2014]
R
2r
2
(a) R
2 rwg
3T
2
(b) R
rw g
6T
rw g
2 3r w g
(d) R
T
T
60. A large number of liquid drops each of radius r coalesce
to from a single drop of radius R. The energy released in
the process is converted into kinetic energy of the big
drop so formed. The speed of the big drop is (given,
surface tension of liquid T, density r)
[Online April 19, 2014, 2012]
2
(c) R
(a)
T æ1 1 ö
r çè r R ÷ø
(b)
2T æ 1 1 ö
r çè r R ÷ø
4T æ 1 1 ö
6T æ 1 1 ö
- ÷
(d)
ç
r èr Rø
r çè r R ÷ø
61. Two soap bubbles coalesce to form a single bubble. If V
is the subsequent change in volume of contained air and S
change in total surface area, T is the surface tension and P
atmospheric pressure, then which of the following relation
is correct?
[Online April 12, 2014]
(a) 4PV + 3ST = 0
(b) 3PV + 4ST = 0
(c) 2PV + 3ST = 0
(d) 3PV + 2ST = 0
62. An air bubble of radius 0.1 cm is in a liquid having surface
tension 0.06 N/m and density 103 kg/m3. The pressure inside
the bubble is 1100 Nm–2 greater than the atmospheric
pressure. At what depth is the bubble below the surface of
the liquid? (g = 9.8 ms–2)
[Online April 11, 2014]
(a) 0.1 m
(b) 0.15 m (c) 0.20 m
(d) 0.25 m
63. A capillary tube is immersed vertically in water and the height
of the water column is x. When this arrangement is taken into
a mine of depth d, the height of the water column is y. If R is
(c)
x
the radius of earth, the ratio y is:
[Online April 9, 2014]
dö
æ
æ 2d ö
(a) ç 1 - ÷
(b) ç 1 - ÷
R
Rø
è
ø
è
æ R -d ö
æR+dö
(c) ç
(d) ç
÷
÷
èR+dø
è R -d ø
64. Wax is coated on the inner wall of a capillary tube and the
tube is then dipped in water. Then, compared to the unwaxed
capillary, the angle of contact q and the height h upto which
water rises change. These changes are :
[Online April 23, 2013]
P-145
Mechanical Properties of Fluids
(a) q increases and h also increases
(b) q decreases and h also decreases
(c) q increases and h decreases
(d) q decreases and h increases
65. A thin tube sealed at both ends is 100 cm long. It lies
horizontally, the middle 20 cm containing mercury and two
equal ends containing air at standard atmospheric pressure. If
the tube is now turned to a vertical position, by what amount
will the mercury be displaced ?
[Online April 23, 2013]
l0
69. Two mercury drops (each of radius ‘r’) merge to form
bigger drop. The surface energy of the bigger drop, if T is
the surface tension, is :
[2011 RS]
(a) 4pr 2T
(c) 28 / 3 pr 2T
(d) 25/ 3 pr 2T
70. A capillary tube (A) is dipped in water. Another identical
tube (B) is dipped in a soap-water solution. Which of the
following shows the relative nature of the liquid columns
in the two tubes?
[2008]
l0
20 cm
100 cm
(Given : cross-section of the tube can be assumed to be
uniform)
(a) 2.95 cm (b) 5.18 cm (c) 8.65 cm
(d) 0.0 cm
66. This question has Statement-1 and Statement-2. Of the four
choices given after the Statements, choose the one that
best describes the two Statetnents.
Statement-1: A capillary is dipped in a liquid and liquid
rises to a height h in it. As the temperature of the liquid is
raised, the height h increases (if the density of the liquid
and the angle of contact remain the same).
Statement-2: Surface tension of a liquid decreases with
the rise in its temperature.
[Online April 9, 2013]
(a) Statement-1 is true, Statement-2 is true; Statement-2 is
not the correct explanation for Statement-1.
(b) Statement-1 is false, Statement-2 is true.
(c) Statement-1 is true, Statement-2 is false.
(d) Statement-1 is true, Statement-2 is true; Statement-2 is
the correct explanation for Statement-1.
67. A thin liquid film formed between a U-shaped wire and a
light slider supports a weight of 1.5 × 10–2 N (see figure).
The length of the slider is 30 cm and its weight is negligible.
The surface tension of the liquid film is
[2012]
FILM
z
W
(a) 0.0125 Nm–1
(b) 0.1 Nm–1
(c) 0.05 Nm–1
(d) 0.025 Nm–1
68. Work done in increasing the size of a soap bubble from a
radius of 3 cm to 5 cm is nearly (Surface tension of soap
solution = 0.03 Nm–1)
[2011]
(a) 0.2 p mJ (b) 2p mJ (c) 0.4p mJ
(d) 4p mJ
(b) 2pr 2T
A
B
A
B
A
B
A
B
(a)
(b)
(c)
(d)
71. A 20 cm long capillary tube is dipped in water. The water
rises up to 8 cm. If the entire arrangement is put in a freely
falling elevator the length of water column in the capillary
tube will be
[2005]
(a) 10 cm
(b) 8 cm (c) 20 cm
(d) 4 cm
72. If two soap bubbles of different radii are connected by a
tube
[2004]
(a) air flows from the smaller bubble to the bigger
(b) air flows from bigger bubble to the smaller bubble till
the sizes are interchanged
(c) air flows from the bigger bubble to the smaller bubble
till the sizes become equal
(d) there is no flow of air.
P-146
1.
Physics
(a) In equilibrium, mg = Fe
FB = V r0 g and mass = volume × density
By volume conservation,
Sx1 + Sx2 = S (2 x f )
4
4
p( R 3 - r 3 )r0 g = pR3rw g
3
3
Given, relative density,
x1 + x2
2
When valve is opened loss in potentail energy occur till
water level become same.
xf =
r0 27
=
rw
8
é æ r ö 3 ù 27
Þ ê1 - ç ÷ ú rw = rw
è Rø ú 8
ëê
û
DU = U i - U f
éæ x 2 x 2 ö
ù
DU = rSg êç 1 + 2 ÷ - x 2f ú
2 ø
ëêè 2
ûú
r3
9
1 r3
2 r3
Þ 1- 3 =
Þ 1- = 3 Þ = 3
27
3 R
3 R
R
Þ
r3
1/3
Þ1-
r3
R3
=
é x2 x2 æ x + x ö 2 ù
= rSg ê 1 + 2 - ç 1 2 ÷ ú
2 è 2 ø úû
êë 2
8
27
8 19
=
27 27
R
8
\ r = 0.89 R = R.
9
(c) Given :
Radius of air bubble = 1 cm,
Upward acceleration of bubble, a = 9.8 cm/s2,
Þ
2.
r æ 2ö
=ç ÷
R è 3ø
3
= 1-
rwater = 1 g cm–3
Volume V =
4p 3 4p
r =
´ (1)3 = 4.19 cm3
3
3
Fbuoyant - mg = ma Þ m =
\m =
3.
Fbuoyant
=
4.
20 cm
Fbuoyant
0°C
a
mg
g +a
= A(100 – 21)r4°c g
21 cm
4°C
x2
Initial potential energy,
x1
x
+ (rSx2 ) g × 2
2
2
Final potential energy,
U f = (rSx f ) g ×
2
´2
r4°c 80
=
= 1.01
r0°c 79
(c) For minimum density of liquid, solid sphere has to
float (completely immersed) in the liquid.
mg = FB (also Vimmersed = Vtotal)
\
xf xf
xf
100 cm
When cylinder is floating in water at 4° C
Net thrust = A(h2 – h1 )r4°c g
(d)
U1 = (rSx1 ) g ×
(c) When cylinder is floating in water at 0°C
Net thrust = A(h2 – h1 )r0°c g
= A(100 – 80)r0°c g
(V rw g ) V rw (4.19) ´ 1 4.19
=
=
=
= 4.15g
a
9.8
g+a
1.01
1+
1+
g
980
x1
ù
rSg é x12 x22
- x1 x2 ú = rSg ( x1 - x2 )2
ê +
2 êë 2
2
úû
4
5.
or ò rdV =
4 3
pR r
l
3
é
ù
æ
r2 ö
(
r
)
1
–
0 < r £ R given ú
r
=
r
ê
0ç
2 ÷
úû
è R ø
ëê
R
æ
r2 ö
4
Þ ò r0 4 p ç 1 – 2 ÷ . r 2 dr = pR 3r l
3
R ø
0
è
P-147
Mechanical Properties of Fluids
R
T' fB
é r3
r5 ù
4
Þ 4pr0 ê – 2 ú = pR3rl
ëê 3 5R ûú 0 3
T
4pr0 R3 2 4 3
´ = pR rl
3
5 3
\ rl =
M
6.
Þ Dl µ F
T = Mg
2r0
5
(d)
r
N
2r
( P2 + P3 )
A
2
F1
5rg
5 1
=
=
=
F2 20rg 20 4
(b) When a body floats then the weight of the body = upthrust
30
\ (50)3 ´
´ (1) ´ g = M cube g
...(i)
100
Let m mass should be placed, then
(50)3 × (1) × g = (Mcube + m)g ...(ii)
Subtracting equation (i) from equation (ii), we get
Þ mg = (50)3 × g (1 – 0.3) = 125 × 0.7 × 103 g
Þ m = 87.5 kg
(a) P1 = P0 + rgd1
P2 = P0 + rgd2
DP = P2 – P1 = rgDd
3.03 × 106 = 103 × 10 × Dd
Þ Dd ; 300 m
\
8.
9.
T = Mg – fB = Mg – r ×rl × g
b
P1 = 0
F1
P2=5rg
F2
P2 = 15 rg
(P + P )
Force on upper part, F1 = 1 2 A
2
7.
...(i)
M
O
Let P1, P2 and P3 be the pressure at points M, N and O
respectively.
Pressure is given by P = rgh
Now, P1 = 0 (Q h = 0)
P2 = rg(5)
P3 = rg(15)
= 15 rg
Force on lower part, F2 =
Mg
Mg
æ 4V ö
(c) Mg = ç ÷ rw g
è 5 ø
æ M ö 4 rw
4rw
or çè ÷ø =
or r =
V
5
5
When block floats fully in water and oil, then
Mg = Fb1 + Fb2
æ
r ö
è
bø
æ
è
l
= ç1 – r ÷ Mg
T=
2ö
= ç1 – ÷ Mg
8
ø
3
Mg
4
From eqn (i)
Dl¢
T¢
3
=
=
[Given: Dl = 4 mm]
Dl
T
4
\ Dl¢ =
11.
3
3
× Dl = ´ 4 = 3 mm
4
4
(d)
4
V = ct or, pr 3 = ct
3
1
Þ r = kt 3
4T
P = P0 + k t1/3
æ 1 ö
P = P0 + c ç 1/3 ÷
èt ø
12. (b) Mass per unit time of the liquid = rav
Momentum per second carried by liquid
= rav × v
Net force due to bounced back liquid,
é1
ù
F1 = 2 ´ ê rav 2 ú
ë4
û
1
2
Net force due to stopped liquid, F2 = rav
4
Total force,
1
1
3
2
2
2
F = F1 + F2 = rav + rav = rav
2
4
4
3 2
Net pressure = rv
4
13. (d) Pressure at interface A must be same from both the
sides to be in equilibrium.
V
æV ö
( rV ) g = ç ÷ roil g + rw g
è 2ø
2
or roil =
10.
3
rw = 0.6rw
5
(a) Using
F
Dl
= Y×
A
l
d2
R
q
q Rcosq
Rsina – Rsin q
d1
Rsinq
R
A
\ (R cos q + R sin q)r2 g = (R cos q - R sin q)r1g
P-148
Physics
Fe ' = T 'sin 30°
d1 cos q + sin q 1 + tan q
Þ d = cos q - sin q = 1 - tan q
2
Þ r1 – r1 tan q = r2 + r2 tan q
Þ (r1 + r2) tan q = r1 – r2
mg = FB + T 'cos 30°
But FB = Buoyant force
ær –r ö
\ q = tan –1 ç 1 2 ÷
è r1 + r2 ø
14. (c) Pressure at interface A must be same from both the
sides to be in equilibrium.
d2
R
a
\ ( R cos a + R sin a )d2 g = ( R cos a - R sin a )d1 g
d1 cos a + sin a 1 + tan a
Þ d = cos a - sin a = 1 - tan a
2
15.
(c) From figure, kx 0 + FB = Mg
KX0
FB
Mg
L
kx 0 + s Ag = Mg
2
[Q mass = density × volume]
Þ kx 0 = Mg - s
L
Ag
2
sLAg
2 = Mg æ1 - LAs ö
Þ x0 =
ç
÷
k è
2M ø
k
Hence, extension of the spring when it is in equilibrium is,
Mg -
Mg æ LAs ö
ç1 ÷
k è
2M ø
16. (b) Oil will float on water so, (b) or (d) is the correct option.
But density of ball is more than that of oil,, hence it will
sink in oil.
x0 =
17.
(c)
T 30°
T cos 30°
Fe
T sin 30°
mg
Fe = T sin 30°
mg = T cos30°
Þ tan 30° =
In liquid ,
Fe
mg
30°
T¢ cos 30°
F¢ e
T¢ sin 30°
mg
m
0.8 mg mg
g=
=
d
1.6
2
mg
+ T ' cos 30°
\ mg =
2
mg
= T 'cos 30°
...(B)
Þ
2
= 0.8
From (A) and (B), tan 30° =
2 Fe'
mg
From (1) and (2)
Fe 2 Fe'
=
mg mg
(2)
Þ Fe = 2Fe'
If K be the dielectric constant, then
Fe' =
Fe
K
2 Fe
ÞK=2
K
18. (d) As liquid 1 floats over liquid 2. The lighter liquid
floats over heavier liquid. So, r1 < r2
Also r3 < r2 because the ball of density r3 does not sink
to the bottom of the jar.
Also r3 > r1 otherwise the ball would have floated in
liquid 1. we conclude that
r1 < r3 < r2.
\ Fe =
19. (d) Using Bernoulli's equation
1
1
P1 + ru12 + rgh1 = P2 + ru 22 + rgh2
2
2
For horizontal pipe, h1 = 0 and h2 = 0 and taking
P
P1 = P, P2 = , we get
2
1 2 P 1
Þ P + ru = + rV 2
2
2 2
Þ
...(1)
T¢
= V (d - r) g = V (1.6 - 0.8) g = 0.8 Vg
A
d1
FB
Rsina
R
a Rcosa
Rsina – Rsin a
...(A)
P 1 2 1
+ ru = rV 2
2 2
2
Þ V = u2 +
P
r
P-149
Mechanical Properties of Fluids
20. (b) According to question, area of cross-section at A,
aA = 40 cm2 and at B, aB = 20 cm2
Let velocity of liquid flow at A, = VA and at B, = VB
Using equation of continuity aAVA = aBVB
40VA = 20VB
Þ 2VA = VB
Now, using Bernoulli’s equation
(
21.
22.
v1 A2 p(4.8) 2
9
=
=
=
Þ
2
16
v2
A1 p(6.4)
(b) Using Bernoullie’s equation
-3
V
5 ´ 10
=
A 3 ´ ´(5 ´ 10-2 )2
10
2
=
m/s
15p 3p
= 0.2 m/s
=
Dvr
h
2
´ 1000
3p
= 2 ´ 104
1
Order of NR = 104
–4
2gh
20 ´ h
1
m = 5 cm
20
740 ´ 740
(Qp2 =10)
24 ´ 24 ´10
4
4 125r 3
( P1 ) pr 3 = ( P2 ) p
3
3
64
5
´ ´ 10 -3 m3
3
\ Velocity of flow of water (v) =
=
\ 10 = 10 ×
–4
74 ´ 74
» 4.8m
2 ´ 24 ´ 24
i.e., The depth of the centre of the opening from the
level of water in the tank is close to 4.8 m
26. (d) Using P1V1 = P2V2
10-4
= 5 × 10–5 m2
2
(b) Rate of flow of water (V) = 100 lit/min
(10 ´ 10-2 ) ´
Qout = Au = 10–4 ×
Þh =
\ A2 =
\ Reynold number (NR) =
Since height of water column is constant therefore, water
inflow rate (Qin)
= water outflow rate
Qin = 10–4 m3s–1
Þ 2gh =
æ
15 ö
2
÷
(1 cm2) (1 m/s) = (A2) ç (1) + 2 ´ 10 ´
100 ø
è
10–4 × 1 = A2 × 2
100 ´ 10
60
Qout
25. (b) Here, volume tric flow rate
0.74
=
= p r 2 v = ( p´ 4 ´10 –4 ) ´ 2gh
60
74 ´100
74 0
Þ 2gh =
Þ 2gh =
240 p
24p
Þ v2 = v12 + 2 gh
Equation of continuity
A1v1 = A2v2
=
h
\h =
Þ v22 = v12 + 2 gh
-3
Q in
)
1
1
1
PA + rVA2 = PB + rVB2 Þ PA - PB = r VB2 - VA2
2
2
2
2ö
æ
1
V
3V 2
Þ DP = 1000 ç VB2 - B ÷ Þ DP = 500 ´ B
ç
2
4 ÷ø
4
è
( DP ) ´ 4 ( 700) ´ 4
m/s = 1.37 ´ 102 cm/s
Þ VB =
=
1500
1500
Volume flow rate Q = aB × vB
= 20 × 100 × VB = 2732 cm3/s » 2720 cm3/s
(a) From the equation of continuity
A1v1= A2v2
Here, v1 and v2 are the velocities at two ends of pipe.
A1 and A2 are the area of pipe at two ends
1
P + (v12 - v22 ) + rgh = P
2
23.
24. (a)
rg (10) + rgh 125
=
rg (10)
64
640 + 64 h = 1250
On solving we get h = 9.5 m
27. (d) The volume of liquid flowing through both the tubes
i.e., rate of flow of liquid is same.
Therefore, V = V1= V2
i.e.,
or
Q
pP1r14 pP2 r24
=
8hl1
8hl2
P1r14 P2r24
=
l1
l2
P2 = 4 P1 and l2 = l1/4
P-150
Physics
P1r14 4P1r24
r4
=
Þ r24 = 1
l1
l1 4
16
28.
r2 = r1 2
\ v 22 - v12 = 2gh
...(1)
According to the equation of continuty
A1v1 = A 2 v2
...(2)
1 2
1
r v1 ∗ r gh < r v22
2
2
A1
6 mm2
=
A 2 10 mm 2
v12 ∗ 2gh < v22
From equation (2),
(a) According to Bernoulli's Principle,
2gH ∗ 2gh < v22
a1v1 = a2v2
...(i)
6
v1
10
Putting this value of v2 in equation (1)
or, v 2 =
pr 2 2gh < p x 2 v 2
r2
2
2
2gh < v 2
æ6 ö
2
3
ç v1 ÷ - ( v1 ) = 2 ´10 ´ 5
è 10 ø
x
Substituting the value of v2 in equation (i)
2gH ∗ 2gh <
29.
r
4
x4
\ Radius, r =
1
ù4
éQ g = 10m / s 2 = 103 cm / s 2 ù
ê
ú
ëêand h = 5 cm
ûú
2
Solving we get v1 =
é
2gh or, x < r ê H ú
êë H ∗ h úû
(d) Given: Diameter of water tap =
1
p
p
cm
´10-2 m
Initially at t = 0 ; h = h
t=t;h=0
2
Reynold’s number, Re =
30.
2
10-2
p
8 cm
r Vr
n
dt = -
54 cm
P
(54–x)
x
Length of the air column above mercury in the tube is,
P + x = P0
Þ P = (76 – x)
Þ 8 × A × 76 = (76 – x) × A × (54 – x)
\ x = 38
Thus, length of air column
= 54 – 38 = 16 cm.
(a) According to Bernoulli’s theorem,
1
1
P1 + rv12 = P2 + rv 22
2
2
A
h
2
a
liquid v = 2 gh ]
Integrating both sides
Hg
31.
A
dh
pa 2 gh
[Q velocity of efflux of
@ 5500
(a)
dh
dt
æ dh ö
Then, A ç - ÷ = pa 2 .v
è dt ø
Þ V = 0.05 m/s
10-3
6 ´ 10
= 7.5cc / s
8
32. (b) Let the rate of falling water level be -
15
æ 1 ö
-4
= 10 3 ´ p ç
÷ ´ 10 V
5 ´ 60
è pø
=
10
8
Therefore the rate at which water flows through the
tube = A1v1 = A 2 v 2 =
dm
= rAV
V
dt
103 ´ 0.5 ´
A1 v 2 6
=
=
A 2 v1 10
t
ò dt = -
0
[ t ]t0 = -
A
2 g pa 2
0
òh
-1 2
dh
h
0
é h1 2 ù
.
ê
ú
2 g pa 2 ëê 1 2 ûú h
A
2A h
pa 2 g
33. (a) For 1 m length of horizontal tube
Mass of water M = density × volume
= 103 × area × length
= 103 × 10–2 × 1 = 10 kg
Dp
Therefore minimum force =
(rate of change of
Dt
momentum)
= 10 × 1.5 = 15 N
t=
P-151
Mechanical Properties of Fluids
34. (c) Pressure difference
(
)
(
1
1
P2 - P1 = r v22 - v12
= ´1.2 (150)2 - (100) 2
2
2
1
= ´ 1.2(22500 - 10000)
2
= 7500 Nm–2
35.
)
37. (a) As A1v1 = A2v2 (Principle of continuity)
or, l 2 2 gh = pr 2 2 g ´ 4h
(Efflux velocity =
l2
l
l2
=
or r =
2p
2p
2p
38. (c) Using Bernoulli's theorem, for horizontal flow
\
(b)
h1
V1
1
1
P0 + rv12 + rgh = P0 + rv22 + 0
2
2
B
V2
v2 = v12 + 2 gh = 0.16 + 2 ´ 10 ´ 0.2 =2.03m/s
H
h2
According to equation of continuity
A2v2 = A1v1
G
p
... (i)
Where v1 = velocity of efflux at A =
(2 gh1 ) and
v2 = velocity of efflux at B = (2 g (H - h2 )
t1 = time of fall water stream through A
=
(
2(H - h1
g
(a) According to Bernoulli’s theorem
1
1
P1 + rv12 = P2 + rv22
...(i)
2
2
From question,
A
P1 - P2 = 3 ´ 105 , 1 = 5
A2
According to equation of continuity
A1 v1 = A2 v2
A1 v2
or, A = v = 5
2
1
Þ v2 = 5v1
From equation (i)
1
P1 - P2 = r v22 - v12
2
1
2
2
or 3 × 105 = ´ 1000 5v1 - v1
2
Þ 600 = 6v1 ´ 4v1
(
Þ v12 = 25
\ v1 = 5 m/s
)
(
v1
= 3.55 × 10–3 m
v2
39. (b) Given, Height of cylinder, h=20 cm Acceleration due
to gravity, g=10 ms–2
Velocity of efflux
v = 2gh
Where h is the height of the free surface of liquid from th e
hole
2h2
g
Putting these values is eqn (i) we get
(H – h1)h1 = (H – h2)h2
or [H – (h1 + h2)][h1 – h2] = 0
Here, H = h1 + h2 is irrelevant because the holes are at
h1
two different heights. Hence h1 = h2 or, h = 1
2
D22
D2
´ v2 = p 1 v1
4
4
Þ D2 = D1
)
t2 = time of fall of the water stream through B =
36.
r2 =
A
i.e. R1 = R2 = R
or, v1 t1 = v2 t2
40.
Þ v = 2 ´10 ´ 20 = 20 m / s
(a) Using, v2 – u2 = 2gh
Þ v 2 - 02 = 2 gh Þ v = 2 gh
Terminal velocity,
2 r 2 (r - s ) g
h
9
After falling through h the velocity should be equal to
terminal velocity
VT =
\ 2 gh =
Þ 2 gh =
Þh=
2 r 2 (r - s ) g
9
h
4 r 4 g 2 (r - s)2
81
h2
2r 4 g (r - s)2
41. (a) 27 ×
81h2
Þ h a r4
4 3 4 3
pr = pr
3
3
R
.
3
Terminal velocity, v µ r2
or r =
)
2gh )
v1 r12
\ v = 2
r2
2
2
2
æ r2 ö
æ R / 3ö
1
or v2 = ç ÷ v1 = çè
÷ø v1 =
R
è r1 ø
9
P-152
Physics
v1
v2 = 9.
42. (c) When a point mass is falling vertically in a viscous
medium, the medium or viscous fluid exerts drag force on
the body to oppose its motion and at one stage body
falling with constant terminal velocity.
43. (b) h = 10–2 poise
W–T –F=0
or
v = 18 km/h =
47.
18000
= 5 m/s
3600
l =5m
v
Strain rate =
l
shearing stress
strain rate
\ Shearing stress = h × strain rate
Coefficient of viscosity, h =
\ VT µ a 2
48. (a) When the ball attains terminal velocity
Weight of the ball = viscous force + buoyant force
\V rg = 6phrv + V rl g
Þ Vg ( r - rl ) = 6phrv
(
'
\ v ' h ' = (r - r l ) ´ v h
(r - r l )
Þ v' =
2
= 1m ´
E=
=
)
'
Also Vg r- ri = 6ph ' rv '
5
= 10 ´ = 10–2 Nm–2
5
(b) Total volume of rain drops, received 100 cm in a year
by area 1 m2
100
m = 1 m3
100
As we know, density of water,
d = 103 kg/m3
Therefore, mass of this volume of water
M = d × v = 103 × 1 = 103 kg
Average terminal velocity of rain drop
v = 9 m/s (given)
Therefore, energy transferred by rain,
2 a 2 (r - s ) g
9h
VT =
-2
44.
2 r 2 (r - s)g
9
h
As in case of upward motion upward force is twice its
effective weight, therefore, it will move with same speed
10 cm/s
(a) Terminal velocity in a viscous medium is given by:
And terminal velocity v =
=
49.
(r - rl' ) vh
´
(r - rl ) h '
(7.8 - 1.2) 10 ´ 8.5 ´ 10-4
´
(7.8 - 1)
13.2
\ v ' = 6.25 ´ 10-4 cm/s
(a) When the ball attains terminal velocity
Weight of the ball = Buoyant force + Viscous force
1
mv2
2
1
× 103 × (9)2
2
1
× 103 × 81 = 4.05 × 104 J
2
(b) According to Toricelli’s theorem,
Velocity of efflex,
=
45.
W=V r1g
\ V r1 g
Veff = 2 gh = 2 ´ 9.8 ´ 5 @ 9.8 ms -1
46.
B=Vr2 g
Fv
(c)
T (upthrust)
F
(viscous
force)
r
W(weight)
Weight of the body
4 3
W = mg = pr rg
3
4 3
T = pr sg
3
and F = 6phvr
When the body attains terminal velocity net force acting
on the body is zero. i.e.,
= V r2 g + kvt2
Þ Vg ( r1 – r2 ) g = kvt2
Vg (r1 - r 2 )
k
(c) Given,
Density of gold, rG = 19.5 kg/m3
Density of silver, r5 = 10.5kg/m3
Density of liquid, s = 1.5kg/m3
Þ vt =
50.
Terminal velocity, vT =
vT
2 r 2 (r - s ) g
9h
(10.5 - 1.5)
9
Þ vT = 0.2 ´
2
0.2 (19.5 - 1.5)
18
\ vT = 0.1 m/s
\
2
=
2
51. (b) From Stoke's law, force of viscosity acting on a
spherical body is
F = 6phrv
hence F is directly proportional to radius & velocity.
P-153
Mechanical Properties of Fluids
52. (101)
Given : Radius of capillary tube,
r = 0.015 cm = 15 × 10–5 mm
h = 15 cm = 15 × 10–2 mm
2T cos q
[cos q = cos0° = 1]
Using, h =
rgr
Surface tension,
rhrg 15 ´ 10-5 ´ 15 ´ 10-2 ´ 900 ´ 10
=
= 101 m i l l i
2
2
newton m–1
53. (c) According to question, pressure inside, 1st soap
bubble,
4T
DP1 = P1 - P0 = 0.01 =
...(i)
R1
56. (b) As we know that
2T cos q
=Rh
rrg
THg
= 7.5
TWater
rHg
rW
T=
And DP2 = P2 - P0 = 0.02 =
Dividing, equation (ii) by (i),
1 R2
=
Þ R1 = 2R2
2 R1
54.
4T
R2
...(ii)
Capillary rise, h =
R Hg
R Water
æ THg ö æ rW ö æ cos qHg ö
=ç
֍
֍
÷
è TW ø è rHg ø è cos qW ø
2T cos q
rrg
Þm ar
m
r
\ 1 =
m2 2r
or m2 = 2m1 = 2m.
T T
58. (d) Here excess pressure, Pexcess = r + r
1
2
ær = R ö
Qç1
÷
è r2 = O ø
59. (a) When the bubble gets detached,
Buoyant force = force due to surface tension
Pexcess =
T
R
R
q
60°
rq
2T cos q
rgr
Access pressure in air bubble =
2 R 4 rw g
2 2rw g
Þ r=R
3T
3T
(d) When drops combine to form a single drop of radius R.
2
Þr =
60.
4 3
2
pR dg – pR 3rg
3
3
2 3
Þ T(2pR) = pR (2d – r) g
3
R=
2T
R
2T
4p R 3
rw g
(p r 2 ) =
R
3T
T .2pR =
R
(2d – r) g Þ
3
T×dl
Force due to excess pressure = upthrust
(d) For the drops to be in equilibrium upward force on
drop = downward force on drop
Þ T=
cos135°
1
=
cos 0°
2
1
1
2
´
= 0.4 =
13.6
5
2
2T cos q
57. (d) We have, h =
r rg
Mass of the water in the capillary
3
2
=
= 0.087 m
667 ´ 10 ´ 0.15 ´ 10-3
2
cos qW
=
= 7.5 ´
2 ´ 0.05 ´
55.
cos qHg
2
m = rV = r × pr2h = r ´ pr ´
4
Volume V = pR 3
3
V1 R13 8R23 8
\ = 3 = 3 =
V2 R2
1
R2
(b) Given,
Angle of contact q = 30°
Surface tension, T = 0.05 Nm–1
Radius of capillary tube, r = 0.15 mm = 0.15 × 10–3m
Density of methylene iodide, r = 667 kg m–3
30°
= 13.6 &
é1 1 ù
Then energy released, E = 4pTR 3 ê - ú
ër R û
If this energy is converted into kinetic energy then
3T
(2d – r) g
1 2
é1 1 ù
mv = 4pR 3T ê - ú
2
ër R û
P-154
Physics
1 é4 3 ù 2
é1 1 ù
´ ê pR rú v = 4pR 3T ê - ú
2 ë3
û
ër R û
v2 =
6T é 1 1 ù
r êë r R úû
v=
6T é 1 1 ù
r êë r R úû
61. (b)
62. (a) Given: Radius of air bubble,
r = 0.1 cm = 10–3 m
Surface tension of liquid,
S = 0.06 N/m = 6 × 10–2 N/m
Density of liquid, r = 103 kg/m3
Excess pressure inside the bubble,
rexe = 1100 Nm–2
Depth of bubble below the liquid surface,
h=?
As we know,
2s
rExcess = hrg +
r
2 ´ 6 ´ 10-2
Þ 1100 = h × 103 × 9.8 +
10-3
Þ 1100 = 9800 h + 120
Þ 9800h = 1100 – 120
980
Þ h=
= 0.1 m
9800
63. (a) Acceleration due to gravity changes with the depth,
dö
æ
g¢ = g ç1 - ÷
R
è
ø
Pressure, P = rgh
Hence ratio,
64.
x æ dö
is ç1 - ÷
y è Rø
(c) Angle of contact q
TSA - TSL
TLA
when water is on a waxy or oily surface
TSA < TSL cos q is negative i.e.,
90° < q < 180°
i.e., angle of contact q increases
And for q > 90° liquid level in capillary tube fall.
i.e., h decreases
65. (b)
cos q =
66. (b) Surface tension of a liquid decreases with the rise in
temperture. At the boiling point of liquid, surface tension
is zero.
Capillary rise h =
2T cos q
rdg
As surface tension T decreases with rise in temperature
hence capillary rise also decreases.
67. (d) Let T is the force due to surface tension per unit length,
then
F = 2lT
l = length of the slider.
At equilibrium, F = W
\ 2Tl = mg
Þ T=
mg
1.5 ´ 10 -2
1.5
=
=
= 0.025 Nm–1
2l 2 ´ 30 ´ 10-2 60
68. (c) Work done = increase in surface area × surface tension
Þ W = 2T4p[(52) – (3)2] × 10–4
= 2 × 0.03 × 4p [25 – 9] × 10–4 J
= 0.4p × 10–3 J = 0.4p mJ
69. (c) As volume remains constant
\Sum of volumes of 2 smaller drops
= Volume of the bigger drop
4
4
2. pr 3 = pR 3 Þ R = 21/ 3 r
3
3
2
Surface energy = Surface tension × Surface area = T .4 pR
= T 4p 22 / 3 r 2 = T .28 / 3 pr 2 .
70. (c) In case of water, the meniscus shape is concave
upwards. From ascent formula h =
2s cos q
r rg
The surface tension (s) of soap solution is less than water.
Therefore height of capillary rise for soap solution should
be less as compared to water. As in the case of water, the
meniscus shape of soap solution is also concave upwards.
71. (c) Water fills the tube entirely in gravityless condition
i.e., 20 cm.
72. (a) Let pressure outside be P0 and r and R be the radius
of smaller bubble and bigger bubble respectively.
2T
\ Pressure P1 For smaller bubble = P0 +
r
2T
P2 For bigger bubble = P0 +
( R > r)
R
\ P1 > P2
hence air moves from smaller bubble to bigger bubble.
10
Thermal Properties
of Matter
TOPIC 1 Thermometer & Thermal
Expansion
1.
Two different wires having lengths L1 and L2 , and
respective temperature coefficient of linear expansion a1
and a 2 , are joined end-to-end. Then the effective
temperature coefficient of linear expansion is :
[Sep. 05, 2020 (II)]
(a)
a1 L1 + a 2 L2
L1 + L2
a + a2
(c) 1
2
2.
3.
4.
5.
6.
7.
(b) 2 a1a 2
aa
L2 L1
(d) 4 1 2
a1 + a 2 ( L2 + L1 ) 2
A bakelite beaker has volume capacity of 500 cc at 30°C.
When it is partially filled with Vm volume (at 30°C) of
mercury, it is found that the unfilled volume of the beaker
remains constant as temperature is varied. If g(beaker) = 6 ×
10–6 °C–1 and g(mercury) = 1.5 × 10–4 °C–1, where g is the
coefficient of volume expansion, then Vm (in cc) is close to
__________.
[NA Sep. 03, 2020 (I)]
When the temperature of a metal wire is increased from
0°C to 10°C, its length increased by 0.02%. The percentage
change in its mass density will be closest to :
[Sep. 02, 2020 (II)]
(a) 0.06
(b) 2.3
(c) 0.008
(d) 0.8
A non-isotropic solid metal cube has coefficients of linear
expansion as: 5 ´ l0–5/°C along the x-axis and 5 ´ 10–6/°C
along the y and the z-axis. If the coefficient of volume
expansion of the solid is C ´ 10–6/°C then the value of C is
¾¾¾ .
[NA 7 Jan. 2020 I]
o
At 40 C, a brass wire of 1 mm radius is hung from the
ceiling. A small mass, M is hung from the free end of the
wire. When the wire is cooled down from 40oC to 20oC it
regains its original length of 0.2 m. The value of M is close
to:
[12 April 2019 I]
(Coefficient of linear expansion and Young’s modulus of
brass are 10–5/oC and 1011 N/m2, respectively; g = 10 ms–2)
(a) 9 kg
(b) 0.5 kg (c) 1.5 kg
(d) 0.9 kg
Two rods A and B of identical dimensions are at
temperature 30°C. If A is heated upto 180°C and B upto
8.
9.
T°C, then the new lengths are the same. If the ratio of the
coefficients of linear expansion of A and B is 4 : 3, then
the value of T is :
[11 Jan. 2019 II]
(a) 230°C
(b) 270°C
(c) 200°C
(d) 250°C
A thermometer graduated according to a linear scale reads
a value x0 when in contact with boiling water, and x0/3
when in contact with ice. What is the temperature of an
object in °C, if this thermometer in the contact with the
object reads x0/2?
[11 Jan. 2019 II]
(a) 25
(b) 60
(c) 40
(d) 35
A rod, of length L at room temperature and uniform area
of cross section A, is made of a metal having coefficient
of linear expansion a/°C. It is observed that an external
compressive force F, is applied on each of its ends,
prevents any change in the length of the rod, when its
temperature rises by DT K. Young’s modulus, Y, for this
metal is:
[9 Jan. 2019 I]
F
F
(a)
(b)
AaDT
A a ( D T - 273)
F
2F
(c)
(d)
2A a D T
AaDT
An external pressure P is applied on a cube at 0oC so that
it is equally compressed from all sides. K is the bulk
modulus of the material of the cube and a is its
coefficient of linear expansion. Suppose we want to bring
the cube to its original size by heating. The temperature
should be raised by :
[2017]
3a
(b) 3PKa
PK
P
P
(c)
(d)
3aK
aK
10. A steel rail of length 5 m and area of cross-section 40
cm2 is prevented from expanding along its length while
the temperature rises by 10°C. If coefficient of linear
expansion and Young’s modulus of steel are 1.2 × 10–5 K–1
and 2 × 1011 Nm–2 respectively, the force developed in
the rail is approximately:
[Online April 9, 2017]
(a) 2 × 107 N
(b) 1 × 105 N
(c) 2 × 109 N
(d) 3 × 10–5 N
(a)
P-156
11.
Physics
A compressive force, F is applied at the two ends of a
long thin steel rod. It is heated, simultaneously, such that
its temperature increases by DT. The net change in its
length is zero. Let l be the length of the rod, A its area of
cross–section, Y its Youn g’s modulus, and a its
coefficient of linear expansion. Then, F is equal to :
[Online April 8, 2017]
(a) l2 Ya DT
(b) lA Ya DT
AY
aDT
The ratio of the coefficient of volume expansion of a glass
container to that of a viscous liquid kept inside the
container is 1 : 4. What fraction of the inner volume of the
container should the liquid occupy so that the volume of
the remaining vacant space will be same at all temperatures ?
[Online April 23, 2013]
(a) 2 : 5
(b) 1 : 4
(c) 1 : 64
(d) 1 : 8
On a linear temperature scale Y, water freezes at – 160° Y
and boils at – 50° Y. On this Y scale, a temperature of 340 K
would be read as : (water freezes at 273 K and boils at 373 K)
[Online April 9, 2013]
(a) – 73.7° Y
(b) – 233.7° Y
(c) – 86.3° Y
(d) – 106.3° Y
A wooden wheel of radius R is made of two semicircular
part (see figure). The two parts are held together by a ring
made of a metal strip of cross sectional area S and length
L. L is slightly less than 2pR. To fit the ring on the wheel, it
is heated so that its temperature rises by DT and it just
steps over the wheel. As it cools down to surrounding
temperature, it presses the semicircular parts together. If
the coefficient of linear expansion of the metal is a, and its
Young's modulus is Y, the force that one part of the wheel
applies on the other part is :
[2012]
(c) A Ya DT
12.
13.
14.
(a)
2pSY aDT
(b)
SY aDT
(c)
pSY aDT
(d)
2SY aDT
(d)
R
TOPIC 2 Calorimetry and Heat
Transfer
15.
Three rods of identical cross-section and lengths are made
of three different materials of thermal conductivity K1, K2
and K3, respectively. They are joined together at their ends
to make a long rod (see figure). One end of the long rod is
maintained at 100ºC and the other at 0ºC (see figure). If the
joints of the rod are at 70ºC and 20ºC in steady state and
there is no loss of energy from the surface of the rod,
the correct relationship between K1, K2 and K3 is :
[Sep. 06, 2020 (II)]
100°C
K1
K2
K3
0°C
70°C 20°C
16.
17.
18.
19.
(a) K1 : K3 = 2 : 3, K1 < K3 = 2 : 5
(b) K1 < K2 < K3
(c) K1 : K2 = 5 : 2, K1 : K3 = 3 : 5
(d) K1 > K2 > K3
A bullet of mass 5 g, travelling with a speed of 210 m/s,
strikes a fixed wooden target. One half of its kinetics energy is converted into heat in the bullet while the other
half is converted into heat in the wood. The rise of temperature of the bullet if the specific heat of its material is
0.030 cal/(g – ºC) (1 cal = 4.2 × 107 ergs) close to :
[Sep. 05, 2020 (I)]
(a) 87.5ºC
(b) 83.3ºC
(c) 119.2ºC
(d) 38.4ºC
The specific heat of water = 4200 J kg–1 K–1 and the latent
heat of ice = 3.4 × 105 J kg–1. 100 grams of ice at 0°C is
placed in 200 g of water at 25°C. The amount of ice that will
melt as the temperature of water reaches 0°C is close to (in
grams) :
[Sep. 04, 2020 (I)]
(a) 61.7
(b) 63.8
(c) 69.3
(d) 64.6
A calorimter of water equivalent 20 g contains 180 g of
water at 25°C. 'm' grams of steam at 100°C is mixed in it till
the temperature of the mixture is 31°C. The value of 'm' is
close to (Latent heat of water = 540 cal g–1, specific heat of
water = 1 cal g–1 °C–1)
[Sep. 03, 2020 (II)]
(a) 2
(b) 4
(c) 3.2
(d) 2.6
Three containers C1, C2and C3 have water at different
temperatures. The table below shows the final temperature
T when different amounts of water (given in liters) are taken
from each container and mixed (assume no loss of heat
during the process)
[8 Jan. 2020 II]
C1
C2
C3
T
1l
2l
—
60°C
–
1l
2l
30°C
2l
—
1l
60°C
1l
1l
1l
q
The value of q (in °C to the nearest integer) is______.
20. M grams of steam at 100°C is mixed with 200 g of ice at its
melting point in a thermally insulated container. If it
produces liquid water at 40°C [heat of vaporization of water
is 540 cal/ g and heat of fusion of ice is 80 cal/g], the value
of M is ________
[NA 7 Jan. 2020 II]
o
21. When M1 gram of ice at –10 C (Specific heat = 0.5 cal g–1 oC–1)
is added to M2 gram of water at 50oC, finally no ice is
left and the water is at 0oC. The value of latent heat of
ice, in cal g–1 is:
[12 April 2019 I]
P-157
Thermal Properties of Matter
(a)
50 M 2
-5
M1
5M1
(b) M - 50
2
(c)
50M 2
M1
(d)
5 M2
-5
M1
22. A massless spring (K = 800 N/m), attached with a mass
(500 g) is completely immersed in 1kg of water. The spring
is stretched by 2cm and released so that it starts vibrating.
What would be the order of magnitude of the change in
the temperature of water when the vibrations stop
completely? (Assume that the water container and spring
receive negligible heat and specific heat of mass = 400 J/kg K,
specific heat of water = 4184 J/kg K)
[9 April 2019 II]
(a) 10–4 K
(b) 10–5 K (c) 10–1 K
(d) 10–3 K
23. Two materials having coefficients of thermal conductivity
‘3K’ and ‘K’ and thickness ‘d’ and ‘3d’, respectively, are
joined to form a slab as shown in the figure. The
temperatures of the outer surfaces are ‘q2’ and ‘q1’
respectively, (q2 > q1). The temperature at the interface is:
[9 April 2019 II]
q1 9q 2
q 2 + q1
+
(b)
10 10
2
q1 5q2
q 2q
+
(d) 1 + 2
(c)
6
6
3
3
24. A cylinder of radius R is surrounded by a cylindrical shell
of inner radius R and outer radius 2R. The thermal
conductivity of the material of the inner cylinder is K1 and
that of the outer cylinder is K2. Assuming no loss of
heat, the effective thermal conductivity of the system for
heat flowing along the length of the cylinder is:
[12 Jan. 2019 I]
K1 + K 2
(a)
(b) K1 + K 2
2
2K1 + 3K 2
K1 + 3K 2
(c)
(d)
5
4
25. Ice at –20°C is added to 50 g of water at 40°C, When the
temperature of the mixture reaches 0°C, it is found that 20
g of ice is still unmelted. The amount of ice added to the
water was close to
[11 Jan. 2019 I]
(Specific heat of water = 4.2J/g/°C
Specific heat of Ice = 2.1 J/g/°C
Heat of fusion of water at 0°C = 334J/g)
(a) 50g
(b) 100 g
(c) 60 g
(d) 40 g
26. When 100 g of a liquid A at 100°C is added to 50 g of a
(a)
liquid B at temperature 75°C, the temperature of the mixture
becomes 90°C. The temperature of the mixture, if 100 g of
liquid A at 100°C is added to 50 g of liquid B at 50°C, will
be :
[11 Jan. 2019 II]
(a) 85°C
(b) 60°C
(c) 80°C
(d) 70°C
27. A metal ball of mass 0.1 kg is heated upto 500°C and
dropped into a vessel of heat capacity 800 JK–1 and
containing 0.5 kg water. The initial temperature of water
and vessel is 30°C. What is the approximate percentage
increment in the temperature of the water? [Specific Heat
Capacities of water and metal are, respectively, 4200 Jkg–
1K–1 and 400 Jkg–1 K–1 ]
[11 Jan. 2019 II]
(a) 15%
(b) 30%
(c) 25%
(d) 20%
28. A heat source at T = 103 K is connected to another heat
reservoir at T = 102 K by a copper slab which is 1 m
thick. Given that the thermal conductivity of copper is
0.1 WK–1 m–1, the energy flux through it in the steady
state is:
[10 Jan. 2019 I]
(a) 90 Wm–2
(b) 120 Wm–2
(c) 65 Wm–2
(d) 200 Wm–2
29. An unknown metal of mass 192 g heated to a temperature
of 100°C was immersed into a brass calorimeter of mass
128 g containing 240 g of water at a temperature of 8.4°C.
Calculate the specific heat of the unknown metal if water
temperature stablizes at 21.5°C. (Specific heat of brass is
394 J kg–1 K–1)
[10 Jan. 2019 II]
–1
–1
(a) 458 J kg K
(b) 1232 J kg–1 K–1
(c) 916 J kg–1 K–1
(d) 654 J kg–1 K–1
30. Temperature difference of 120°C is maintained between
two ends of a uniform rod AB of length 2L. Another bent
3L
, is
2
connected across AB (See figure). In steady state,
temperature difference between P and Q will be close to:
[9 Jan. 2019 I]
rod PQ, of same cross-section as AB and length
L
4
A
L
2
P
L
Q
B
(a) 45°C
(b) 75°C
(c) 60°C
(d) 35°C
31. A copper ball of mass 100 gm is at a temperature T. It is
dropped in a copper calorimeter of mass 100 gm, filled with
170 gm of water at room temperature. Subsequently, the
temperature of the system is found to be 75°C. T is given
by (Given : room temperature = 30° C, specific heat of
copper = 0.1 cal/gm°C
[2017]
(a) 1250°C (b) 825°C (c) 800°C
(d) 885° C
32. In an experiment a sphere of aluminium of mass 0.20 kg
is heated upto 150°C. Immediately, it is put into water of
volume 150 cc at 27°C kept in a calorimeter of water
equivalent to 0.025 kg. Final temperature of the system
is 40°C. The specific heat of aluminium is :
(take 4.2 Joule=1 calorie)
[Online April 8, 2017]
(a) 378 J/kg – °C
(b) 315 J/kg – °C
(c) 476 J/kg – °C
(d) 434 J/kg – °C
P-158
Physics
33. An experiment takes 10 minutes to raise the temperature
of water in a container from 0ºC to 100ºC and another 55
minutes to convert it totally into steam by a heater
supplying heat at a uniform rate. Neglecting the specific
heat of the container and taking specific heat of water
to be 1 cal / g ºC, the heat of vapourization according
to this experiment will come out to be :
[Online April 11, 2015]
(a) 560 cal/ g
(b) 550 cal/ g
(c) 540 cal/ g
(d) 530 cal/ g
34. Three rods of Copper, Brass and Steel are welded together
to form a Y shaped structure. Area of cross - section of
each rod = 4 cm2. End of copper rod is maintained at 100ºC
where as ends of brass and steel are kept at 0ºC. Lengths
of the copper, brass and steel rods are 46, 13 and 12 cms
respectively. The rods are thermally insulated from
surroundings excepts at ends. Thermal conductivities of
copper, brass and steel are 0.92, 0.26 and 0.12 CGS units
respectively. Rate of heat flow through copper rod is:[2014]
(a) 1.2 cal/s
(b) 2.4 cal/s
(c) 4.8 cal/s
(d) 6.0 cal/s
35. A black coloured solid sphere of radius R and mass M is
inside a cavity with vacuum inside. The walls of the cavity
are maintained at temperature T0. The initial temperature
of the sphere is 3T0. If the specific heat of the material of
the sphere varies as aT3 per unit mass with the temperature
T of the sphere, where a is a constant, then the time taken
for the sphere to cool down to temperature 2T0 will be (s is
Stefan Boltzmann constant)
[Online April 19, 2014]
(a)
(c)
36.
37.
Ma
æ3ö
In ç ÷
4pR s è 2 ø
(b)
2
Ma
æ 16 ö
In ç ÷
16pR s è 3 ø
2
(d)
Ma
æ 16 ö
In ç ÷
4pR s è 3 ø
2
Ma
æ3ö
In ç ÷
16pR s è 2 ø
2
Water of volume 2 L in a closed container is heated with a
coil of 1 kW. While water is heated, the container loses
energy at a rate of 160 J/s. In how much time will the
temperature of water rise from 27°C to 77°C? (Specific heat
of water is 4.2 kJ/kg and that of the container is negligible).
[Online April 9, 2014]
(a) 8 min 20 s
(b) 6 min 2 s
(c) 7 min
(d) 14 min
Assume that a drop of liquid evaporates by decrease in its
surface energy, so that its temperature remains
unchanged.What should be the minimum radius of the
drop for this to be possible? The surface tension is T,
density of liquid is r and L is its latent heat of
vaporization.
[2013]
(a) rL/T
(b)
T / rL (c) T/rL
(d) 2T/rL
38. A mass of 50g of water in a closed vessel, with
surroundings at a constant temperature takes 2 minutes to
cool from 30°C to 25°C. A mass of 100g of another liquid in
an identical vessel with identical surroundings takes the
same time to cool from 30° C to 25° C. The specific heat of
the liquid is :
(The water equivalent of the vessel is 30g.)
[Online April 25, 2013]
(a) 2.0 kcal/kg
(b) 7 kcal/kg
(c) 3 kcal/kg
(d)
0.5 kcal/kg
39. 500 g of water and 100 g of ice at 0°C are in a calorimeter
whose water equivalent is 40 g. 10 g of steam at 100°C is
added to it. Then water in the calorimeter is : (Latent heat
of ice = 80 cal/g, Latent heat of steam = 540 cal/ g)
[Online April 23, 2013]
(a) 580 g
(b) 590 g
(c) 600 g
(d) 610 g
40. Given that 1 g of water in liquid phase has volume 1 cm 3
and in vapour phase 1671 cm3 at atmospheric pressure
and the latent heat of vaporization of water is 2256 J/g; the
change in the internal energy in joules for 1 g of water at
373 K when it changes from liquid phase to vapour phase
at the same temperature is :
[Online April 22, 2013]
(a) 2256
(b) 167
(c) 2089
(d) 1
41. A large cylindrical rod of length L is made by joining two
L
identical rods of copper and steel of length æç ö÷ each.
è 2ø
The rods are completely insulated from the surroundings.
If the free end of copper rod is maintained at 100°C and
that of steel at 0°C then the temperature of junction is
(Thermal conductivity of copper is 9 times that of steel)
[Online May 19, 2012]
(a) 90°C
(b) 50°C
(c) 10°C
(d) 67°C
42. The heat radiated per unit area in 1 hour by a furnace
whose temperature is 3000 K is (s = 5.7 × 10–8 W m–2 K–4)
[Online May 7, 2012]
(a) 1.7 × 1010 J
(b) 1.1 × 1012 J
(c) 2.8 × 108 J
(d) 4.6 × 106 J
43. 100g of water is heated from 30°C to 50°C. Ignoring the
slight expansion of the water, the change in its internal
energy is (specific heat of water is 4184 J/kg/K): [2011]
(a) 8.4 kJ (b) 84 kJ
(c) 2.1 kJ (d) 4.2 kJ
44. The specific heat capacity of a metal at low temperature
(T) is given as
3
æ T ö
C p (kJK -1kg -1 ) = 32 ç
è 400 ÷ø
A 100 gram vessel of this metal is to be cooled from 20ºK
to 4ºK by a special refrigerator operating at room
temperature (27°C). The amount of work required to
cool the vessel is
[2011 RS]
(a) greater than 0.148 kJ
(b) between 0.148 kJ and 0.028 kJ
(c) less than 0.028 kJ
(d) equal to 0.002 kJ
P-159
Thermal Properties of Matter
45. A long metallic bar is carrying heat from one of its ends
to the other end under steady–state. The variation of
temperature q along the length x of the bar from its hot end
is best described by which of the following figures?[2009]
q
q
(a)
r1
(b)
x
x
q
r2
q
(c)
One end of a thermally insulated rod is kept at a temperature
T1 and the other at T2. The rod is composed of two sections
of length l1 and l2 and thermal conductivities K1 and K2
respectively. The temperature at the interface of the two
section is
[2007]
T1 l1
K1
47.
x
l2
T2
K2
(a)
( K1l1T1 + K 2l2T2 )
( K1l1 + K 2l2 )
(b)
( K 2l2T1 + K1l1T2 )
( K1l1 + K 2 l2 )
(c)
( K 2l1T1 + K1l2T2 )
( K 2 l1 + K1l2 )
(d)
( K1l2T1 + K 2l1T2 )
( K1l2 + K 2 l1 )
(a)
(c)
r02 R2 s
T
T4
r
2
(b)
4
(b)
(r2 - r1 )
(r1 r2 )
(c)
( r2 - r1 )
(d)
r1 r2
(r2 - r1 )
50. If the temperature of the sun were to increase from T to
2T and its radius from R to 2R, then the ratio of the radiant
energy received on earth to what it was previously will
be
[2004]
(a) 32
(b) 16
(c) 4
(d) 64
51. The temperature of the two outer surfaces of a composite
slab, consisting of two materials having coefficients of
thermal conductivity K and 2K and thickness x and 4x,
respectively, are T2 and T1 (T2 > T1 ) . The rate of heat
transfer through the slab, in a steady state is
æ A(T2 - T1 ) K ö
çè
÷ø f , with f equal to
x
(d)
R2 s
T
r
7
T0
3
5
(d) T f = T0
2
(b) T f =
[2004]
4x
2K
T1
4
temperature T0, while Box contains one mole of helium at
æ 7ö
temperature çè ÷ø T0 . The boxes are then put into thermal
3
contact with each other, and heat flows between them until
the gases reach a common final temperature (ignore the
heat capacity of boxes). Then, the final temperature of
the gases, Tf in terms of T0 is
[2006]
3
T0
7
3
(c) T f = T0
2
K
2
Two rigid boxes containing different ideal gases are placed
on a table. Box A contains one mole of nitrogen at
(a) T f =
x
T4
4pr
r2
where r0 is the radius of the Earth and s is Stefan's constant.
48.
2
pr02 R2 s
T2
ær ö
ln ç 2 ÷
è r1 ø
Assuming the Sun to be a spherical body of radius R at a
temperature of TK, evaluate the total radiant powerd
incident of Earth at a distance r from the Sun
[2006]
4pr02 R 2s
T1
(a)
(d)
x
46.
49. The figure shows a system of two concentric spheres of
radii r1 and r2 are kept at temperatures T1 and T2,
respectively. The radial rate of flow of heat in a substance
between the two concentric spheres is proportional to
[2005]
2
1
1
(b)
(c) 1
(d)
3
2
3
52. The earth radiates in the infra-red region of the spectrum.
The spectrum is correctly given by
[2003]
(a) Rayleigh Jeans law
(b) Planck’s law of radiation
(c) Stefan’s law of radiation
(d) Wien’s law
53. Heat given to a body which raises its temperature by 1°C
is
[2002]
(a) water equivalent
(b) thermal capacity
(c) specific heat
(d) temperature gradient
(a)
P-160
Physics
62. A hot body, obeying Newton’s law of cooling is cooling
down from its peak value 80°C to an ambient temperature
of 30°C. It takes 5 minutes in cooling down from 80°C to
40°C. How much time will it take to cool down from 62°C to
32°C?
(Given In 2 = 0.693, In 5 = 1.609) [Online April 11, 2014]
(a) 3.75 minutes
(b) 8.6 minutes
(c) 9.6 minutes
(d) 6.5 minutes
63. If a piece of metal is heated to temperature q and then
allowed to cool in a room which is at temperature q0, the
graph between the temperature T of the metal and time t
will be closest to
[2013]
T
(a)
(c)
T
q0
(a)
60.
(d)
T
q0
(b)
0
t
t
(d)
A body takes 10 minutes to cool from 60°C to 50°C. The
temperature of surroundings is constant at 25°C. Then,
the temperature of the body after next 10 minutes will
be approximately
[Online April 15, 2018]
(a) 43°C
(b) 47°C
(c) 41°C
(d) 45°C
61. Hot water cools from 60°C to 50°C in the first 10 minutes
and to 42°C in the next 10 minutes. The temperature of
the surroundings is:
[Online April 12, 2014]
(a) 25°C
(b) 10°C
(c) 15°C
(d) 20°C
0
(d)
t
loge (q – q0)
loge (q – q0)
(b)
(c)
(c)
t
O
t
t
O
O
64. A liquid in a beaker has temperature q(t) at time t and q0
is temperature of surroundings, then according to
Newton's law of cooling the correct graph between
loge(q – q0) and t is :
[2012]
0
(a)
T
q0
loge (q – q0)
A metallic sphere cools from 50°C to 40°C in 300 s. If
atmospheric temperature around is 20°C, then the sphere's
temperature after the next 5 minutes will be close to :
[Sep. 03, 2020 (II)]
(a) 31°C
(b) 33°C
(c) 28°C
(d) 35°C
59. Two identical beakers A and B contain equal volumes of
two different liquids at 60°C each and left to cool down.
Liquid in A has density of 8 × 102 kg/m3 and specific heat of
2000 J kg–1 K–1 while liquid in B has density of 103 kg m–3
and specific heat of 4000 J kg–1 K–1. Which of the following
best describes their temperature versus time graph
schematically ? (assume the emissivity of both the beakers
to be the same)
[8 April 2019 I]
t
O
TOPIC 3 Newton's Law of Cooling
58.
(b)
loge (q – q0)
54. Infrared radiation is detected by
[2002]
(a) spectrometer
(b) pyrometer
(c) nanometer
(d) photometer
55. Which of the following is more close to a black body?
[2002]
(a) black board paint
(b) green leaves
(c) black holes
(d) red roses
56. If mass-energy equivalence is taken into account, when
water is cooled to form ice, the mass of water should[2002]
(a) increase
(b) remain unchanged
(c) decrease
(d) first increase then decrease
57. Two spheres of the same material have radii 1 m and 4 m
and temperatures 4000 K and 2000 K respectively. The
ratio of the energy radiated per second by the first sphere
to that by the second is
[2002]
(a) 1 : 1
(b) 16 : 1
(c) 4 : 1
(d) 1 : 9.
0
t
65. According to Newton’s law of cooling, the rate of cooling
of a body is proportional to (Dq)n , where Dq is the
difference of the temperature of the body and the
surroundings, and n is equal to
[2003]
(a) two
(b) three
(c) four
(d) one
P-161
Thermal Properties of Matter
1.
(a) Let L'1 and L'2 be the lengths of the wire when
temperature is changed by DT °C .
At T °C,
Leq = L1 + L2
4.
Volume increase by 0.06% therefore density decrease by
0.06%.
(60.00) Volume, V = Ibh
5.
DV Dl Db Dh
=
+
+
V
l
b
h
(g = coefficient of volume expansion)
Þ g = 5 × 10–5 + 5 × 10–6 + 5 × 10–6
= 60 × 10–6/°C
\ Value of C = 60.00
(Bonus) Dtemp = Dload and A = pr2 = p(10–3)2 = p × 10–6
\ g=
At T + D°C
L'eq = L'1 + L'2
\ Leq (1 + a eq DT ) = L1 (1 + a1DT ) + L2 (1 + a 2 DT )
[Q L ' = L (1 + a DT )]
Þ ( L1 + L2 )(1 + a eq DT ) = L1 + L2 + L1a1DT + L2a 2 DT
Þ a eq =
2.
a1 L1 + a 2 L2
L1 + L2
Vb1 = Vm + Vm g m DT
f
= 2p = 6.28 kg
g
(a) Change in length in both rods are same i.e.
Dl1 =D l 2
6.
Unfilled volume (V0 - Vm ) = (Vb - Vm1 )
la1D q1 = la 2 Dq 2
Þ V0 g beaker = Vm g M
a1 Dq 2
=
a 2 Dq1
V0 g beaker
gM
or, Vm =
500 ´ 6 ´ 10 -6
q = 230°C
7.
(a) Let required temperature = T°C
M.P.
o
o
0C
Dl = l aDT
Here, a = Coefficient of linear expansion
Here, Dl = 0.02%, DT = 10ºC
TC
B.P.
x0
x0
2
Þ a = 2 ´ 10-5
x0
6
-5
M
Qr =
V
DV
-5
-2
´ 100 = gDT = (6 ´ 10 ´ 10 ´ 100) = 6 ´ 10
V
x 0 x 0 x0
– =
2 3 6
x ö
æ
& ç x 0 – 0 ÷ = (100 – 0°C)
3 ø
è
Þ T° C =
o
100 C
x0
3
Dl
0.02
=
l DT 100 ´ 10
Volume coefficient of expansion, g = 3a = 6 ´ 10
é a1 4 ù
êQ = ú
ë a2 3 û
4 q – 30
=
3 180 – 30
= 20 cc.
1.5 ´ 10-4
(a) Change in length of the metal wire (Dl) when its
temperature is changed by DT is given by
\a =
( p ´ 10 -6 ) ´ 1011
\ F = 20p N \ m =
When beaker is partially filled with Vm volume of mercury,
\Vm =
F ´ 0.2
or 0.2 × 10–5 × 20 =
(20.00)
Volume capacity of beaker, V0 = 500 cc
Vb = V0 + V0 g beaker DT
3.
FL
AY
L a DT =
P-162
Physics
2x 0
300
=100 Þ x 0 =
3
2
x 0 150
Þ T° C = =
= 25° C
6
6
340 - 273 °Y - ( -160)
=
373 - 273 -50 - ( -160)
Þ
8.
Þ
F/ A
stress
= A Dl / l
(
)
strain
Using, coefficient of linear expansion,
(a) Young’s modulus Y =
a=
14.
(c)
K=
Y=
F
A ( aDT )
As we know, Bulk modulus
DP
æ -DV ö
çè
÷
V ø
Þ
Y=
Þ
DV P
=
V
K
g=3 a
F
æ DL
ö
= a D q÷
çèQ
ø
A. a. Dq
L
Force developed in the rail F = YAaDt
= 2 × 1011 × 40 × 10–4 × 1.2 × 10–5 × 10
= 9.6 × 104 = 1 × 105 N
11. (c) Due to thermal exp., change in length (Dl) = l a DT ...
(i)
Normal stress
Young’s modulus (Y) =
Longitudinal strain
FA
Dl
F
Y=
Þ
=
Dl l
l
AY
Fl
AY
From eqn (i),
F = AY a DT
12. (b) When there is no change in liquid level in vessel
then g¢real = g¢ vessel
Change in volume in liquid relative to vessel
DVapp = Vg 'app Dq = V(g 'real - g 'vessel )
13.
Reading on any scale – LFP
(c)
UFP - LFP
= constant for all scales
… (ii)
F
Þ F = Y .S.aDT
S .aDT
\ The ring is pressing the wheel from both sides, Thus
Fnet = 2F = 2YSaDT
(a) As the rods are identical, so they have same length (l)
and area of cross-section (A). They are connected in series.
So, heat current will be same for all rods.
Y =
15.
æ DQ ö
æ DQ ö
æ DQ ö
Heat current = ç
÷ =ç
÷ =ç
÷
D
t
D
t
è
ø AB è
ø BC è Dt øCD
Þ
(100 - 70) K1 A (70 - 20) K 2 A (20 - 0) K 3 A
=
=
l
l
l
Þ K1 (100 - 70) = K 2 (70 - 20) = K3 (20 - 0)
Þ K1 (30) = K 2 (50) = K3 (20)
Þ
Fl
= l a DT
AY
DR
R DT
R
1
DR
=
= a.DT Þ
DR aDT
R
From equation (i) and (ii)
P
P
P
= gDt Þ Dt =
=
K
gK 3aK
Thermal stress F A
(b) Young's modulus =
=
DL L
Strain
Y=
…(i)
Þ
\
Dl =
F
´ 2 pR
S2 pDR
FR
Y=
S.DR
The coefficient of linear expansion a =
V = V0 (1 + gDt)
DV
= gDt
V0
10.
\ Y = – 86.3° Y
(d) The Young modulus is given as
stress
F/S
=
strain DL / L
Here, DL = 2p DR L = 2p R
Dl
Dl
Þ
=aDT
lD T
l
\Y =
9.
67 y + 160
=
100
110
K1 K 2 K3
=
=
10
6
15
Þ K1 : K 2 : K3 = 10 : 6 :15
Þ K1 : K3 = 2 : 3.
16. (a) According to question, one half of its kinetic energy
is converted into heat in the wood.
1 2 1
mv ´ = ms DT
2
2
v2
210 ´ 210
=
= 87.5°C
4 ´ s 4 ´ 4.2 ´ 0.3 ´ 1000
17. (a) Here ice melts due to water.
Let the amount of ice melts = mice
Þ DT =
mw sw Dq = mice Lice
P-163
Thermal Properties of Matter
\ mice =
=
18.
mw sw Dq
Lice
22. (b)
0.2 ´ 4200 ´ 25
3.4 ´ 105
= 0.0617 kg = 61.7 g
\ DT = 1 ´ 10 -5 K
(a) Heat given by water = mw Cw (Tmix - Tw )
= 200 ´1´ (31 - 25)
Heat taken by steam = m Lstem + m Cw (Ts – Tmix)
= m × 540 + m (1) × (100 –31)
= m × 540 + m (1) × (69)
From the principal of calorimeter,
Heat lost = Heat gained
\ (200)(31 - 25) = m ´ 540 + m(1)(69)
Þ 1200 = m(609) Þ m » 2.
19. (50.00)
Let Q1, Q2, Q3 be the temperatures of container C1,C2 and
C3 respectively.
Using principle of calorimetry in container C1, we have
(q1 – 60) = 2 ms(60 – q)
Þ q1 – 60 = 120 – 2q
Þ q1 = 180 – 2q
...(i)
For container C2
ms (q2 – 30) = 2ms (30 – q)
Þ q2= 90 – 2q3
...(ii)
For container C3
2ms (q1 – 60) = ms (60 – q)
Þ 2q1–120 = 60 – q
Þ 2q1 + q = 180
...(iii)
Also, q1 + q2 + q3 = 3q
...(iv)
Adding (i), (ii) and (iii)
3q1 + 3q2 + 3q3 = 450
Þ q1 + q2 + q3 = 150
Þ 3q = 150 Þ q = 50 ºC
20. (40) Using the principal of calorimetry
Mice Lf + mice (40 – 0) Cw
= mstream Lv + mstream (100 – 40) Cw
Þ M (540) + M × 1 × (100 – 40)
= 200 × 80 + 200 × 1 × 40
Þ 600 M = 24000
Þ M = 40g
21. (a) M1Cice × (10) + M1L = M2Cw (50)
or M1 × Cice (=0.5) × 10 + M1L = M2 × 1 × 50
50M 2
Þ L = M -5
1
1 2
.kx = mC (DT ) + mw Cw DT
2
1
2
or ´ 800 ´ 0.02 = 0.5 ´ 400 ´ DT + 1 ´ 4184 ´ DT
2
d
23. (a) H1 = H2 q 2 3k
q 3d
k q1
æ q2 - q ö
æ q - q1 ö
= kA ç
or (3k ) A ç
÷
è d ø
è 3d ÷ø
æ q + 9q 2 ö
or q = ç 1
è 10 ÷ø
24. (d) Effective thermal conductivity of system
K eq =
=
K1A1 + K 2 A 2
A1 + A 2
A2
K1pR 2 + K 2 [p (2R)2 - pR 2 ]
K1A1
p(2R)2
K1(pR 2 ) + K 2 (3pR 2 )
K2
K1 + 3K 2
4
4pR
25. (d) Let m gram of ice is added.
From principal of calorimeter
heat gained (by ice) = heat lost (by water)
\ 20 × 2.1 × m + (m – 20) × 334
= 50 × 4.2 × 40
376 m = 8400 + 6680
m = 40.1
26. (c) Heat loss = Heat gain = mSDq
So, mASADqA= mBSBDqB
Þ 100 × SA × (100 – 90) = 50 × SB × (90 – 75)
=
2
=
3
S
4 B
Now, 100 × SA × (100 – q) = 50× SB × (q – 50)
2SA = 1.5SB Þ SA =
æ 3ö
2 × çè ÷ø × (100 – q) = (q – 50)
4
300 – 3q = 2q – 100
400 = 5q Þ q = 80°C
27. (d) Assume final temperature = T°C
Heat lass = Heat gain = msDT
ÞmB sB DTB = mw swDTw
0.1 × 400 × (500 – T)
= 0.5 × 4200 × (T – 30) + 800 (T – 30)
Þ 40 (500 – T) = (T – 30) (2100 + 800)
Þ 20000 – 40T = 2900 T– 30 × 2900
Þ 20000 + 30 × 2900 = T(2940)
T = 30.4°C
P-164
Physics
and P × 55 × 60 = mL
...(ii)
Dividing equation (i) by (ii) we get
10 C ´100
=
55
L
\ L = 550 cal./g.
34. (c) Rate of heat flow is given by,
DT
6.4
´100 =
´100 = 21%,
T
30
so the closest answer is 20%.
Temp. of
Temp. of
heat source heat reservoir
1m
28. (a)
3
2
10 K
10 K
Q=
æ dQ ö kADT
çè ÷ø =
dt
l
Where, K = coefficient of thermal conductivity
l = length of rod and A = area of cross-section of rod
1 æ dQ ö
kD T
Energy flux, çè ÷ø =
A dt
l
29.
100°C
( 0.1)( 900)
= 90 W/m 2
1
(c) Let specific heat of unknown metal be ‘s’ According
to principle of calorimetry, Heat lost
= Heat gain m × sDq = m1sbrass (Dq1 + m2 swater + Dq2)
Þ 192 × S × (100 – 21.5)
= 128 × 394 × (21.5 – 8.4)
Solving we get,+ 240 × 4200 × (21.5 – 8.4)
S = 916 Jkg–1k–1
=
30. (a)
Copper
B
120
R/2
R/4
32.
33.
Brass
0°C
0.92 ´ 4(100 - T )
46
0.26 ´ 4 ´ (T - 0)
0.12 ´ 4 ´ (T - 0)
+
13
12
Þ 200 – 2T = 2T + T
Þ T = 40°C
=
R
P
R/4
L/4
Q
R/2
B
120 ´ 5 3
360
DTPQ =
= 45°C
× R=
8R
5
8
(d) According to principle of calorimetry,
Heat lost = Heat gain
100 × 0.1(T – 75) = 100 × 0.1 × 45 + 170 × 1 × 45
10 T– 750 = 450 + 7650 = 8100
Þ T – 75 = 810
T = 885°C
(d) According to principle of calorimetry,
Qgiven = Qused
0.2 × S × (150 – 40) = 150 × 1 × (40 – 27) + 25 × (40 – 27)
0.2 × S × 110 = 150 × 13 + 25 × 13
Specific heat of aluminium
13 ´ 25 ´ 7
S=
= 434 J/kg-°C
0.2 ´ 110
(b) As Pt = mCDT
So, P × 10 × 60 = mC 100 ...(i)
0.92 ´ 4 ´ 60
= 4.8 cal/s
46
(c) In the given problem, fall in temperature of sphere,
\
O
In steady state temperature difference between P and
Q,
31.
Steel
If the junction temperature is T, then
QCopper = QBrass + QSteel
L
L/4
T
0°C
DTAB 120 120 ´ 5
=
=
8
R AB
8R
R
5
A
KA(q1 - q 2 )
l
35.
QCopper =
dT = ( 3T0 - 2T0 ) = T0
Temperature of surrounding, Tsurr = T0
Initial temperature of sphere, Tinitial = 3T0
Specific heat of the material of the sphere varies as,
c = aT 3 per unit mass (a = a constant)
Applying formula,
(
dT sA 4
4
=
T - Tsurr
dt McJ
Þ
)
T0
s 4pR 2 é
=
( 3T )4 - ( T0 )4 ùûú
dt Ma ( 3T )3 J ëê 0
0
Þ dt =
Ma 27T04 J
s 4 pR 2 ´ 80T04
Solving we get,
Time taken for the sphere to cool down temperature 2T 0,
t=
Ma
æ 16 ö
ln ç ÷
16pR s è 3 ø
2
P-165
Thermal Properties of Matter
36. (a) From question,
In 1 sec heat gained by water
= 1 KW – 160 J/s
= 1000 J/s – 160 J/s
= 840 J/s
Total heat required to raise the temperature of water (volume 2L) from 27°c to 77°c
= mwater ×sp. ht × Dq
= 2 × 103 × 4.2 × 50 [Q mass = density × volume]
And, 840 × t = 2 × 103 × 4.2 × 50
2 ´103 ´ 4.2 ´ 50
840
= 500 s = 8 min 20s
37. (d) When radius is decrease by DR,
or, t =
4pR 2 DRrL = 4pT[R 2 - (R - DR) 2 ]
Þ rR2 DRL = T[R 2 - R 2 + 2RDR - DR 2 ]
Þ rR 2 DRL = T2RDR
[ DR is very small]
2T
rL
38. (d) As the surrounding is identical, vessel is identical
time taken to cool both water and liquid (from 30°C to 25°C)
is same 2 minutes, therefore
ÞR=
æ dQ ö
æ dQ ö
=ç ÷
çè ÷ø
dt water è dt ø liquid
(mw C w + W)DT (ml Cl + W) D T
=
t
t
(W = water equivalent of the vessel)
or , m w C w = m l C l
=
L
L
+ k ´ 0´
2
2
L
L
9k ´ + k ´
2
2
9k ´ 100 ´
=
900
kL
= 2
= 90°C
10kL
2
42. (a) According to Stefan’s law
E = s T4
Heat radiated per unit area in 1 hour (3600s) is
= 5.7× 10–8 × (300)4 × 3600 = 1.7 × 1010 J
43. (a) DU = DQ = mcDT
100
× 4184 (50 – 30) » 8.4 kJ
1000
44. (d) Required work = energy released
=
ò
Here, Q = mc dT
4
æ T3 ö
= ò 0.1 ´ 32 ´ ç
dT =
3÷
è
400
ø
20
4
òT
3
4
3.2
ò 64 ´ 106 T
3
dT
20
dT = 0.002kJ
Therefore, required work = 0.002 kJ
45. (a) Let Q be the temperature at a distance x from hot end
of bar. Let Q is the temperature of hot end.
The heat flow rate is given by
mWCW
ml
dQ kA(q1 - q)
=
dt
x
39. (b) As 1g of steam at 100°C melts 8g of ice at 0°C.
10 g of steam will melt 8× 10 g of ice at 0°C
Water in calorimeter = 500 + 80 + 10g = 590g
40. (c)
41. (a)
L
Copper
K copper lsteel + Ksteel lcopper
20
50 ´ 1
= 0.5 kcal / kg
100
100°C
K copper qcopper lsteel + K steel qsteel lcopper
q=
= 5 ´ 10 –8
or,
\ Specific heat of liquid , Cl =
From formula temperature of junction;
Steel
0°C
L/2
L/2
Let conductivity of steel Ksteel = k then from question
Conductivity of copper Kcopper = 9k
qcopper = 100°C
qsteel = 0°C
L
lsteel = lcopper =
2
x dQ
kA dt
Thus, the graph of Q versus x is a straight line with a positive
intercept and a negative slope.
The above equation can be graphically represented by
option (a).
46. (d) Let T be the temperature of the interface.
In the steady state, Q1 = Q2
Þ q - q = x dQ
1
kA dt
T1
\
Þ q = q1 -
1
2
K1
K2
T2
K1 A(T1 - T ) K 2 A(T - T2 )
,
=
l1
l2
P-166
Physics
where A is the area of cross-section.
Þ
K1 A(T1 - T )l 2 = K 2 A(T - T2 )l1
Þ
K1T1l 2 - K1T l 2 = K 2T l1 - K 2T2 l1
Þ
( K 2 l1 + K1l 2 )T = K1T1l 2 + K2T2 l1
K1T1l 2 + K 2T2 l1
K 2 l1 + K1l 2
Þ T =
=
K1l 2T1 + K 2 l1T2
.
K1l 2 + K 2 l1
47. (b) From stefan's law, total power radiated by Sun, E = sT4 ×
4pR2
The intensity of power Per unit area incident on earth's
surface
=
4
sT ´ 4pR
2
E
× Cross – Section area of earth facing the
4pr 2
4 2
sun =
dQ
KA[(T - dT ) - T ] - KAdT
=
=
dt
dr
dr
dT
(Q A = 4pr 2 )
dr
Since the area of the surface through which heat will
flow is not constant. Integrating both sides between the
limits of radii and temperatures of the two shells, we get
2
= -4pKr
æ dQ ö
çè
÷
dt ø
æ dQ ö
çè
÷
dt ø
r2
1
T2
ò r 2 dr = -4pK ò dT
r1
T1
r2
T2
r1
T1
-2
ò r dr = -4pK
ò dT
2
4pr
Total power received by Earth
E' =
The radial rate of flow of heat through this elementary
shell will be
sT R
2
dQ é 1 1 ù
ê - ú = -4pK [T2 - T 1 ]
dt ë r1 r2 û
or
(pr02 )
r
48. (c) When two gases are mixed to gether then
Heat lost by He gas = Heat gained by N2 gas
n1Cv1 DT1 = n2Cv2 DT2
\
dQ -4pKr1r2 (T2 - T1 )
=
dt
(r2 - r1 )
r r
dQ
µ 1 2
dt
(r2 - r1 )
50. (d) From stefan's law, energy radiated by sun per second
E = sAT 4 ;
3 é7
ù 5
R T0 - T f ú = R éëT f - T0 ùû
2 ëê 3
û 2
\ A µ R2
7T0 - 3T f = 5T f - 5T0
\ E µ R 2T 4
Þ 12T0 = 8T f Þ T f =
Þ Tf =
12
T0
8
\
3
T0 ..
2
E2 R22 T24
=
E1 R12 T14
put R2 = 2R, R1 = R ; T2 = 2T, T1 = T
49. (d)
T - dT
dr
·
T1
r1
r
T2
r2
Consider a thin concentric shell of thickness (dr) and of
radius (r) and let the temperature of inner and outer
surfaces of this shell be T and (T – dT) respectively.
E
(2R )2 (2T ) 4
Þ 2 =
= 64
E1
R 2T 4
51. (d) The thermal resistance is given by
x
4x
x
2 x 3x
+
=
+
=
KA 2KA KA KA KA
Amount of heat flow per second,
(T - T1 ) KA
dQ D T
=
= 2
x
3
dt
3x
KA
=
1 ì A(T2 - T1 ) K ü
í
ý
3î
x
þ
\f =
1
3
P-167
Thermal Properties of Matter
52. (d) Wein’s law correctly explains the spectrum
53. (b) Heat required for raising the temperature of a body
through 1ºC is called its thermal capacity.
54. (b) Pyrometer is used to detect infra-red radiation.
55. (a) Black body is one which absorb all incident radiation.
Black board paint is quite approximately equal to black
bodies.
56. (c) When water is cooled at 0°C to form ice, energy is
released from water in the form of heat. As energy is
equivalent to mass, therefore, when water is cooled to
ice, its mass decreases.
57. (a) From stefan's law, the energy radiated per second is
given by E = esT4 A
Here, T = temperature of the body
A = surface area of the body
For same material e is same. s is stefan's constant
Let T1 and T2 be the temperature of two spheres. A1 and
A2 be the area of two spheres.
\
=
E1 T14 A1 T14 4pr12
=
=
E2 T24 A2 T24 4pr22
(4000)4 ´12
(2000)4 ´ 42
=
Þ 40 - T =
\T =
T
Þ 200 - 5T = T
5
200
= 33.3°C
6
æ dT ö
4
59. (b) Rate of Heat loss = mS çè ÷ø = esAT
dt
æ dT ö
æ dT ö
Þ ç- ÷ > ç- ÷
è dt ø A è dt ø B
So, A cools down at faster rate.
60. (a) According to Newton’s law of cooling,
æ q1 - q2 ö
æ q1 + q2
ö
- q0 ÷
çè
÷ø = K çè
ø
t
2
æ 60 - 50 ö
æ 60 + 50
ö
- 25÷
çè
÷ø = K çè
ø
10
2
..... (i)
..... (ii)
q1 - q2
é q + q2
ù
= -K ê 1
- q0 ú
t
2
ë
û
where q0 is the temperature of surrounding.
Now, hot water cools from 60°C to 50°C in 10 minutes,
...(i)
Let T be the temperature of sphere after next 5 minutes.
Then
40 - T 40 + T - 40 T
=
=
10
50 + 40 - 40 50
rB S B
103
4000
´
=
´
r A S A 8 ´ 102 2000
61. (b) By Newton’s law of cooling
Here, T1 = 50°C, T2 = 40°C
an d To = 20°C, t = 600S = 5 minutes
Dividing eqn. (ii) by (i), we get
æ dT ö
çè - ÷ø
dt B
=
10
60
=
Þ q = 42.85°C @ 43°C
(50 - q) q
T1 - T2
éT + T
ù
= K ê 1 2 - T0 ú
t
2
ë
û
40 - T
æ 40 + T
ö
= Kç
- 20 ÷
5
è 2
ø
æ dT ö
çè - ÷ø
dt A
Dividing eq. (i) by (ii),
1
1
50 - 40
æ 50 + 40
ö
= Kç
- 20 ÷
5 Min
è 2
ø
dT es ´ A ´ T 4
dT
1
=
Þµ
dt
r ´ Vol. ´ S
dt rS
æ 50 - q ö
æ 50 + q
ö
- 25÷
and, çè
÷ø = K çè
ø
10
2
58. (b) From Newton's Law of cooling,
Þ
-
...(ii)
60 - 50
é 60 + 50
ù
= -K ê
- q0 ú
10
2
ë
û
...(i)
Again, it cools from 50°C to 42°C in next 10 minutes.
50 - 42
é 50 + 42
ù
= -K ê
- q0 ú
10
ë 2
û
...(ii)
Dividing equations (i) by (ii) we get
55 - q0
1
=
0.8 46 - q0
10 55 - q 0
=
8 46 - q 0
460 - 10q0 = 440 - 8q0
2q0 = 20
q0 = 10°C
P-168
Physics
62. (b) From Newton’s law of cooling,
t=
63.
æ q - q0 ö
1
log e ç 2
÷
k
è q1 - q0 ø
64.
1
(40 - 30)
log e
k
(80 - 30)
And, t =
1
(32 - 30)
log e
(62 - 30)
k
...(1)
Þ
...(2)
Þ
Dividing equation (2) by (1),
1
log e
t k
=
5 1
log e
k
(a) According to newton's law of cooling
dq
= - k(q - q0 )
dt
From question and above equation,
5=
(c) According to Newton’s law of cooling, the
temperature goes on decreasing with time non-linearly.
(32 - 30)
(62 - 30)
(40 - 30)
(80 - 30)
On solving we get, time taken to cool down from 62°C
to 32°C, t = 8.6 minutes.
dq
= - kdt
(q - q0 )
q
ò
q0
t
dq
= - k ò dt
(q - q 0 )
q
Þ
log(q - q0 ) = -kt + c
Which represents an equation of straight line.
Thus the option (a) is correct.
65. (d) From Newton’s law of cooling -
dQ
µ (Dq)
dt
11
P-169
Thermodynamics
Thermodynamics
TOPIC 1 First Law of Thermodynamics
1.
A gas can be taken from A to B via two different processes
ACB and ADB.
(c)
5.
6.
2.
3.
4.
When path ACB is used 60 J of heat flows into the system
and 30J of work is done by the system. If path ADB is
used work done by the system is 10 J. The heat Flow into
the system in path ADB is :
[9 Jan. 2019 I]
(a) 40 J
(b) 80 J (c) 100 J
(d)
20 J
200g water is heated from 40°C to 60°C. Ignoring the slight
expansion of water, the change in its internal energy is
close to (Given specific heat of water = 4184 J/kgK):
[Online April 9, 2016]
(a) 167.4 kJ (b) 8.4 kJ (c) 4.2 kJ (d) 16.7 kJ
A gas is compressed from a volume of 2m3 to a volume of
1m3 at a constant pressure of 100 N/m2. Then it is heated at
constant volume by supplying 150 J of energy. As a result,
the internal energy of the gas:
[Online April 19, 2014]
(a) increases by 250 J
(b) decreases by 250 J
(c) increases by 50 J
(d) decreases by 50 J
An insulated container of gas has two chambers separated
by an insulating partition. One of the chambers has volume
V1 and contains ideal gas at pressure P1 and temperature
T1. The other chamber has volume V2 and contains
ideal gas at pressure P2 and temperature T2. If the partition
is removed without doing any work on the gas, the
final equilibrium temperature of the gas in the container
will be
[2008]
(a)
T1T2 ( PV
1 1 + P2V2 )
PV
1 1T2 + P2V2T1
(b)
PV
1 1T1 + P2V2T2
PV
1 1 + P2V2
PV
1 1T2 + P2V2T1
PV
1 1 + P2V2
(d)
T1T2 ( PV
1 1 + P2V2 )
PV
T
1 1 1 + P2V2T2
When a system is taken from state i to state f along the
path iaf, it is found that Q =50 cal and W = 20 cal. Along the
path ibf Q = 36 cal. W along the path ibf is
[2007]
a
f
i
b
(a) 14 cal (b) 6 cal
(c) 16 cal (d) 66 cal
A system goes from A to B via two processes I and II as
shown in figure. If DU1 and DU2 are the changes in internal
energies in the processes I and II respectively, then [2005]
p
II
A
B
I
v
(a) relation between DU1 and DU 2 can not be determined
7.
(b)
DU1 = DU 2
(c)
DU 2 < DU1
(d) DU 2 > DU1
Which of the following is incorrect regarding the first law
of thermodynamics?
[2005]
(a) It is a restatement of the principle of conservation of
energy
(b) It is not applicable to any cyclic process
(c) It does not introduces the concept of the entropy
(d) It introduces the concept of the internal energy
TOPIC 2
8.
Specific Heat Capacity and
Thermodynamical Processes
Three different processes that can occur in an ideal
monoatomic gas are shown in the P vs V diagram. The
paths are lebelled as A ® B, A ® C and A ® D. The change
P-170
Physics
in internal energies during these process are taken as EAB,
EAC and EAD and the workdone as WAB, WAC and WAD.
The correct relation between these parameters are :
[5 Sep. 2020 (I)]
D T1>T2
C
P
B
T1
A
T2
V
9.
(a) EAB = EAC < EAD, WAB > 0, WAC = 0, WAD < 0
(b) EAB = EAC = EAD, WAB > 0, WAC = 0, WAD > 0
(c) EAB < EAC < EAD, WAB > 0, WAC > WAD
(d) EAB > EAC > EAD, WAB < WAC < WAD
In an adiabatic process, the density of a diatomic gas
becomes 32 times its initial value. The final pressure of the
gas is found to be n times the initial pressure. The value of
n is :
[5 Sep. 2020 (II)]
1
32
10. Match the thermodynamic processes taking place in a
system with the correct conditions. In the table : DQ is the
heat supplied, DW is the work done and DU is change in
internal energy of the system.
[4 Sep. 2020 (II)]
Process
Condition
(I) Adiabatic
(A) DW = 0
(II) Isothermal
(B) DQ = 0
(a) 32
(III) Isochoric
(b) 326
(c) 128
(d)
(a)
(b)
(c)
(d)
14. Starting at temperature 300 K, one mole of an ideal diatomic
gas (g = 1.4) is first compressed adiabatically from volume
V1
. It is then allowed to expand isobarically to
16
volume 2V2. If all the processes are the quasi-static then
the final temperature of the gas (in °K) is (to the nearest
integer) ______.
[9 Jan. 2020 II]
15. A thermodynamic cycle xyzx is shown on a V-T diagram.
V1 to V2 =
(C) DU ¹ 0, DW ¹ 0,
DQ ¹ 0
(IV) Isobaric
(D) DU = 0
(a) (I)-(A), (II)-(B), (III)-(D), (IV)-(D)
(b) (I)-(B), (II)-(A), (III)-(D), (IV)-(C)
(c) (I)-(A), (II)-(A), (III)-(B), (IV)-(C)
(d) (I)-(B), (II)-(D), (III)-(A), (IV)-(C)
11. A balloon filled with helium (32°C and 1.7 atm.) bursts.
Immediately afterwards the expansion of helium can be
considered as :
[3 Sep. 2020 (I)]
(a) irreversible isothermal (b) irreversible adiabatic
(c) reversible adiabatic
(d) reversible isotherm7al
12. An engine takes in 5 mole of air at 20°C and 1 atm, and
compresses it adiabaticaly to 1/10th of the original
volume. Assuming air to be a diatomic ideal gas made up
of rigid molecules, the change in its internal energy during
this process comes out to be X kJ. The value of X to the
nearest integer is ________.
[NA 2 Sep. 2020 (I)]
13. Which of the following is an equivalent cyclic process
corresponding to the thermodynamic cyclic given in the
figure?
where, 1 ® 2 is adiabatic.
(Graphs are schematic and are not to scale) [9 Jan. 2020 I]
The P-V diagram that best describes this cycle is:
(Diagrams are schematic and not to scale)
[8 Jan. 2020 I]
(a)
(b)
(c)
(d)
16. A litre of dry air at STP expands adiabatically to a volume
of 3 litres. If g = 1.40, the work done by air is:
(31.4 = 4.6555) [Take air to be an ideal gas]
[7 Jan. 2020 I]
(a) 60.7 J (b) 90.5 J (c) 100.8 J (d) 48 J
P-171
Thermodynamics
17. Under an adiabatic process, the volume of an ideal gas gets
doubled. Consequently the mean collision time between
the gas molecule changes from t1 to t2. If
Cp
Cv
= g for this
t2
gas then a good estimate for t is given by:
1
[7 Jan. 2020 I]
(a) 2
(b)
1
2
(c)
æ 1ö
çè ÷ø
2
g
(d)
æ 1ö
çè ÷ø
2
g +1
2
18. A sample of an ideal gas is taken through the cyclic process
abca as shown in the figure. The change in the internal
energy of the gas along the path ca is – 180 J, The gas
absorbs 250 J of heat along the path ab and 60 J along the
path bc. The work down by the gas along the path abc is:
[12 Apr. 2019 I]
(a) 120 J
(b) 130 J (c) 100 J
(d) 140 J
19. A cylinder with fixed capacity of 67.2 lit contains helium
gas at STP. The amount of heat needed to raise the
temperature of the gas by 20oC is : [Given that R = 8.31 J
mol – 1 K – 1]
[10 Apr. 2019 I]
(a) 350 J
(b) 374 J (c) 748 J
(d) 700 J
(a) DQA < DQB, DUA < DUB
(b) DQA > DQB, DUA > DUB
(c) DQA > DQB, DUA = DUB
(d) DQA = DQB; DUA = DUB
23. A thermally insulted vessel contains 150 g of water at 0°C.
Then the air from the vessel is pumped out adiabatically. A
fraction of water turns into ice and the rest evaporates at
0°C itself. The mass of evaporated water will be closed to:
(Latent heat of vaporization of water = 2.10 × 106 J kg–1 and
Latent heat of Fusion of water = 3.36 × 10 5 J kg–1)
[8 April 2019 I]
(a) 150 g
(b) 20 g
(c) 130 g
(d)
35 g
24. The given diagram shows four processes i.e., isochoric,
isobaric, isothermal and adiabatic. The correct assignment
of the processes, in the same order is given by :
[8 Apr. 2019 II]
(a) a d b c
(b) d a c b (c) a d c b
(d) d a b c
25. For the given cyclic process CAB as shown for gas, the
work done is:
[12 Jan. 2019 I]
6.0
5
20. n moles of an ideal gas with constant volume heat capacity
CV undergo an isobaric expansion by certain volume. The
ratio of the work done in the process, to the heat supplied
is:
[10 Apr. 2019 I]
nR
CV + nR
nR
(b) C - nR
V
4nR
(c)
CV - nR
4nR
(d) C + nR
V
(a)
21. One mole of an ideal gas passes through a process where
é 1 æ V ö2 ù
P
=
P
ê1 - ç 0 ÷ ú .
0
pressure and volume obey the relation
êë 2 è V ø úû
Here Po and Vo are constants. Calculate the charge in the
temperature of the gas if its volume changes from Vo to 2Vo.
[10 Apr. 2019 II]
1 Po Vo
5 Po Vo
3 Po Vo
1 Po Vo
(b)
(c)
(d)
2 R
4 R
4 R
4 R
22. Following figure shows two processes A and B for a gas.
If DQA and DQB are the amount of heat absorbed by the
system in two cases, and DUA and DUB are changes in
internal energies, respectively, then: [9 April 2019 I]
(a)
A
C
4
p(Pa) 3
2
B
1
1
2
3
4
5
V(m3)
(a) 30 J
(b) 10 J
(c) 1 J
(d) 5 J
26. A rigid diatomic ideal gas undergoes an adiabatic process
at room temperature. The relation between temperature and
volume for this process is TVx = constant, then x is:
[11 Jan. 2019 I]
3
2
2
5
(b)
(c)
(d)
5
5
3
3
27. Half mole of an ideal monoatomic gas is heated at constant
pressure of 1 atm from 20°C to 90°C. Work done by gas is
close to: (Gas constant R = 8.31 J/mol-K) [10 Jan. 2019 II]
(a) 581 J
(b) 291 J (c) 146 J
(d) 73 J
28. One mole of an ideal monoatomic gas is taken along the
path ABCA as shown in the PV diagram. The maximum
temperature attained by the gas along the path BC is given
by
[Online April 16, 2018]
(a)
P-172
Physics
V
P
3P0
b
C
P0 A
V0
2V0
P
P
d
(a)
9P0 V0
9P0 V0
9P0 V0
3P0 V0
(b)
(c)
(d)
2nR
nR
4nR
2nR
31. The ratio of work done by an ideal monoatomic gas to the
heat supplied to it in an isobaric process is :
[Online April 9, 2016]
(a)
2
3
3
2
(a)
(b)
(c)
(d)
5
2
5
3
32. Consider an ideal gas confined in an isolated closed
chamber. As the gas undergoes an adiabatic expansion,
the average time of collision between molecules increases
as Vq, where V is the volume of the gas. The value of q
Cp ö
æ
is : ç g =
[2015]
÷
Cv ø
è
3g + 5
3g - 5
g +1
g -1
(a)
(b)
(c)
(d)
6
6
2
2
33. Consider a spherical shell of radius R at temperature T. The
black body radiation inside it can be considered as an ideal
gas of photons with internal energy per unit volume u =
U
1æUö
µ T 4 and pressure p = ç ÷ . If the shell now
V
3è V ø
undergoes an adiabatic expansion the relation between T
and R is :
[2015]
Tµ
1
(c) T µ e–R (d) T µ e–3R
R3
34. An ideal gas goes through a reversible cycle a®b®c®d
has the V - T diagram shown below. Process d®a and
b®c are adiabatic.
a
c
a
(b)
b
b
c
d
V
V
P
P
d
c
a
b
B
V0 2V0 V
(b)
T
(c)
P0
a
The corresponding P - V diagram for the process is (all
figures are schematic and not drawn to scale) :
[Online April 10, 2015]
A
2P0
1
R
d
V
(a) 25 P0 V0 (b) 25 P0 V0 (c) 25 P0 V0 (d) 5 P0V0
8 R
4 R
16 R
8 R
29. One mole of an ideal monoatomic gas is compressed
isothermally in a rigid vessel to double its pressure at room
temperature, 27°C. The work done on the gas will be:
[Online April 15, 2018]
(a) 300R ln 6
(b) 300R
(c) 300R ln 7
(d) 300R ln 2
30. 'n' moles of an ideal gas undergoes a process A ® B as
shown in the figure. The maximum temperature of the gas
during the process will be :
[2016]
P
(a) T µ
c
B
(d)
V
a
b
d
c
V
35. One mole of a diatomic ideal gas undergoes a cyclic process
ABC as shown in figure. The process BC is adiabatic. The
temperatures at A, B and C are 400 K, 800 K and 600 K
respectively. Choose the correct statement:
[2014]
B
800 K
P
A
400 K
600 k
C
V
(a) The change in internal energy in whole cyclic process
is 250 R.
(b) The change in internal energy in the process CA is 700 R.
(c) The change in internal energy in the process AB is 350 R.
(d) The change in internal energy in the process BC is –
500 R.
36. An ideal monoatomic gas is confined in a cylinder by a spring
loaded piston of cross section 8.0 × 10–3 m2. Initially the gas
is at 300 K and occupies a volume of 2.4 × 10–3 m3 and the
spring is in its relaxed state as shown in figure. The gas is
heated by a small heater until the piston moves out slowly
by 0.1 m. The force constant of the spring is 8000 N/m and
the atmospheric pressure is 1.0 × 105 N/m2. The cylinder and
P-173
Thermodynamics
the piston are thermally insulated. The piston and the spring
are massless and there is no friction between the piston and
the cylinder. The final temperature of the gas will be:
(Neglect the heat loss through the lead wires of the heater.
The heat capacity of the heater coil is also negligible).
[Online April 11, 2014]
41. Helium gas goes through a cycle ABCDA (consisting of
two isochoric and isobaric lines) as shown in figure. The
efficiency of this cycle is nearly : (Assume the gas to be
close to ideal gas)
[2012]
B
2P0
(a) 15.4 %
(b) 9.1 %
P0
Po To R
(a)
Po - a
Po To R
(b)
Po + a
(c) PoToRIn 2
(d) PoToR
39. A certain amount of gas is taken through a cyclic process
(A B C D A) that has two isobars, one isochore and one
isothermal. The cycle can be represented on a P-V indicator
diagram as :
[Online April 22, 2013]
P
(a)
A
B
C
B
P
D
B
A
A
)
(
B
C
(d)
A
(b)
)
D
V
V
40. An ideal gas at atmospheric pressure is adiabatically
compressed so that its density becomes 32 times of its
initial value. If the final pressure of gas is 128 atmospheres,
the value of ‘g’of the gas is :
[Online April 22, 2013]
(a) 1.5
(b) 1.4
(c) 1.3
(d) 1.6
)
(
a 2V 2 2
m -1
2
(
)
a 2
aV 2 2
m -1
(d)
m -1
2
2
44. n moles of an ideal gas undergo a process A ® B as shown
in the figure. Maximum temperature of the gas during the
process is
[Online May 12, 2012]
(c)
A
2P0
D
C
D
(
aV 2
m -1
2
B
P0
V
P
(c)
(a)
C
V
P
(d) 12.5 %
2V0
V0
42. An ideal monatomic gas with pressure P, volume V and
temperature T is expanded isothermally to a volume 2V
and a final pressure Pi. If the same gas is expanded
adiabatically to a volume 2V, the final pressure is Pa. The
P
ratio a is
[Online May 26, 2012]
Pi
(a) 2–1/3
(b) 21/3
(c) 22/3
(d) 2–2/3
43. The pressure of an ideal gas varies with volume as P = aV,
where a is a constant. One mole of the gas is allowed to
undergo expansion such that its volume becomes ‘m’ times
its initial volume. The work done by the gas in the process
is
[Online May 19, 2012]
P
(b)
D
A
(c) 10.5%
(a) 300 K
(b) 800 K (c) 500 K
(d) 1000 K
37. During an adiabatic compression, 830 J of work is done on
2 moles of a diatomic ideal gas to reduce its volume by
50%. The change in its temperature is nearly:
(R = 8.3 JK–1 mol–1)
[Online April 11, 2014]
(a) 40 K
(b) 33 K (c) 20 K
(d) 14 K
38. The equation of state for a gas is given by PV = nRT + aV,
where n is the number of moles and a is a positive constant.
The initial temperature and pressure of one mole of the gas
contained in a cylinder are To and Po respectively. The
work done by the gas when its temperature doubles
isobarically will be:
[Online April 9, 2014]
C
V0 2V
0
3P0V0
9 P0V0
9 P0V0
(b)
(c)
(d)
nR
2 nR
2 nR
4 nR
45. This question has Statement 1 and Statement 2. Of the four
choices given after the Statements, choose the one that
best describes the two Statements.
Statement 1: In an adiabatic process, change in internal
energy of a gas is equal to work done on/by the gas in the
process.
Statement 2: The temperature of a gas remains constant
in an adiabatic process.
[Online May 7, 2012]
(a) Statement 1 is true, Statement 2 is true, Statement 2 is a
correct explanation of Statement 1.
(b) Statement 1 is true, Statement 2 is false.
(a)
9P0V0
V
P-174
Physics
(c) Statement 1 is false, Statement 2 is true.
(d) Statement 1 is false, Statement 2 is true, Statement 2 is
not a correct explanation of Statement 1.
46. A container with insulating walls is divided into equal parts
by a partition fitted with a valve. One part is filled
with an ideal gas at a pressure P and temperature T, whereas
the other part is completly evacuated. If the valve is
suddenly opened, the pressure and temperature of the gas
will be :
[2011 RS]
T
P
P T
,T
(b) P, T
(c) P,
(d)
,
2
2
2 2
Directions for questions 47 to 49: Questions are based on
the following paragraph.
Two moles of helium gas are taken over the cycle ABCDA, as
shown in the P-T diagram.
[2009]
(a)
5
2 × 10
A
B
D
C
300K
500K
P (Pa)
1 × 10
5
T
47. Assuming the gas to be ideal the work done on the gas in
taking it from A to B is
(a) 300 R
(b) 400 R (c) 500 R
(d) 200 R
48. The work done on the gas in taking it from D to A is
(a) + 414 R (b) – 690 R (c) + 690 R
(d) – 414 R
49. The net work done on the gas in the cycle ABCDA is
(a) 279 R
(b) 1076 R (c) 1904 R
(d) zero
50. The work of 146 kJ is performed in order to compress one
kilo mole of gas adiabatically and in this process the
temperature of the gas increases by 7°C. The gas is [2006]
(R = 8.3 J mol–1 K–1)
(a) diatomic
(b) triatomic
(c) a mixture of monoatomic and diatomic
(d) monoatomic
51. Which of the following parameters does not characterize
the thermodynamic state of matter?
[2003]
(a) Temperature
(b) Pressure
(c) Work
(d) Volume
Carnot Engine, Refrigerators
TOPIC 3 and Second Law of
Thermodynamics
52. An engine operates by taking a monatomic ideal gas through
the cycle shown in the figure. The percentage efficiency of
the engine is close is ______.
[NA 6 Sep. 2020 (II)]
3 PO
B
C
A
D
2 PO
PO
VO
2VO
53. If minimum possible work is done by a refrigerator in
converting 100 grams of water at 0°C to ice, how much heat
(in calories) is released to the surroundings at temperature
27°C (Latent heat of ice = 80 Cal/gram) to the nearest integer?
[NA 3 Sep. 2020 (II)]
54. A heat engine is involved with exchange of heat of 1915 J,
– 40 J, +125 J and – Q J, during one cycle achieving an
efficiency of 50.0%. The value of Q is :
[2 Sep. 2020 (II)]
(a) 640 J
(b) 40 J
(c) 980 J
(d) 400 J
1
55. A Carnot engine having an efficiency of
is being used
10
as a refrigerator. If the work done on the refrigerator is 10
J, the amount of heat absorbed from the reservoir at lower
temperature is:
[8 Jan. 2020 II]
(a) 99 J
(b) 100 J (c) 1 J
(d) 90 J
56. A Carnot engine operates between two reservoirs of
temperatures 900 K and 300 K. The engine performs 1200 J
of work per cycle. The heat energy (in J) delivered by the
engine to the low temperature reservoir, in a cycle, is
_______.
[NA 7 Jan. 2020 I]
57. Two ideal Carnot engines operate in cascade (all heat
given up by one engine is used by the other engine to
produce work) between temperatures, T1 and T2. The
temperature of the hot reservoir of the first engine is T1
and the temperature of the cold reservoir of the second
engine is T2. T is temperature of the sink of first engine
which is also the source for the second engine. How is
T related to T1 and T2, if both the engines perform equal
amount of work ?
[7 Jan. 2020 II]
2T1T2
T1 + T2
(a) T = T + T
(b) T =
2
1
2
(c) T = T1T2
(d) T = 0
58. A Carnot engine has an efficiency of 1/6. When the
temperature of the sink is reduced by 62oC, its efficiency
is doubled. The temperatures of the source and the sink
are, respectively.
[12 Apr. 2019 II]
o
o
o
(a) 62 C, 124 C
(b) 99 C, 37oC
o
o
(c) 124 C, 62 C
(d) 37oC, 99oC
59. Three Carnot engines operate in series between a heat
source at a temperature T1 and a heat sink at temperature
T4 (see figure). There are two other reservoirs at temperature
T2 and T3, as shown, with T1 > T2 > T3 > T(4) The three
engines are equally efficient if:
[10 Jan. 2019 I]
P-175
Thermodynamics
( )
= (T T ) ; T = (T T )
= (T T ) ; T = (T T )
= ( T T ) ; T = (T T )
(a) T2 = ( T1 T4 )
1/2
(b) T2
(c) T2
(d) T2
2
1 4
2
1 4
3
1 4
; T3 = T12 T4
1/3
3
2
1 4
3
2
1
4
3
1
3
4
1/3
1/3
(ii) Sequentially keeping in contact with 8 reservoirs such
that each reservoir supplies same amount of heat.
1/3
In both the cases body is brought from initial temperature
100°C to final temperature 200°C. Entropy change of the
body in the two cases respectively is :
1/3
1/4
1/4
60. Two Carnot engines A and B are operated in series. The
first one, A receives heat at T1 (= 600 K) and rejects to a
reservoir at temperature T2. The second engine B receives
heat rejected by the first engine and in turn, rejects to a
heat reservoir at T3 (= 400 K). Calculate the temperature
T2 if the work outputs of the two engines are equal:
[9 Jan. 2019 II]
(a) 600 K
(b) 400 K (c) 300 K
(d) 500 K
61. A Carnot's engine works as a refrigerator between 250 K
and 300 K. It receives 500 cal heat from the reservoir at the
lower temperature.The amount of work done in each cycle
to operate the refrigerator is: [Online April 15, 2018]
(a) 420 J
(b) 2100 J (c) 772 J
(d) 2520 J
62. Two Carnot engines A and B are operated in series. Engine
A receives heat from a reservoir at 600K and rejects heat to
a reservoir at temperature T. Engine B receives heat rejected
by engine A and in turn rejects it to a reservoir at 100K. If
the efficiencies of the two engines A and B are represented
h
by hA and hB respectively, then what is the value of A
hB
[Online April 15, 2018]
12
12
5
7
(b)
(c)
(d)
7
5
12
12
63. An engine operates by taking n moles of an ideal gas
through the cycle ABCDA shown in figure. The thermal
efficiency of the engine is : (Take Cv =1.5 R, where R is gas
constant)
[Online April 8, 2017]
(a)
2P0
(a) 0.24
(b) 0.15
B
C
P
P0
A
D
(c) 0.32
V0
2V0
V
(d) 0.08
64. A Carnot freezer takes heat from water at 0°C inside it and
rejects it to the room at a temperature of 27°C. The latent
heat of ice is 336 × 103 J kg–1. If 5 kg of water at 0°C is
converted into ice at 0°C by the freezer, then the energy
consumed by the freezer is close to :
[Online April 10, 2016]
(a) 1.51 × 105 J
(b) 1.68 × 106 J
(c) 1.71 × 107 J
(d) 1.67 × 105 J
65. A solid body of constant heat capacity 1 J/°C is being heated
by keeping it in contact with reservoirs in two ways : [2015]
(i) Sequentially keeping in contact with 2 reservoirs such
that each reservoir supplies same amount of heat.
(a) ln2, 2ln2
(b) 2ln2, 8ln2
(c) ln2, 4ln2
(d) ln2, ln2
66. A Carnot engine absorbs 1000 J of heat energy from a
reservoir at 127°C and rejects 600 J of heat energy during
each cycle. The efficiency of engine and temperature of
sink will be:
[Online April 12, 2014]
(a) 20% and – 43°C
(b) 40% and – 33°C
(c) 50% and – 20°C
(d) 70% and – 10°C
67.
p
2p0
p0
A
D
v0
B
C
2v0 v
The above p-v diagramrepresents the thermodynamic cycle
of an engine, operating with an ideal monatomic gas. The
amount of heat, extracted from the source in a single cycle
is
[2013]
(a) p 0 v 0
æ 13 ö
(b) ç ÷ p0 v0
è2ø
æ 11 ö
(c) ç ÷ p0 v0
è2ø
(d) 4p0v0
68. A Carnot engine, whose efficiency is 40%, takes in heat
from a source maintained at a temperature of 500K. It is
desired to have an engine of efficiency 60%. Then, the
intake temperature for the same exhaust (sink) temperature
must be :
[2012]
(a) efficiency of Carnot engine cannot be made larger than 50%
(b) 1200 K
(c) 750 K
(d) 600 K
69. The door of a working refrigerator is left open in a
well insulated room. The temperature of air in the room
will
[Online May 26, 2012]
(a) decrease
(b) increase in winters and decrease in summers
(c) remain the same
(d) increase
70. This question has Statement 1 and Statement 2. Of the four
choices given after the Statements, choose the one that
best describes the two Statements.
Statement 1: An inventor claims to have constructed an
engine that has an efficiency of 30% when operated
between the boiling and freezing points of water. This is
not possible.
P-176
Physics
Statement 2: The efficiency of a real engine is always
less than the efficiency of a Carnot engine operating
between the same two temperatures.
[Online May 19, 2012]
(a) Statement 1 is true, Statement 2 is true, Statement 2 is
not the correct explanation of Statement 1.
(b) Statement 1 is true, Statement 2 is false.
(c) Statement 1 is false, Statement 2 is true.
(d) Statement 1 is true, Statement 2 is true, Statement 2 is
the correct explanation of Statement 1.
71. A Carnot engine operating between temperatures T1 and T2
has efficiency
increases to
75.
76.
1
. When T2 is lowered by 62 K its efficiency
6
1
. Then T1 and T2 are, respectively:
3
[2011]
(a) 372 K and 310 K
(b) 330 K and 268 K
(c) 310 K and 248 K
(d) 372 K and 310 K
72. A diatomic ideal gas is used in a Carnot engine as the
working substance. If during the adiabatic expansion part
of the cycle the volume of the gas increases from V to 32
V, the efficiency of the engine is
[2010]
(a) 0.5
(b) 0.75
(c) 0.99
(d) 0.25
73. A Carnot engine, having an efficiency of h = 1/10 as heat
engine, is used as a refrigerator. If the work done on the
system is 10 J, the amount of energy absorbed from the
reservoir at lower temperature is
[2007]
(a) 100 J
(b) 99 J
(c) 90 J
(d) 1 J
74. The temperature-entropy diagram of a reversible engine
cycle is given in the figure. Its efficiency is
[2005]
T
2T0
T0
S0
2S0
S
1
2
1
1
(b)
(c)
(d)
3
2
4
3
Which of the following statements is correct for any
thermodynamic system ?
[2004]
(a) The change in entropy can never be zero
(b) Internal energy and entropy are state functions
(c) The internal energy changes in all processes
(d) The work done in an adiabatic process is always zero.
“Heat cannot by itself flow from a body at lower temperature
to a body at higher temperature” is a statement or
consequence of
[2003]
(a) second law of thermodynamics
(b) conservation of momentum
(c) conservation of mass
(d) first law of thermodynamics
A Carnot engine takes 3 × 106 cal of heat from a reservoir at
627°C, and gives it to a sink at 27°C. The work done by the
engine is
[2003]
(a) 4.2 × 106 J
(b) 8.4 × 106 J
(c) 16.8 × 106 J
(d) zero
Which statement is incorrect?
[2002]
(a) All reversible cycles have same efficiency
(b) Reversible cycle has more efficiency than an
irreversible one
(c) Carnot cycle is a reversible one
(d) Carnot cycle has the maximum efficiency in all cycles
Even Carnot engine cannot give 100% efficiency because
we cannot
[2002]
(a) prevent radiation
(b) find ideal sources
(c) reach absolute zero temperature
(d) eliminate friction
(a)
77.
78.
79.
P-177
Thermodynamics
1.
2.
5.
(a)
DU remains same for both paths ACB and ADB
DQACB = DWACB + DUACB
Þ 60 J = 30 J + DUACB
Þ UACB = 30 J
\ DUADB = DUACB = 30 J
DQADB = DUADB + DWADB
= 10 J + 30 J = 40 J
(d) Volume of water does not change, no work is done on
or by the system (W = 0)
According to first law of thermodynamics
Q < ΧU ∗ W
3.
For Isochoric process Q < ΧU
DU = mcdT = 2 × 4184 × 20 = 16.7 kJ.
(a) As we know,
(Ist law of thermodynamics)
DQ = D u + D w
6.
7.
8.
Þ DQ = D u + P D v
or 150 = Du + 100 (1 - 2 )
4.
= Du - 100
\ Du = 150 + 100 = 250J
Thus the internal energy of the gas increases by 250 J
(a) Here Q = 0 and W = 0. Therefore from first law of
thermodynamics DU = Q + W = 0
Internal energy of first vessle + Internal energy of second
vessel = Internal energy of combined vessel
n1Cv T1 + n2 Cv T2 = (n1 + n2 )Cv T
\T =
n1T1 + n2 T2
n1 + n2
For first vessel n1 =
PV
1 1 and for second vessle
RT1
PV
n2 = 2 2
RT2
PV
1 1 ´ T + P2V2 ´ T
1
2
RT1
RT2
\T=
PV
1 1 + P2V2
RT1 RT2
=
T1T2 ( PV
1 1 + P2V2 )
PV
T
1 1 2 + P2V2T1
(b) For path iaf,
Q1 = 50 cal, W1 = 20 cal
By first law of thermodynamics, a
f
DU = Q1 – W1 = 50 – 20 = 30 cal.
For path ibf
Q2 = 36 cal
i
b
W2 = ?
DUibf = Q2 – W2
Since, the change in internal energy does not depend on the
path, therefore DUiaf = DUibf
DUiaf = DUibf
Þ 30 = Q2 – W2
Þ W2 = 36 – 30 = 6 cal.
(b) Change in internal energy is independent of path taken
by the process. It only depends on initial and final states i.e.,
DU1 = DU2
(b, c) First law is applicable to a cyclic process. Concept of
entropy is introduced by the second law of thermodynamics.
(b) Temperature change DT is same for all three processes
A ® B; A ® C and A ® D
DU = nCv DT = same
E AB = E AC = E AD
Work done, W = P ´ DV
AB ® volume is increasing Þ WAB > 0
AD ® volume is decreasing Þ WAD < 0
9.
AC ® volume is constant Þ WAC = 0
(c) In adiabatic process
PV g = constant
g
æ mö
\ P ç ÷ = constant
è rø
As mass is constant
æ
mö
çèQ V = r ø÷
\ P µ rg
If Pi and Pf be the initial and final pressure of the gas and
ri and r f be the initial and final density of the gas. Then
g
ærf ö
= ç ÷ = (32)7 / 5
Pi è ri ø
nP
Þ i = (25 )7 /5 = 27
Pi
Pf
Þ n = 27 = 128.
10. (d)
(I) Adiabatic process : No exchange of heat takes
place with surroundings.
Þ DQ = 0
(II) Isothermal process : Temperature remains constant
P-178
Physics
f
nR DT Þ DU = 0
2
No change in internal energy [DU = 0].
(III) Isochoric process volume remains constant
\ DT = 0 Þ DU =
DV = 0 Þ W = ò P × dV = 0
Hence work done is zero.
(IV) In isobaric process pressure remains constant.
W = P × DV ¹ 0
DU =
f
f
nR DT = [ P DV ] ¹ 0
2
2
\ DQ = nC p DT ¹ 0
11. (b) Bursting of helium balloon is irreversible and in this
process DQ = 0 , so adiabatic.
12. (46)
For adiabatic process, TV g -1 = constant
or, T1V1g -1 = T2V2g -1
T1 = 20°C + 273 = 293 K , V2 =
æV ö
T1 (V1 ) g -1 = T2 ç 1 ÷
è 10 ø
æ 1ö
Þ 293 = T2 ç ÷
è 10 ø
7
V1
and g =
5
10
g -1
2/5
14. (1818) For an adiabatic process,
TVg–1 = constant
\ T1V1g –1 = T2V2g –1
1.4 -1
æ ö
çV ÷
Þ T2 = (300) ´ ç 1 ÷
çç V1 ÷÷
è 16 ø
0.4
Þ T2=300×(16)
Ideal gas equation, PV = nRT
nRT
\ V=
P
Þ V = kT (since pressure is constant for isobaric
process)
So, during isobaric process
V2 = kT2
...(i)
2V2 = kTf
...(ii)
Dividing (i) by (ii)
1 T2
=
2 Tf
Tf = 2T2 = 300 × 2 × (16)0.4 =1818 K
15. (a) From the corresponding V-T graph given in question,
Process xy ® Isobaric expansion,
Process yz ® Isochoric (Pressure decreases)
Process zx ® Isothermal compression
Therefore, corresponding PV graph is as shown in figure
Þ T2 = 293(10) 2/ 5 ; 736 K
DT = 736 - 293 = 443 K
During the process, change in internal energy
DU = NCV DT = 5 ´
5
´ 8.3 ´ 443 ; 46 ´ 103 J = X kJ
2
\ X = 46 .
13. (c) For process 3 ® 1 volume is constant
\ Graph given in option (d) is wrong.
And process 1 ® 2 is adiabatic \ graph in option (1) is
wrong
Q v = constant
P ­, T ­
For Process 2 ® 3 Pressure constant i.e., P = constant
\ V¯T¯
Hence graph (c) is the correct V – T graph of given
P – V graph
V
2
3
16. (b) Given, V1 = 1 litre, P1 = 1 atm
V2 = 3 litre, g = 1.40,
g
g
Using, PVr = constant Þ PV
1 1 = P2V2
1.4
æ1ö
Þ P2 = P1 ´ ç ÷
è 3ø
T
1
atm
4.6555
PV – P V
\ Work done, W = 1 1 2 2
g –1
1
æ
ö
´ 3 ÷1.01325 ´ 105 ´ 10 –3
ç1´1 –
4.6555 ø
è
=
= 90.1 J
0.4
Closest value of W = 90.5 J
17. (Bonus) We know that Relaxation time,
Tµ
1
=
V
...(i)
T
Equation of adiabatic process is
TVg–1 = constant
Þ
1
T µ g-1
V
P-179
Thermodynamics
Þ T µ V 1+
Þ T µV
Þ
g –1
2
using (i)
1+g
2
Tf
1+g
2
æ 2V ö
=ç
÷
Ti è V ø
= (2)
1+g
2
18. (b) DUac = – (DUca) = – (– 180) = 180 J
Q = 250 + 60 = 310 J
Now Q = DU + W
or 310 = 180 + W
or W = 130 J
19. (c) As the process is isochoric so,
67.2 3R
´
´ 20 = 90R = 90 ´ 8.31 ; 748 j.
22.4 2
20. (a) At constant volume
Work done (W) = nRDT
Heat given Q = CvDT + nRDT
Q = nc v DT =
So, \
W
nRDT
nR
=
=
Q Cv DT + nRDT C V + nR
21. (b) We have given,
é 1 æ V ö2 ù
P = P0 ê1 - ç 0 ÷ ú
êë 2 è V ø úû
When V1 = V0
é 1 ù P0
Þ P1 = P0 ê1 - ú =
ë 2û 2
When V2 = 2V0
é 1 æ 1 ö ù æ 7 P0 ö
Þ P2 = P0 ê1 - ç ÷ ú = ç
÷
ë 2 è 4 øû è 8 ø
PV P V é
PV ù
DT = T2 - T1 = 1 1 - 2 2 êQ T =
nR
nR ë
nR úû
æ 1 ö
æ 1 ö æ P0V0 7 P0V0 ö
DT = ç
1 1 - P2V2 ) = ç
÷ ( PV
֍
4 ÷ø
è nR ø
è nR ø è 2
5P V
5P V
= 0 0 = 0 0
(Q n = 1)
4 nR
4R
22. (c) Internal energy depends only on initial and final state
So, DUA = DUB
Also DQ = DU + W
As WA > WB Þ DQA > DQB
23. (b) Suppose amount of water evaporated be M gram.
Then (150 – M) gram water converted into ice.
so, heat consumed in evoporation = Heat released in fusion
M × Lv = (150 – M) × Ls
M × 2.1 × 106 = (150 – M) × 3.36 × 105
Þ M – 20 g
24. (d) a ® Isobasic, b ® Isothermal, c ® Adiabatic,
d ® Isochoric
25. (b) Total work done by the gas during the cycle is equal
to area of triangle ABC.
1
\ DW = ´ 4 ´ 5 = 10 J
2
26. (b) Equation of adiabatic change is
TVg-1 = constant
7
7
Put g = , we get: g - 1 = - 1
5
5
2
\x =
5
27. (b) Work done,
1
W = PDV = nRDT = ´ 8.31´ 70 ; 291J
2
28. (a) Equation of the BC
2P
P = P0 - 0 (V - 2V0 )
V0
using PV = nRT
2P V 2
+ 4P0 V
P0 V - 0
V0
Temperature, T =
1´ R
(Q n = 1 mole given)
P é
2V 2 ù
T = 0 ê5V ú
F ëê
V0 ûú
dT
4V
5
=0Þ5= 0 Þ V = V0
dV
V0
4
P é 5V
2 25 2 ù 25 P0 V0
T = 0 ê5 ´ 0 ´ V0 ú =
Rë
4
V0 16
û 8 R
æ pf ö
29. (d) Work done on gas = nRT ln çç
÷÷ = R(300) ln(2)
è p1 ø
æ Pf
ö
= 2 given ÷÷
= 300 Rln2 ççQ
è pi
ø
30. (c) The equation for the line is
P
3Po
c
2Po
Po
q
Po
q
Vo
Vo
2Vo
V
-P0
- P0
P = V V + 3P
[slope = V , c = 3P0]
0
0
PV0 + P0V = 3P0V0
But pV = nRT
nRT
\P=
V
nRT
From (i) & (ii)
V0 + P0V = 3P0V0
V
...(i)
...(ii)
P-180
Physics
nRT
V0 + P0V = 3P0V0
V
2
\ nRT V0 + P0V = 3P0V0V
...(iii)
From (i) & (ii)
dT
For temperature to be maximum
=0
dV
Differentiating e.q. (iii) by ‘V’ we get
dT
nRV0
+ P0(2V) = 3P0V0
dV
dT
\ nRV0
= 3P0V0 – 2 P0V
dV
dT
3P0 V0 - 2P0 V
=0
=
dV
nRV0
3V0
V=
2
\
3P
P= 0
2
[From (i)]
9P0 V0
[From (iii)]
4nR
(a) Efficiency of heat engine is given by
C
w
R
R
2
h = = 1- V =
=
=
5R
Q
CP
Cp
5
2
(Q Cp – Cv = R)
5
For monoatomic gas C P = R .
2
1
(a) t =
æ N ö 3RT
2pd2 ç ÷
èVø M
\ Tmax =
8.
9.
t µ
V
T
As, TVg–1 = K
So, t µ Vg + 1/2
Therefore, q =
g+ 1
2
1æU ö
10. (a) As, P = ç ÷
3èV ø
U
= KT 4
But
V
1
4
So, P = KT
3
uRT 1
= KT 4 [As PV = u RT]
or
V
3
4
3 3
p R T = constant
3
1
Therefore, Tµ
R
11. (b) In VT graph
ab-process : Isobaric, temperature increases.
bc process : Adiabatic, pressure decreases.
cd process : Isobaric, volume decreases.
da process : Adiabatic, pressure increases.
The above processes correctly represented in P-V diagram (b).
12. (d) In cyclic process, change in total internal energy is zero.
DUcyclic = 0
5R
DT
2
Where, Cv = molar specific heat at constant volume.
For BC, DT = –200 K
\ DUBC = –500R
13. (c)
14. (c) Given : work done, W = 830 J
No. of moles of gas, m = 2
For diatomic gas g = 1.4
Work done during an adiabatic change
DUBC = nCv DT = 1 ´
W=
mR (T1 - T2 )
g -1
Þ 830 =
Þ DT =
2 ´ 8.3( DT ) 2 ´ 8.3(DT )
=
1.4 - 1
0.4
830 ´ 0.4
= 20 K
2 ´ 8.3
15. (a)
16. (c) P-V indicator diagram for isobaric
P
slope
dP
=0
dV
V
P-V indicator diagram for isochoric process
P
slope
dP
=¥
dV
V
P-V indicator diagram for isothermal process
P
slope
dP -P
=
=
dV V
V
17. (b) Volume of the gas
m
v=
and
d
Using PV g = constant
g
P' V æ d'ö
=
=ç ÷
P V' è d ø
or 128 = (32)g
7
\ g = = 1.4
5
18. (a) The efficiency
output work
h=
heat given to the system
P-181
Thermodynamics
3
3
3
= n RDT = V0 DP = P0V0
2
2
2
n
n
Wi = ( P0V0 ) + (2 P0V0 ) + 2 P0V0
2
2
Heat given in going B to C = nCpDT
5
æ5 ö
= n ç R ÷ DT = (2 P0 )DV
2
è2 ø
= 5P0V0
and W0 = area under PV diagram P0V0
2
W
PV
h= = 0 0 =
13
Q
P0V0 13
2
Efficiency in %
2
200
´ 100 =
; 15.4%
13
13
42. (d) For isothermal process :
PV = Pi .2V
P = 2Pi
...(i)
For adiabatic process
PVg = Pa (2V)g
h=
(Q for monatomic gas g= 5 3 )
5
or,
5
[From (i)]
2Pi V 3 = Pa (2V) 3
Pa
2
= 5
Pi
2 3-2
Pa
=2 3
Þ
Pi
43. (d) Given P = aV
ò
PdV
V
=
ò
aVdV =
V
Vacuum
It is the free expansion
\ So, T remains constant
Þ PV
1 1 = P2V2
Þ P
V
= P2 (V )
2
æ Pö
P2 = ç ÷
è 2ø
47. (b) The process A ® B is isobaric.
\ work done WAB = nR(T2 – T1)
= 2R (500 - 300) = 400 R
48. (a) The process D to A is isothermal as temperature is
constant.
P
Work done, WDA = 2.303nRT log10 D
PA
= 2.303 ´ 2 R ´ 300
log10
1 ´ 105
– 414R.
2 ´ 105
Therefore, work done on the gas is +414 R.
49. (a) The net work in the cycle ABCDA is
W = W AB + WBC + WCD + WDA
PB
PC
= 2.303 ´ 2R ´ 500log
mV
mV
P, T
= 400R + 2.303nRT log
Þ
Work done, w =
46. (d)
aV 2
(m 2 - 1) .
2
44. (b) Work done during the process A ® B
= Area of trapezium (= area bounded by indicator diagram
with V-axis)
1
3
2 P0 + P0 ) ( 2V0 - V0 ) = P0V0
(
2
2
Ideal gas eqn : PV = nRT
=
PV 3P0V0
=
nR
2nR
45. (b) In an adiabatic process, dH = 0
And according to first law of thermodynamics
dH = dU + W
\ W = – dU
+ (-400R) - 414R
2 ´ 105
1 ´ 105
- 414 R
= 693.2 R – 414 R
= 279.2 R
50. (a) Work done in adiabatic compression is given by
nRDT
W=
1- g
1000 ´ 8.3 ´ 7
Þ -146000 =
1- g
58.1
58.1
or 1 - g = Þ g = 1+
= 1.4
146
146
Hence the gas is diatomic.
51. (c) Work is not a state function. The remaining three
parameters are state function.
P
52. (19)
3P0
B
C
Þ T=
P0
A
V0
D
2V0
V
P-182
Physics
From the figure,
Also,
Work, W = 2 P0V0
Heat given, Qin = WAB + WBC = n × CV DTAB + nCP DTBC
3R
n5 R
=n
(TB - TA ) +
(TC - TB )
2
2
3R
5R ö
æ
and CP =
çèQ Cv =
÷
2
2ø
3
5
= ( PBVB - PAV A ) + ( PCVC - PBVB )
2
2
3
5
= ´ [3P0V0 - P0V0 ] + [6 P0V0 - 3P0V0 ]
2
2
15
21
= 3 P0V0 + P0V0 =
P0V0
2
2
2 P0V0
W
4
=
=
21
21
Qin
P0V0
2
400
h% =
» 19.
21
53. (8791)
Given,
Heat absorbed, Q2 = mL = 80 × 100 = 8000 Cal
Temperature of ice, T2 = 273 K
Temperature of surrounding,
T1 = 273 + 27 = 300 K
1
w
=
10 Q1
Þ Q1 = w × 10 = 100 J
So, Q1 – Q2 = w
Þ Q2 = Q1– w
Þ 100 – 10 = Q2 = 90 J
Þ
56. (600.00) Given; T1 = 900 K, T2 = 300K, W = 1200 J
Using, 1 –
Þ 1–
Efficiency, h =
Efficiency =
Þ
w Q1 - Q2 T1 - T2
300 - 273
=
=
=
Q2
Q2
T2
273
Q1 - 8000 27
=
Þ Q1 = 8791 Cal
8000
273
54. (c) Efficiency, h =
=
Work done
W
=
Heat absorbed SQ
Q1 + Q2 + Q3 + Q4
= 0.5
Q1 + Q3
Here, Q1 = 1915 J, Q2 = – 40 J and Q3 = 125 J
\
1915 - 40 + 125 + Q4
= 0.5
1915 + 125
Þ
Þ Q4 = -Q = -980 J
Þ Q = 980 J
55. (d) For carnot refrigerator
Efficiency =
Q1 – Q2
Q1
Where,
Q1 = heat lost from sorrounding
Q2 = heat absorbed from reservoir at low temperature.
T2 W
=
T1 Q1
300 1200
=
900
Q1
2 1200
=
Þ Q1 = 1800
3
Q1
Therefore heat energy delivered by the engine to the low
temperature reservoir, Q2 = Q1 – W = 1800 – 1200 =
600.00 J
57. (b) Let QH = Heat taken by first engine
QL = Heat rejected by first engine
Q2 = Heat rejected by second engine
Work done by 1st engine = work done by 2nd engine
W = QH – QL = QL – Q2 Þ 2QL = QH + Q2
qH q2
+
qL qL
Let T be the temperature of cold reservoir of first engine.
Then in carnot engine.
QH T1
Q
T
= and L =
QL T
Q2 T2
2=
Þ 2=
Þ
T1 T2
+
T T
2T = T1 + T2
58. (b) Using, n = 1 -
Þ 1915 - 40 + 125 + Q4 = 1020
Þ Q4 = 1020 - 2000
Q1 – Q2
w
=
Q1
Q1
n=
using (i)
Þ T=
T1 + T2
2
T2
T1
T2
1
= 1- T
6
1
T2 - 62
T
and 3 = 1 - T
1
On solving, we get
T1 = 99°C and T2 = 37°C
59. (b) According to question, h1 = h2 = h3
T2
T
T
= 1– 3 =1– 4
T1
T2
T3
[Q Three engines are equally efficient]
\1–
P-183
Thermodynamics
Þ
T2 T3 T4
= =
T1 T2 T3
Thermal efficiency of engine (h) =
64. (d) DH = mL = 5 × 336 × 103 = Qsink
Þ T2 = T1T3
...(i)
T3 = T2 T4
From (i) and (ii)
...(ii)
T2 = (T12 T4 )
1
2
T3 = (T1 T4 )
1
60. (d) hA =
Qsink
T
< sink
Qsource Tsource
Tsource
´ Qsink
Tsink
Energy consumed by freezer
[ Qsource <
3
æ Tsource ö÷
ç
,1÷÷
[ w output < Qsource , Qsink < Qsink çç
çè Tsink
ø÷
3
T1 – T2 w A
=
Tl
Q1
Given: Tsource < 27°C ∗ 273 < 300K,
T –T W
and, hB = 2 3 = B
T2
Q2
According to question,
WA = WB
Q T T - T3 T1
\ 1= 1´ 2
=
Q 2 T2 T1 - T2 T2
\T2 =
Tsink < 0°C ∗ 273 < 273 k
ö
3 æ 300
,1÷÷ < 1.67 ´105 J
Woutput = 5´336´10 ççç
è 273 ø÷
Tl + T3
2
600 + 400
2
= 500K
61. (a) Given: Temperature of cold body, T 2 = 250 K
temperature of hot body; T1 = 300 K
Heat received, Q2 = 500 cal work done, W = ?
=
Efficiency = 1 –
W=
T2
W
=
T1 Q2 + W
Þ 1–
W
250
=
300 Q2 + W
Q2 500 ´ 4.2
=
J = 420 J
5
5
62. (d) Efficiency of engine A, nA =
and nB =
W
2
=
= 0.15
Q given 13
T1 - T2
T1
T2 - T3
T +T
; T2 = 1 3 = 350 K
T2
2
600 - 350
nA
7
600
=
=
or
350
100
nB
12
350
63. (b) Work-done (W) = P0V0
According to principle of calorimetry
Heat given = QAB = QBC
= nCVdTAB + nCPdTBC
3
5
= (nRTB - nRTA ) + (nRTC - nRTB )
2
2
3
5
= (2P0 V0 - P0 V0 ) + (4P0 V0 - 2 P0V)
2
2
13
= P0 V0
2
65. (d) The entropy change of the body in the two cases is
same as entropy is a state function.
66. (b) Given :
Q1 = 1000 J
Q2 = 600 J
T1 = 127°C = 400 K
T2 = ?
h=?
Efficiency of carnot engine,
h=
W
´100%
Q1
or, h =
Q2 - Q1
´100%
Q1
or, h =
1000 - 600
´ 100%
1000
h = 40%
Q 2 T2
Now, for carnot cycle Q = T
1
1
T
600
= 2
1000 400
600 ´ 400
1000
= 240 K
= 240 – 273
T2 =
\ T2 = -33°C
67. (b) Heat is extracted from the source in path DA and AB is
DQ =
Þ
3 æ P0V0 ö 5 æ 2 P0V0 ö
R
+ R
2 çè R ÷ø 2 çè R ÷ø
æ 13ö
3
5
P0V0 + 2 P0V0 = ç ÷ P0V0
è 2ø
2
2
P-184
Physics
68. (c) The efficiency of the carnot’s heat engine is given as
æ T ö
h = ç 1 - 2 ÷ ´100
è T1 ø
When efficiency is 40%,
T1 = 500 K; h = 40
T2 ö
æ
40 = ç 1 ÷ ´ 100
500
è
ø
T
40
Þ
= 1- 2
500
100
60
T2
=
Þ T2 = 300 K
Þ
100
500
When efficiency is 60%, then
Þ
2
T
\ 1 = (32) 5 Þ T1 = 4T2
T2
T2
T1
T2
T1
T2
3
1
= 1 - = = 0.75.
4 4
4T2
73. (c) The efficiency (h) of a Carnot engine and the
coefficient of performance (b) of a refrigerator are related
as
= 1-
b=
1- h
h
Also, b =
Q2
W
1 – n Q2
=
n
W
1
Q
10
= 2.
\ b=
æ 1ö W
ç ÷
è 10 ø
is independent of path taken by the process.
1-
T2 5
=
T1 6
When T2 is lowered by 62K, then
T2 - 62
Again, h2 = 1 T1
Þ
....(i)
Q2
10
Þ Q2 = 90 J.
Þ 9=
1
74. (d) Q1 = area under BC = T0 S0 + T0 S0
2
Q2 = area under AC = T0(2S0 – S0) = T0S0
and Q3 = 0
....(ii)
5
× 372 = 310 K
6
T1
Now, efficiency = 1 -
\b=
where, T1 = temperature of source
T2 = temperature of sink
71. (d) Efficiency of engine
T
1
h1 = 1 - 2 =
T1 6
72. (b) P
7
5
2
5
\ g -1 =
Efficiency of Carnot's engine, n = 1 –
T1 = 372 K and T2 =
T1
= (32)g–1
T2
For diatomic gas, g =
æ 300 ö
300
60
40
= ç1 =
÷Þ
T2 ø
T2
100 è
100
100 ´ 300
Þ T2 =
Þ T2 = 750 K
40
69. (d) In a refrigerator, the heat dissipated in the
atmosphere is more than that taken from the cooling
chamber, therefore the room is heated. If the door of
a refrigerator is kept open.
70. (d) According to Carnot's theorem - no heat engine working
between two given temperatures of source and sink can be
more efficient than a perfectly reversible engine i.e. Carnot
engine working between the same two temperatures.
T 62 1
=1– 2 +
=
T1 T1 3
Solving (i) and (ii), we get,
For adiabatic expansion T1V1g-1 = T2V2g-1
Þ T1V g - 1 = T2 (32V )g - 1
Efficiency, h =
T
(V, T1)
(32 V, T2)
B
2T0
Q1
Q3
T0
T2
W Q1 - Q2
=
Q1
Q1
A
Q2
C
V
S0
2S0
P-185
Thermodynamics
Q2
TS
1
= 1- 0 0 =
3
Q1
3
T S
2 0 0
75. (b) Internal energy and entropy are state function, they are
independent of path taken.
76. (a) This is a consequence of second law of
thermodynamics
77. (b) Here, T1 = 627 + 273 = 900 K
T2 = 27 + 273 = 300 K
= 1-
Efficiency, h = 1 = 1-
T2
T1
300
1 2
= 1- =
900
3 3
But h =
W
Q
2
2
W 2
= Þ W = ´ Q = ´ 3 ´ 106
Q 3
3
3
= 2 × 106 cal
= 2 × 106 × 4.2 J = 8.4 × 106 J
78. (a) All reversible engines have same efficiencies if they
are working for the same temperature of source and sink.
If the temperatures are different, the efficiency is
different.
79. (c) In Carnot’s cycle we assume frictionless piston,
absolute insulation and ideal source and sink (reservoirs).
\
The efficiency of carnot’s cycle h = 1 -
T2
T1
The efficiency of carnot engine will be 100% when its
sink (T2) is at 0 K.
The temperature of 0 K (absolute zero) cannot be realised
in practice so, efficiency is never 100%.
12
P-186
Physics
Kinetic Theory
TOPIC 1
1.
2.
3.
4.
Kinetic Theory of an Ideal
Gas and Gas Laws
Initially a gas of diatomic molecules is contained in a
cylinder of volume V1 at a pressure P1 and temperature
250 K. Assuming that 25% of the molecules get
dissociated causing a change in number of moles. The
pressure of the resulting gas at temperature 2000 K,
when contained in a volume 2V1 is given by P2. The ratio
P2/P1 is ______.
[NA Sep. 06, 2020 (I)]
The change in the magnitude of the volume of an ideal gas
when a small additional pressure DP is applied at a constant
temperature, is the same as the change when the
temperature is reduced by a small quantity DT at constant
pressure. The initial temperature and pressure of the gas
were 300 K and 2 atm. respectively. If | DT |= C | DP | ,
then value of C in (K/atm.) is __________.
[NA Sep. 04, 2020 (II)]
The number density of molecules of a gas depends on
their distance r from the origin as , n(r) = n0e–ar4. Then
the total number of molecules is proportional to :
[12 April 2019 II]
–3/4
(a) n0a
(b) n0 a1/2
(c) n0 a1/4
(d) n0 a–3
A vertical closed cylinder is separated into two parts by a
frictionless piston of mass m and of negligible thickness.
The piston is free to move along the length of the cylinder.
The length of the cylinder above the piston is l1, and that
below the piston is l2, such that l1 > l2. Each part of the
cylinder contains n moles of an ideal gas at equal temperature
T. If the piston is stationary, its mass, m, will be given by:
(R is universal gas constant and g is the acceleration due to
gravity)
[12 Jan. 2019 II]
(a)
RT é l1 - 3l2 ù
ê
ú
ng ë l1 I 2 û
RT é 2l1 + l2 ù
(b) g ê l I ú
ë 1 2 û
nRT é l1 - l2 ù
nRT é 1 1 ù
ê
ú
(d)
ê + ú
g ë l1 l2 û
g ë l2 l1 û
The temperature of an open room of volume 30 m3 increases
from 17°C to 27°C due to sunshine. The atmospheric pressure
in the room remains 1 × 105 Pa. If ni and nf are the number of
molecules in the room before and after heating, then nf – ni
will be :
[2017]
(a) 2.5 × 1025
(b) –2.5 × 1025
(c) –1.61 × 1023
(d) 1.38 × 1023
For the P-V diagram given for an ideal gas,
(c)
5.
6.
1
P
P=
Constant
V
2
V
out of the following which one correctly represents the
T-P diagram ?
[Online April 9, 2017]
2
2
T
(a) T
(b)
1
1
P
P
T
T
2
(c)
7.
1
1
2
(d)
P
Chamber I
ideal
gas
1
P
Chamber II
real
gas
2
3
4
P-187
Kinetic Theory
8.
There are two identical chambers, completely thermally
insulated from surroundings. Both chambers have a
partition wall dividing the chambers in two compartments.
Compartment 1 is filled with an ideal gas and
Compartment 3 is filled with a real gas. Compartments 2
and 4 are vacuum. A small hole (orifice) is made in the
partition walls and the gases are allowed to expand in
vacuum.
Statement-1: No change in the temperature of the gas
takes place when ideal gas expands in vacuum. However,
the temperature of real gas goes down (cooling) when it
expands in vacuum.
Statement-2: The internal energy of an ideal gas is only
kinetic. The internal energy of a real gas is kinetic as
well as potential.
[Online April 9, 2013]
(a) Statement-1 is false and Statement-2 is true.
(b) Statement-1 and Statement-2 both are true.
Statement-2 is the correct explanation of Statement-1.
(c) Statement-1 is true and Statement-2 is false.
(d) Statement-1 and Statement-2 both are true.
Statement-2 is not correct explanation of Statement-1.
Cooking gas containers are kept in a lorry moving with
uniform speed. The temperature of the gas molecules
inside will
[2002]
(a) increase
(b) decrease
(c) remain same
(d) decrease for some, while increase for others
(a) 104 N/m2
(b) 108 N/m2
3
2
(c) 10 N/m
(d) 1016 N/m2
13. The temperature, at which the root mean square velocity
of hydrogen molecules equals their escape velocity from
the earth, is closest to :
[8 April 2019 II]
–23
[Boltzmann Constant kB = 1.38 × 10 J/K
Avogadro Number NA = 6.02 × 1026 /kg
Radius of Earth : 6.4 × 106 m
Gravitational acceleration on Earth = 10 ms–2]
(a) 800 K
(b) 3 × 105 K
4
(c) 10 K
(d) 650 K
14. A mixture of 2 moles of helium gas (atomic mass = 4u), and
1 mole of argon gas (atomic mass = 40u) is kept at 300 K in
a container. The ratio of their rms speeds
é Vrms ( helium ) ù
ê
ú is close to :
[9 Jan. 2019 I]
ë Vrms ( argon ) û
(a) 3.16
(b) 0.32
(c) 0.45
(d) 2.24
15. N moles of a diatomic gas in a cylinder are at a temperature
T. Heat is supplied to the cylinder such that the temperature remains constant but n moles of the diatomic gas get
converted into monoatomic gas. What is the change in
the total kinetic energy of the gas ?
[Online April 9, 2017]
1
nRT
(b) 0
2
3
5
nRT
nRT
(c)
(d)
2
2
16. In an ideal gas at temperature T, the average force that a
molecule applies on the walls of a closed container
depends on T as Tq. A good estimate for q is:
[Online April 10, 2015]
1
(a)
(b) 2
2
1
(c) 1
(d)
4
17. A gas molecule of mass M at the surface of the Earth has
kinetic energy equivalent to 0°C. If it were to go up
straight without colliding with any other molecules, how
high it would rise? Assume that the height attained is much
less than radius of the earth. (kB is Boltzmann constant).
[Online April 19, 2014]
273k B
(a) 0
(b)
2Mg
(a)
TOPIC 2
Speed of Gas, Pressure
and Kinetic Energy
Number of molecules in a volume of 4 cm 3 of a perfect
monoatomic gas at some temperature T and at a pressure
of 2 cm of mercury is close to? (Given, mean kinetic energy of a molecule (at T) is 4 × 10–14 erg, g = 980 cm/s2,
density of mercury = 13.6 g/cm3) [Sep. 05, 2020 (I)]
(a) 4.0 × 1018
(b) 4.0 × 1016
16
(c) 5.8 × 10
(d) 5.8 × 1018
10. Nitrogen gas is at 300°C temperature. The temperature
(in K) at which the rms speed of a H2 molecule would be
equal to the rms speed of a nitrogen molecule, is
_____________. (Molar mass of N2 gas 28 g);
[NA Sep. 05, 2020 (II)]
11. For a given gas at 1 atm pressure, rms speed of the
molecules is 200 m/s at 127°C. At 2 atm pressure and at
227°C, the rms speed of the molecules will be:
[9 April 2019 I]
(a) 100 m/s
(b) 80 5 m/s
(c) 100 5 m/s
(d) 80 m/s
12. If 1022 gas molecules each of mass 10–26 kg collide with a
surface (perpendicular to it) elastically per second over
an area 1 m2 with a speed 104 m/s, the pressure exerted
by the gas molecules will be of the order of :
[8 April 2019 I]
9.
819k B
546k B
(d)
2Mg
3Mg
18. At room temperature a diatomic gas is found to have an
r.m.s. speed of 1930 ms–1. The gas is:
[Online April 12, 2014]
(a) H2
(b) Cl2
(c) O2
(d) F2
(c)
P-188
Physics
19. In the isothermal expansion of 10g of gas from volume
V to 2V the work done by the gas is 575J. What is the
root mean square speed of the molecules of the gas at
that temperature?
[Online April 25, 2013]
(a) 398m/s
(b) 520m/s
(c) 499m/s
(d) 532m/s
20. A perfect gas at 27°C is heated at constant pressure so as
to double its volume. The final temperature of the gas will
be, close to
[Online May 7, 2012]
(a) 327°C
(b) 200°C
(c) 54°C
(d) 300°C
21. A thermally insulated vessel contains an ideal gas of
molecular mass M and ratio of specific heats g. It is
moving with speed v and it's suddenly brought to rest.
Assuming no heat is lost to the surroundings, its
temperature increases by:
[2011]
(a)
22.
23.
24.
25.
( g - 1) Mv 2 K
2 gR
The total internal energy, U of a mole of this gas, and the
æ Cp ö
value of g ç =
÷ are given, respectively, by:
è Cv ø
[Sep. 06, 2020 (I)]
(a) U =
( g - 1)
( g - 1)
Mv 2 K
2R
(a)
n1T1 + n2T2 + n3T3
n1 + n2 + n3
(b)
(c)
n12T12 + n22T22 + n32T32
n1T1 + n2T2 + n3T3
(d)
(b) U = 5RT and g =
7
5
5
7
6
RT and g =
(d) U = 5RT and g =
2
5
5
27. In a dilute gas at pressure P and temperature T, the mean
time between successive collisions of a molecule varies
with T is :
[Sep. 06, 2020 (II)]
(c) U =
(a) T
gM 2v
(b)
K
2R
(c)
5
6
RT and g =
2
5
(b)
1
T
1
(d) T
T
Match the Cp/Cv ratio for ideal gases with different type
of molecules :
[Sep. 04, 2020 (I)]
Column-I
Column-II
Molecule Type
Cp/Cv
(A) Monatomic
(I) 7/5
(B) Diatomic rigid molecules (II) 9/7
(C) Diatomic non-rigid molecules(III) 4/3
(D) Triatomic rigid molecules (IV) 5/3
(a) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
(b) (A)-(III), (B)-(IV), (C)-(II), (D)-(I)
(c) (A)-(IV), (B)-(I), (C)-(II), (D)-(III)
(d) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
A closed vessel contains 0.1 mole of a monatomic ideal
gas at 200 K. If 0.05 mole of the same gas at 400 K is
added to it, the final equilibrium temperature (in K) of
the gas in the vessel will be close to _________.
[NA Sep. 04, 2020 (I)]
(c)
2
(d) 2( g + 1) R Mv K
Three perfect gases at absolute temperatures T1, T2 and T3
are mixed. The masses of molecules are m1, m2 and m3 and
the number of molecules are n1, n2 and n3 respectively.
Assuming no loss of energy, the final temperature of the
mixture is :
[2011]
28.
n1T12 + n2T22 + n3T32
n1T1 + n2T2 + n3T3
(T1 + T2 + T3 )
3
One kg of a diatomic gas is at a pressure of
8 × 104N/m2. The density of the gas is 4kg/m3. What is
the energy of the gas due to its thermal motion?[2009]
(a) 5 × 104 J
(b) 6 × 104 J
4
(c) 7 × 10 J
(d) 3 × 104 J
The speed of sound in oxygen (O2) at a certain temperature
is 460 ms–1. The speed of sound in helium (He) at the same
temperature will be (assume both gases to be ideal)
[2008]
–1
–1
(a) 1421 ms
(b) 500 ms
(c) 650 ms–1
(d) 330 ms–1
At what temperature is the r.m.s velocity of a hydrogen
molecule equal to that of an oxygen molecule at 47°C?
[2002]
(a) 80 K
(b) –73 K
(c) 3 K
(d) 20 K
29.
30.
Consider a gas of triatomic molecules. The molecules are
assumed to be triangular and made of massless rigid rods
whose vertices are occupied by atoms. The internal energy
of a mole of the gas at temperature T is :
[Sep. 03, 2020 (I)]
Degree of Freedom, Specific
TOPIC 3 Heat Capacity, and Mean
Free Path
26.
Molecules of an ideal gas are known to have three
translational degrees of freedom and two rotational
degrees of freedom. The gas is maintained at a
temperature of T.
(a)
5
RT
2
(b)
3
RT
2
9
RT
(d) 3RT
2
31. To raise the temperature of a certain mass of gas by 50°C
at a constant pressure, 160 calories of heat is required.
When the same mass of gas is cooled by 100°C at constant
(c)
P-189
Kinetic Theory
32.
33.
34.
35.
36.
volume, 240 calories of heat is released. How many
degrees of freedom does each molecule of this gas have
(assume gas to be ideal)?
[Sep. 03, 2020 (II)]
(a) 5
(b) 6
(c) 3
(d) 7
A gas mixture consists of 3 moles of oxygen and 5 moles of
argon at temperature T. Assuming the gases to be ideal
and the oxygen bond to be rigid, the total internal energy
(in units of RT) of the mixture is : [Sep. 02, 2020 (I)]
(a) 15
(b) 13
(c) 20
(d) 11
An ideal gas in a closed container is slowly heated. As its
temperature increases, which of the following statements
are true?
[Sep. 02, 2020 (II)]
(1) The mean free path of the molecules decreases
(2) The mean collision time between the molecules
decreases
(3) The mean free path remains unchanged
(4) The mean collision time remains unchanged
(a) (2) and (3)
(b) (1) and (2)
(c) (3) and (4)
(d) (1) and (4)
Consider two ideal diatomic gases A and B at some
temperature T. Molecules of the gas A are rigid, and have
a mass m. Molecules of the gas B have an additional
m
vibrational mode, and have a mass
. The ratio of the
4
B
specific heats (CA
V and CV ) of gas A and B, respectively
is:
[9 Jan 2020 I]
(a) 7 : 9
(b) 5 : 9
(c) 3 : 5
(d) 5 : 7
Two gases-argon (atomic radius 0.07 nm, atomic weight
40) and xenon (atomic radius 0.1 nm, atomic weight 140)
have the same number density and are at the same
temperature. The ratio of their respective mean free
times is closest to:
[9 Jan 2020 II]
(a) 3.67
(b) 1.83
(c) 2.3
(d) 4.67
The plot that depicts the behavior of the mean free time t
(time between two successive collisions) for the molecules
of an ideal gas, as a function of temperature (T),
qualitatively, is: (Graphs are schematic and not drawn to
scale)
[8 Jan. 2020 I]
37. Consider a mixture of n moles of helium gas and 2n
moles of oxygen gas (molecules taken to be rigid) as an
ideal gas. Its CP/CV value will be:
[8 Jan. 2020 II]
(a) 19/13
(b) 67/45
(c) 40/27
(d) 23/15
Cp 5
38. Two moles of an ideal gas with C = are mixed with 3
3
V
Cp 4
moles of another ideal gas with C = . The value of
3
V
Cp
39.
40.
41.
42.
for the mixture is:
[7 Jan. 2020 I]
CV
(a) 1. 45
(b) 1.50
(c) 1.47
(d) 1.42
Two moles of helium gas is mixed with three moles of
hydrogen molecules (taken to be rigid). What is the molar
specific heat of mixture at constant volume?
(R = 8.3 J/mol K)
[12 April 2019 I]
(a) 19.7 J/mol L
(b) 15.7 J/mol K
(c) 17.4 J/mol K
(d) 21.6 J/mol K
A diatomic gas with rigid molecules does 10 J of work
when expanded at constant pressure. What would be the
heat energy absorbed by the gas, in this process ?
[12 April 2019 II]
(a) 25 J
(b) 35 J
(c) 30 J
(d) 40 J
A 25×10 – 3 m3 volume cylinder is filled with 1 mol of O2
gas at room temperature (300 K) . The molecular diameter
of O2, and its root mean square speed, are found to be 0.3
nm and 200 m/s, respectively. What is the average collision
rate (per second) for an O2 molecule?
[10 April 2019 I]
(a) ~1012
(b) ~1011
(c) ~1010
(d) ~1013
When heat Q is supplied to a diatomic gas of rigid
molecules, at constant volume its temperature increases
by DT. The heat required to produce the same change in
temperature, at a constant pressure is :
[10 April 2019 II]
(a)
2
Q
3
(b)
5
Q
3
7
3
Q
Q
(d)
5
2
An HCl molecule has rotational, translational and
vibrational motions. If the rms velocity of HCl molecules
in its gaseous phase is v , m is its mass and k B is
Boltzmann constant, then its temperature will be:
[9 April 2019 I]
(c)
(a)
43.
t
t
(b)
1
T
T
t
(a)
t
(c)
(d)
1
T
T
mv 2
6kB
mv 2
(c)
7k B
(b)
mv 2
3k B
mv 2
(d)
5k B
P-190
44.
45.
46.
47.
48.
49.
50.
Physics
The specific heats, C p and Cv of a gas of diatomic
molecules, A, are given (in units of J mol–1 k–1) by 29 and
22, respectively. Another gas of diatomic molecules, B,
has the corresponding values 30 and 21. If they are treated
as ideal gases, then:
[9 April 2019 II]
(a) A is rigid but B has a vibrational mode.
(b) A has a vibrational mode but B has none.
(c) A has one vibrational mode and B has two.
(d) Both A and B have a vibrational mode each.
An ideal gas occupies a volume of 2 m 3 at a pressure of 3
× 106 Pa. The energy of the gas:
[12 Jan. 2019 I]
(a) 9 × 106 J
(b) 6 × 104 J
(c) 108 J
(d) 3 × 102 J
An ideal gas is enclosed in a cylinder at pressure of 2 atm
and temperature, 300 K. The mean time between two
successive collisions is 6 × 10–8 s. If the pressure is doubled
and temperature is increased to 500 K, the mean time
between two successive collisions wiil be close to:
[12 Jan. 2019 II]
–7
–8
(a) 2 × 10 s
(b) 4 × 10 s
(c) 0.5 × 10–8 s
(d) 3 × 10–6 s
51. Two moles of an ideal monoatomic gas occupies a volume
V at 27°C. The gas expands adiabatically to a volume 2 V.
Calculate (1) the final temperature of the gas and (2) change
in its internal energy.
[2018]
(a) (1) 189 K
(2) 2.7 kJ
(b) (1) 195 K
(2) –2.7 kJ
(c) (1) 189 K
(2) –2.7 kJ
(d) (1) 195 K
(2) 2.7 kJ
52. Two moles of helium are mixed with n with moles of
3
C
hydrogen. If P = for the mixture, then the value of n
CV 2
is
[Online April 16, 2018]
(a) 3/2
(b) 2
(c) 1
(d) 3
53. Cp and Cv are specific heats at constant pressure and
constant volume respectively. It is observed that
Cp – Cv = a for hydrogen gas
Cp – Cv = b for nitrogen gas
The correct relation between a and b is :
[2017]
(a) a = 14 b
(b) a = 28 b
A gas mixture consists of 3 moles of oxygen and 5 moles
of argon at temperature T. Considering only translational
and rotational modes, the total internal energy of the
system is :
[11 Jan. 2019 I]
(a) 15 RT
(b) 12 RT
(c) 4 RT
(d) 20 RT
In a process, temperature and volume of one mole of an
ideal monoatomic gas are varied according to the relation
VT = K, where K is a constant. In this process the
temperature of the gas is increased by DT. The amount of
heat absorbed by gas is (R is gas constant) :
[11 Jan. 2019 II]
1
1
KRΔT
(a) RΔT
(b)
2
2
2K
3
ΔT
RΔT
(c)
(d)
3
2
Two kg of a monoatomic gas is at a pressure of 4 × 104
N/m2. The density of the gas is 8 kg/m3. What is the
order of energy of the gas due to its thermal motion?
[10 Jan 2019 II]
3
5
(a) 10 J
(b) 10 J
(c) 104 J
(d) 106 J
A 15 g mass of nitrogen gas is enclosed in a vessel at
a temperature 27°C. Amount of heat transferred to the
gas, so that rms velocity of molecules is doubled, is
about: [Take R = 8.3 J/K mole]
[9 Jan. 2019 II]
(a) 0.9 kJ
(b) 6 kJ
(c) 10 kJ
(d) 14 kJ
54.
1
b
(d) a = b
14
An ideal gas has molecules with 5 degrees of freedom.
The ratio of specific heats at constant pressure (C p) and
at constant volume (Cv) is : [Online April 8, 2017]
7
(a) 6
(b)
2
7
5
(c)
(d)
5
2
An ideal gas undergoes a quasi static, reversible process
in which its molar heat capacity C remains constant. If
during this process the relation of pressure P and volume
V is given by PVn = constant, then n is given by (Here CP
and CV are molar specific heat at constant pressure and
constant volume, respectively) :
[2016]
CP - C
C - CV
(a) n =
(b) n =
C - CV
C - CP
C – CP
CP
(c) n =
(d) n =
C – CV
CV
Using equipartition of energy, the specific heat
(in J kg–1 K–1) of aluminium at room temperature can
be estimated to be (atomic weight of aluminium = 27)
[Online April 11, 2015]
(a) 410
(b) 25
(c) 1850
(d) 925
Modern vacuum pumps can evacuate a vessel down to a
pressure of 4.0 × 10–15 atm. at room temperature (300
K). Taking R = 8.0 JK–1 mole–1, 1 atm = 105 Pa and
NAvogadro = 6 × 1023 mole–1, the mean distance between
molecules of gas in an evacuated vessel will be of the
order of:
[Online April 9, 2014]
(a) 0.2 mm
(b) 0.2 mm
(c) 0.2 cm
(d) 0.2 nm
(c) a =
55.
56.
57.
P-191
Kinetic Theory
58.
Figure shows the variation in temperature (DT) with the
amount of heat supplied (Q) in an isobaric process
corresponding to a monoatomic (M), diatomic (D) and a
polyatomic (P) gas. The initial state of all the gases are the
same and the scales for the two axes coincide. Ignoring
vibrational degrees of freedom, the lines a, b and c
respectively correspond to :
[Online April 9, 2013]
oxygen. The ratio
a
b
Q
c
DT
59.
(a) P, M and D
(b) M, D and P
(c) P, D and M
(d) D, M and P
A given ideal gas with g =
Cp
= 1.5 at a temperature T. If
Cv
the gas is compressed adiabatically to one-fourth of its
initial volume, the final temperature will be
[Online May 12, 2012]
(a) 2 2T
(c) 2 T
60. If CP and CV denote the specific heats of nitrogen per unit
mass at constant pressure and constant volume
respectively, then
[2007]
(a) CP – CV = 28R
(b) CP – CV = R/28
(c) CP – CV = R/14
(d) CP – CV = R
61. A gaseous mixture consists of 16 g of helium and 16 g of
(b) 4 T
(d) 8 T
Cp
Cv
of the mixture is
[2005]
(a) 1.62
(b) 1.59
(c) 1.54
(d) 1.4
62. One mole of ideal monatomic gas (g = 5/3) is mixed with
one mole of diatomic gas (g = 7/5). What is g for the
mixture? g Denotes the ratio of specific heat at constant
pressure, to that at constant volume
[2004]
(a) 35/23
(b) 23/15
(c) 3/2
(d) 4/3
63. During an adiabatic process, the pressure of a gas is found
to be proportional to the cube of its absolute temperature.
The ratio CP/CV for the gas is
[2003]
(a)
4
3
(b) 2
(c)
5
3
(d)
3
2
P-192
1.
Physics
(5) Using ideal gas equation, PV = nRT
æ 1
Þ PV
1 1 = nR ´ 250
[Q T1 = 250 K]
...(i)
5n
R ´ 2000
4
Dividing eq. (i) by (ii),
[Q T2 = 2000 K]
...(ii)
P2 (2V1 =
nRT æ l1 – l 2 ö
5.
P1
P 1
4 ´ 250
=
Þ 1 =
2 P2 5 ´ 2000
P2 5
P2
= 5.
P1
(150) In first case,
From ideal gas equation
PV = nRT
\
2.
DP
V
...(i)
P
In second case, using ideal gas equation again
P DV = -nR DT
DV = -
nR DT
P
Equating (i) and (ii), we get
DV = -
6.
8.
4
0
4.
r
òr
2
4
(e -ar )dr
0
µ n0 a–3/4
(d) Clearly from figure,
P2A =
P1A + mg
or,
7.
DT 300 K
V
=
=
= 150 K/atm
nR DP 2 atm
-ar
´ 4pr 2 dr = 4p n
= ò n0 e
0
9.
nRT × A
nRT × A
+ mg
Al 2 = Al1
l1
n
l2
n
T
T
1 ´ 105 ´ 30
1 ö
æ 1
´ 6.023 ´ 10 23 ç
÷
8.314
300
290
è
ø
= – 2.5 × 1025
(c) From P-V graph,
1
, T = constant and Pressure is increasing from 2
V
to 1 so option (3) represents correct T-P graph.
(a) In ideal gases the molecules are considered as point
particles and for point particles, there is no internal
excitation, no vibration and no rotation. For an ideal gas
the internal energy can only be translational kinetic energy
and for real gas both kinetic as well as potential energy.
(c) The centre of mass of gas molecules also moves with
lorry with uniform speed. As there is no relative motion of
gas molecule. So, kinetic energy and hence temperature
remain same.
3
kT = 4 ´ 10 -14
2
P = 2 cm of Hg, V = 4 cm3
(c) Given : K.E.mean =
N=
PV PrgV 2 ´ 13.6 ´ 980 ´ 4
=
; 4 ´ 1018
8
KT
KT
´ 10 -14
3
10. (41) Room mean square speed is given by
P1A
P2A
P0V0 æ 1 1 ö
- ÷N
ç
R è T f Ti ø 0
Pµ
(a) N = ò r(dv)
r
PV
(N )
RT 0
=
...(ii)
Comparing the above equation with | DT | = C | DP | , we
have
3.
n=
\ nf – ni =
DP
nR DT
V
=V Þ DT = DP
P
P
nR
C=
\ m = g ç l ×l ÷
è 1 2 ø
(b) Given: Temperature Ti = 17 + 273 = 290 K
Temperature Tf = 27 + 273 = 300 K
Atmospheric pressure, P0 = 1 × 105 Pa
Volume of room, V0 = 30 m3
Difference in number of molecules, nf – ni = ?
Using ideal gas equation, PV = nRT(N0),
N0 = Avogadro's number
Þ
(As temperature is constant)
P D V + V DP = 0
1ö
Þ nRT ç l – l ÷ = mg
è 2
1ø
mg
vrms =
3RT
M
P-193
Kinetic Theory
Here, M = Molar mass of gas molecule
T = temperature of the gas molecule
We have given vN2 = vH2
3RTN2
\
M N2
Þ
11.
TH2
2
=
3RTH 2
M H2
573
Þ TH2 = 41 K
28
(c) Vrms =
v1
=
v2
=
3RT
M
T1 (273 + 127)
=
=
T2 (273 + 237)
400
=
500
4
2
=
5
5
5
5
v1 =
´ 200 = 100 5 m/s.
2
2
(Bouns) Rate of change of momentum during collision
\ v2 =
12.
=
mv – (– mv )
2mv
=
N
Dt
Dt
so pressure P = N ´ (2mv)
Dt ´ A
1022 ´ 2 ´ 10–26 ´ 104
= 2 N / m2
1´ 1
(c) vrms = ve
=
13.
3RT
M
3 ´ 8.314 ´ 300
2
(1930 ) =
M
3 ´ 8.314 ´ 300
M=
» 2 ´10 -3 kg
1930 ´1930
The gas is H2.
18. (a) Q
3RT
= 11.2 ´ 103
M
or
or
14.
3kT
= 11.2 ´ 103
m
3 ´ 1.38 ´ 10-23 T
2 ´ 10-3
V1rms
(a) Using V
=
2rms
= 11.2 ´ 10
3
\ v = 104 K
M2
M1
Vrms ( He )
15.
M Ar
40
=
M He = 4 = 3.16
Vrms ( Ar )
(a) Energy associated with N moles of diatomic gas,
5
Ui = N RT
2
Energy associated with n moles of monoatomic gas
3
= n RT
2
Total energy when n moles of diatomic gas converted into
3
5
monoatomic (Uf) = 2n RT + (N - n) RT
2
2
=
1
5
nRT + NRT
2
2
Now, change in total kinetic energy of the gas
1
DU = Q = nRT
2
1 mN 2
V rm s
16 . (c) Pressure, P =
3 V
(mN )T
or, P =
V
If the gas mass and temperature are constant then
P µ (Vrms)2 µ T
So, force µ (Vrms)2 µ T
i.e., Value of q = 1
17. (d) Kinetic energy of each molecule,
3
K.E. = K B T
2
In the given problem,
Temperature, T = 0°C = 273 K
Height attained by the gas molecule, h = ?
819K B
3
K.E. = K B ( 273) =
2
2
K.E. = P.E.
819K B
= Mgh
Þ
2
819K B
or h =
2Mg
C=
3rv
mass of the gas
20. (a) Given, V1 = V
V2 = 2V
19. (c) v rms =
T1 = 27° + 273 = 300 K
T2 = ?
From charle’s law
V1 V2
=
Q Pressure is constant )
T1 T2 (
V
2V
or, 300 = T
2
\ T2 = 600 K = 600 – 273 = 327°C
21. (c) As, work done is zero.
So, loss in kinetic energy = heat gain by the gas
1 2
R
mv = nCv DT = n
DT
2
g -1
1 2 m R
mv =
DT
2
M g -1
P-194
Physics
\ DT =
22.
Mv 2 ( g - 1)
K
2R
And g =
n1
(a) Number of moles of first gas = N
A
But vrms =
n1 + n2 + n3
RTmix
NA
(a) Given, mass = 1 kg
Density = 4 kg m–3
g=
\
vO2
vHe
=
M O2
´
\ g = 1+
2 9
=
7 7
(D) Triatomic rigid molecules, f = 6
\ g = 1+
gRT
M
M He
g He
1.4
4
´
= 0.3237
32 1.67
vO 2
460
\ vHe =
=
= 1421 m / s
0.3237 0.3237
(d) RMS velocity of a gas molecule is given by
3RT
M
Let T be the temperature at which the velocity of hydrogen
molecule is equal to the velocity of oxygen molecule.
2 4
=
6 3
29. (266.67) Here work done on gas and heat supplied to the
gas are zero.
Let T be the final equilibrium temperature of the gas in the
vessel.
Total internal energy of gases remain same.
\ g = 1+
i.e., u1 + u2 = u '1 + u '2
or, n1Cv DT1 + n2Cv DT2 = (n1 + n2 )CvT
Þ (0.1)Cv (200) + (0.05)Cv (400) = (0.15)CvT
Vrms =
3RT
3R ´ (273 + 47)
=
2
32
Þ T = 20K
(c) Total degree of freedom f = 3 + 2 = 5
\
26.
Total energy, U =
nfRT 5RT
=
2
2
2
, where f = degree of freedom
f
2 7
=
5 5
(C) Diatomic non-rigid molecules, f = 7
[Q PM = dRT ]
=
25.
Cv
= 1+
2 5
=
3 3
(B) Diatomic rigid molecules, f = 5
[As R and T is constant]
g O2
Cp
\ g = 1+
(a) The speed of sound in a gas is given by v =
g
M
3kT
R
(A) Monatomic, f = 3
=
\ vµ
vrms
l
1
Þtµ
3kT
T
m
28. (c) As we know,
mass
1
= m3
Volume =
density 4
Internal energy of the diatomic gas
24.
l
\t =
n1T1 + n2T2 + n3T3
n1 + n2 + n3
5
5
1
PV = ´ 8 ´ 104 ´ = 5 ´ 10 4 J
2
2
4
Alternatively:
5
5m
5 m PM
RT =
´
K.E = nRT =
d
2
2M
2M
4
5 mP 5 1 ´ 8 ´ 10
=
= ´
= 5 ´ 104 J
2 d
2
4
1
But, mean time of collision, t =
n1
n
n
RT1 + 2 RT2 + 3 RT3
NA
NA
NA
23.
2
2 7
= 1+ =
f
5 5
2pnd 2
where, d = diameter of the molecule
n = number of molecules per unit volume
If there is no loss of energy then
P1V1 + P2V2 + P3V3 = PV
Tmix =
= 1+
Cv
27. (b) Mean free path, l =
n2
Number of moles of second gas = N
A
n3
Number of moles of third gas = N
A
=
Cp
800
= 266.67 K
3
(d) Here degree of freedom, f = 3 + 3 = 6 for triatomic nonlinear molecule.
Internal energy of a mole of the gas at temperature T,
\T =
30.
f
6
nRT = RT = 3RT
2
2
31. (b) Let Cp and Cv be the specific heat capacity of the gas
at constant pressure and volume.
U=
P-195
Kinetic Theory
At constant pressure, heat required
DQ1 = nC p DT
Þ 160 = nC p × 50
...(i)
At constant volume, heat required
DQ2 = nCv DT
Þ 240 = nCv ×100
...(ii)
Dividing (i) by (ii), we get
Cp 4
160 C p 50
=
×
Þ
=
Cv 3
240 Cv 100
g=
Cp
Cv
=
4
2
= 1+
3
f
(Here, f = degree of freedom)
32.
Þ f = 6.
(a) Total energy of the gas mixture,
33.
f1n1 RT1 f 2 n2 RT2
+
Emix =
2
2
5
5
= 3 ´ RT + ´ 3RT = 15RT
2
2
(a) As we know mean free path
1
æ Nö
2 ç ÷ pd 2
èV ø
l=
Here,
N = no. of molecule
V = volume of container
d = diameter of molecule
But PV = nRT = nNKT
N
P
Þ =
=n
V KT
1 KT
l=
2 pd 2 P
For constant volume and hence constant number density
P
is constant.
T
So mean free path remains same.
As temperature increases no. of collision increases so
relaxation time decreases.
(d) Specific heat of gas at constant volume
n of gas molecules
34.
5
R
5
7
= 2 =
B
B
C
=
R
\
Hence C
7
v
7
v
2
R
2
CvA
35. (Bonus) Mean free path of a gas molecule is given by
1
l=
2pd 2 n
Here, n = number of collisions per unit volume
d = diameter of the molecule
If average speed of molecule is v then
l
Mean free time, t =
v
M
1
1
Þ t=
=
2
2 3RT
2pnd v
2pnd
æ
3RT ö
çQ v = M ÷
è
ø
\ tµ
2
M \ t1 = M1 ´ d2
t2
M2
d12
d2
2
=
40 æ 0.1 ö
´ç
÷ = 1.09
140 è 0.07 ø
36. (c) Relaxation time (t ) µ
mean free path
1
Þ tµ
speed
v
and, v µ T
1
\tµ
T
Hence graph between t v/s
1
is a straight line which
T
is correctly depicted by graph shown in option (c).
37. (a) Helium is a monoatomic gas and Oxygen is a diatomic
gas.
For helium, CV1 =
3
5
R and CP1 = R
2
2
For oxygen, CV2 =
g=
5
7
R and CP = R
2
2
2
N1CP1 + N 2 CP2
N1CV1 + N 2 CV2
Cv =
1
fR; f = degree of freedom
2
For gas A (diatomic)
f = 5 (3 translational + 2 rotational)
5
7
n. R + 2n. R
2 = 19nR ´ 2
g= 2
Þ
3
5
2(13nR )
n. R + 2n. R
2
2
5
R
2
For gas B (diatomic) in addition to (3 translational + 2
rotational) 2 vibrational degree of freedom.
æC ö
19
\ç P ÷
=
C
è V ø mixture 13
\C A =
v
P-196
38.
Physics
(d) Using, gmixture=
n1C p + n2C p
1
n1Cv + n2Cv
1
Þ
Þ
2
2
n1
n
n +n
+ 2 = 1 2
g 1 –1 g 2 –1 g m –1
3
2
5
=
g m –1
+
4
5
–1
–1
3
3
9 2´3
5
5
Þ +
=
Þ g m –1 =
g m –1
1
2
12
17
= 1.42
12
n1[Cv1 ] + n2 [Cv2 ]
(c) [Cv]min =
n1 + n2
Þ gm =
39.
3R
5R ù
é
ê2´ 2 + 3´ 2 ú
ú
= êê
2+3
ú
ë
û
= 2.1 R = 2.1 × 8.3 = 17.4 J/mol–k
40.
Cv 1
1
5
(b) F = C = r = (7 / 5) =
p
7
or
W
5 2
= 1- =
Q
7 7
7
7 ´ 10
W =
= 35 J
2
2
(c) V = 25 × 10–3 m3, N = 1 mole of O2
T = 300 K
Vrms = 200 m/s
\ l=
1
2Npr 2
1 <V>
= 200.Npr 2 . 2
Average time =
t
l
42.
23
2 ´ 200 ´ 6.023 ´ 10
.p ´ 10-18 ´ 0.09
25 ´ 10 -3
The closest value in the given option is = 1010
(c) Amount of heat required (Q) to raise the temperature
at constant volume
Q = nCvDT
...(i)
Amount of heat required (Q1) at constant pressure
Q1 = nCPDT
...(ii)
Dividing equation (ii) by (i), we get
=
\
Q1 C p
=
Q Cv
æ 7ö
Þ Q1 = (Q) ç ÷
è 5ø
or T =
mv 2
6k B
gA =
44. (b)
CP 29
=
= 1.32 < 1.4 (diatomic)
Cv 22
30 10
=
= 1.43 > 1.4
21 7
Gas A has more than 5-degrees of freedom.
45. (a) Energy of the gas, E
f
f
= nRT = PV
2
2
f
= (3 ´ 106 )(2) = f ´ 3 ´ 106
2
Considering gas is monoatomic i.e., f = 3
Energy, E = 9 × 106 J
1
46. (b) Using, t=
2npd 2 Vavg
and g B =
\t µ
or,
T
P
no.of molecules ù
é
êë\n =
úû
Volume
t1
500
P
=
´
» 4´10–8
–8
6 ´10
2P
300
f1
f
n1RT + 2 n 2RT
2
2
Considering translational and rotational modes, degrees
of freedom f1 = 5 and f2 = 3
47. (a) U =
or Q =
41.
1
mv 2 = 3k BT
2
43. (a)
C p 7ö
æ
çQ g = C = 5 ÷
è
ø
v
5
3
\ u = (3RT) + ´ 5RT
2
2
U = 15RT
48. (a) According to question VT = K
we also know that PV = nRT
æ PV ö
ÞT = ç
è nR ÷ø
æ PV ö
2
ÞVç
÷ = k Þ PV = K
è nR ø
R
Q C=
+ CV (For polytropic process)
1– x
R 3R R
+
=
1– 2 2 2
\DQ = nC DT
R
= ´DT [here, n = 1 mole]
2
49. (c) Thermal energy of N molecule
C=
æ3 ö
= N ç kT ÷
è2 ø
P-197
Kinetic Theory
50.
=
N 3
3
3
RT = ( nRT ) = PV
NA 2
2
2
=
3 æmö 3 æ 2ö
Pç ÷= Pç ÷
2 è r ø 2 è8ø
3
2
4
4
= ´ 4´10 ´ = 1.5 ´10 J
2
8
therefore, order = 104 J
(c) Heat transferred,
Q = nCv DT as gas in closed vessel
To double the rms speed, temperature should be 4 times
i.e., T' = 4T as vrms =
3RT / M
15 5 ´ R
´
´ ( 4T - T )
28
2
7
é CP
ù
êë\ CV = γ diatomic = 5 & C p - Cv = R úû
or, Q = 10000 J = 10 kJ
(c) In an adiabatic process
TVg–1 = Constant
or, T1V1g–1 = T2V2g–1
5
For monoatomic gas g =
3
300
(300)V2/3 = T2(2V)2/3 Þ T2 =
(2) 2/3
T2 = 189 K (final temperature)
f
Change in internal energy DU = n R DT
2
æ 3 öæ 25 ö
= 2 ç ÷ç ÷ (-111) = -2.7 kJ
è 2 øè 3 ø
(b) Using formula,
n1g 1 n 2 g 2
+
æ Cp ö
g1 - 1 g 2 - 1
g mixture = ç ÷
=
n1
n
è Cv ø mix
+ 2
g1 - 1 g 2 - 1
54. (d) The ratio of specific heats at constant pressure (C p)
and constant volume (Cv)
Cp
æ 2ö
= g = ç1 + ÷
è fø
Cv
where f is degree of freedom
Cp æ 2 ö 7
= ç1 + ÷ =
Cv è 5 ø 5
55. (d) For a polytropic process
R
R
C = Cv +
\ C - Cv =
1- n
1- n
R
R
\ 1- n =
\ 1=n
C - Cv
C - Cv
\Q=
51.
52.
æ Cp ö
3
Putting the value of n1 = 2, n2 = n, ç ÷
=
è C v ø mix 2
53.
5
7
g 1 = , g 2 = and solving we get, n = 2
3
5
(a) As we know, Cp – Cv = R where Cp and Cv are molar
specific heat capacities
or, Cp – Cv =
R
M
For hydrogen (M = 2) Cp – Cv = a =
R
2
For nitrogen (M = 28) Cp – Cv = b =
R
28
\
a
= 14
b
or, a = 14b
\ n=
C - Cp
(Q C p - C v = R )
C - Cv
(d) Using equipartition of energy, we have
=
56.
C - C v - R C - C v - Cp + C v
=
C - Cv
C - Cv
6
KT = mCT
2
C=
3 ´1.38 ´10 –23 ´ 6.02 ´1023
27 ´10–3
\ C = 925 J/kgK
57. (b)
58. (b) On giving same amount of heat at constant pressure,
there is no change in temperature for mono, dia and
polyatomic.
æ
No. of molecules ö
( DQ) P = mC p DT ç m =
è
Avogedro 's no. ÷ø
or
DT µ
1
no. of molecules
59. (c) TV g -1 = constant
g -1
T1V1g -1 = T2V2
Þ T (V )
1
2
æV ö
=T2 ç ÷
è 4ø
1
2
Vù
é
êëQ g = 1.5, T1 = T ,V1 = V and V2 = 4 úû
1
æ 4V ö 2
\ T2 = ç ÷ T = 2T
èV ø
60. (b) According to Mayer's relationship
CP – CV = R, as per the question (CP – CV) M = R
Þ CP – CV = R/28
Here M = 28 = mass of 1 unit of N2
61. (a) For mixture of gas specific heat at constant volume
Cv =
n1Cv1 + n2Cv2
n1 + n2
P-198
Physics
No. of moles of helium,
mHe
16
n1 = M
=4
=
4
He
Number of moles of oxygen,
16 1
=
n2 =
32 2
3
1 5
5
4 ´ R + ´ R 6R + R
2
2
2
4
=
=
9
1ö
æ
\ Cv
çè 4 + ÷ø
2
2
29 R ´ 2 29 R
=
and
9´ 4
18
Specific heat at constant pressure
=
Cp =
n1C p1 + n2C p2
(n1 + n2 )
7
10 R + R
4 = 47 R
=
9
18
2
5R 1 7 R
+ ´
4´
2 2 2
=
1ö
æ
çè 4 + ÷ø
2
Þ
\
Cp
Cv
=
47 R 18
´
= 1.62
18 29 R
5
7
g =
3 2 5
n1 = 1, n2 = 1
62. (c) g 1 =
n1 + n2
n
n
= 1 + 2
g -1
g1 - 1 g 2 - 1
1 +1
1
1
3 5
=
+
= + =4
g - 1 5 - 1 7 -1 2 2
3
5
2
3
\
=4 Þ g=
g -1
2
Þ
63. (d) P µ T 3 Þ PT -3 = constant
....(i)
But for an adiabatic process, the pressure temperature
relationship is given by
P1-g T g = constant
Þ PT
g
1-g
= constt.
From (i) and (ii)
....(ii)
g
3
= -3 Þ g = -3 + 3g Þ g =
1- g
2
13
P-199
Oscillations
Oscillations
TOPIC 1
1.
2.
3.
Displacement, Phase, Velocity
and Acceleration in S.H.M.
5.
The position co-ordinates of a particle moving in a 3-D
coordinate system is given by
[9 Jan 2019, II]
x = a coswt
y = a sinwt
and z = awt
The speed of the particle is:
(a)
(c)
2 aw (b) aw
3 aw (d) 2aw
Two simple harmonic motions, as shown, are at right angles.
They are combined to form Lissajous figures.
x(t) = A sin (at + d)
y(t) = B sin (bt)
Identify the correct match below
[Online April 15, 2018]
p
(a) Parameters: A = B, a = 2b; d = ; Curve: Circle
2
p
(b) Parameters: A = B, a = b; d = ; Curve: Line
2
p
(c) Parameters: A ¹ B, a = b; d = ; Curve: Ellipse
2
(d) Parameters: A ¹ B, a = b; d = 0; Curve: Parabola
The ratio of maximum acceleration to maximum velocity in
a simple harmonic motion is 10 s–1. At, t = 0 the displacement
is 5 m. What is the maximum acceleration ? The initial phase
is p
[Online April 8, 2017]
4
(a)
500 m/s2
(b)
500
2
m/s2
Two particles are performing simple harmonic motion in a
straight line about the same equilibrium point. The
amplitude and time period for both particles are same
and equal to A and T, respectively. At time t = 0 one particle
has displacement A while the other one has displacement
-A
and they are moving towards each other. If they cross
2
each other at time t, then t is:
[Online April 9, 2016]
(a)
6.
7.
5T
6
(b)
T
3
(c)
T
4
(d)
T
6
A simple harmonic oscillator of angular frequency 2 rad
s–1 is acted upon by an external force F = sin t N. If the
oscillator is at rest in its equilibrium position at t = 0, its
position at later times is proportional to :
[Online April 10, 2015]
1
1
(a) sin t + cos 2t
(b) cos t - sin 2t
2
2
1
1
(c) sin t - sin 2t
(d) sin t + sin 2t
2
2
x and y displacements of a particle are given as x(t) = a sin
wt and y (t) = a sin 2wt. Its trajectory will look like :
[Online April 10, 2015]
y
y
x
(a)
750 m/s2
4.
(c)
(d)
750 2 m/s2
A particle performs simple harmonic mition with amplitude
A. Its speed is trebled at the instant that it is at a distance
x
(b)
2A
from equilibrium position. The new amplitude of the
3
motion is :
[2016]
7A
A
41 (d) 3A
(a) A 3
(b)
(c)
3
3
y
y
(c)
x
(d)
x
P-200
8.
9.
10.
Physics
A body is in simple harmonic motion with time period half
second (T = 0.5 s) and amplitude one cm (A = 1 cm). Find
the average velocity in the interval in which it moves form
equilibrium position to half of its amplitude.
[Online April 19, 2014]
(a) 4 cm/s (b) 6 cm/s (c) 12 cm/s (d) 16 cm/s
Which of the following expressions corresponds to simple
harmonic motion along a straight line, where x is the
displacement and a, b, c are positive constants?
[Online April 12, 2014]
2
(a) a + bx – cx
(b) bx2
(c) a – bx + cx2
(d) – bx
A particle which is simultaneously subjected to two
perpendicular simple harmonic motions represented by;
x = a1 cos wt and y = a2 cos 2 wt traces a curve given by:
[Online April 9, 2014]
(a)
y
y
a2
a2
a1
x
O
(b)
12. Two particles are executing simple harmonic motion of
the same amplitude A and frequency w along the x-axis.
Their mean position is separated by distance X0(X0 > A).
If the maximum separation between them is (X0 + A), the
phase difference between their motion is:
[2011]
(a)
(c)
11.
x
O
(d)
a1
y
(c)
(d)
[2011]
1
(a)
M +m
M
(c)
æ M + mö 2
çè
÷
M ø
(b)
æ M ö2
çè
÷
M + mø
(d)
M
M +m
1
is written as a = A cos(wt + d ) ,then
a2
x
t
y
t
p
2
[2007]
A = x0 w 2 , d = 3p / 4 (b) A = x0, d = -p / 4
2
O
(b)
(d)
æ A1 ö
(c) A = x0 w 2 , d = p / 4 (d) A = x0 w , d = -p / 4
15. A coin is placed on a horizontal platform which undergoes
vertical simple harmonic motion of angular frequency w.
The amplitude of oscillation is gradually increased. The
coin will leave contact with the platform for the first time
(a) at the mean position of the platform
[2006]
g
(b) for an amplitude of 2
w
(c) for an amplitude of
16.
t
p
6
law x = x0 cos(wt - p / 4) . If the acceleration of the particle
The displacement y(t) = A sin (wt + f) of a pendulum for
2p
is correctly represented by
f=
3
[Online May 19, 2012]
y
y
(a)
(c)
of ç A ÷ is:
è 2ø
x
y
a2
p
4
13. A mass M, attached to a horizontal spring, executes S.H.M.
with amplitude A1. When the mass M passes through its
mean position then a smaller mass m is placed over it and
both of them move together with amplitude A2. The ratio
(a)
a1
(b)
14. A point mass oscillates along the x-axis according to the
O a1
y
p
3
t
g2
w2
(d) at the highest position of the platform
The maximum velocity of a particle, executing simple
harmonic motion with an amplitude 7 mm, is 4.4 m/s. The
period of oscillation is
[2006]
(a) 0.01 s (b) 10 s
(c) 0.1 s
(d) 100 s
17. The function sin 2 (wt ) represents
[2005]
(a) a periodic, but not simple harmonic motion with a
p
period
w
(b) a periodic, but not simple harmonic motion with a
2p
period
w
(c) a simple harmonic motion with a period
p
w
(d) a simple harmonic motion with a period
2p
w
P-201
Oscillations
18.
Two simple harmonic motions are represented by the
pö
æ
equations y1 = 0.1 sin ç100pt + ÷ and y 2 = 0.1 cos pt .
3ø
è
19.
The phase difference of the velocity of particle 1 with
respect to the velocity of particle 2 is
[2005]
p
-p
p
-p
(a)
(b)
(c)
(d)
6
6
3
3
Two particles A and B of equal masses are suspended from
two massless springs of spring constants k 1 and k 2,
respectively. If the maximum velocities, during
oscillation, are equal, the ratio of amplitude of A and B
is
[2003]
k2
k1
k1
k2
(b) k
(c)
(d) k
k2
k1
1
2
The displacement of a particle varies according to the
(a)
20.
relation x = 4(cos pt + sin pt ). The amplitude of the
particle is
[2003]
(a) – 4
TOPIC 2
(d) 8
Energy in Simple Harmonic
Motion
The displacement time graph of a particle executing S.H.M.
is given in figure : (sketch is schematic and not to scale)
displacement
21.
(c) 4 2
(b) 4
O
2T
4
T
4
3T T
4
time (s)
5T
4
Which of the following statements is/are true for this
motion?
[Sep. 02, 2020 (II)]
3T
4
(2) The acceleration is maximum at t = T
(1) The force is zero at t =
(3) The speed is maximum at t =
T
4
(4) The P.E. is equal to K.E. of the oscillation at t =
22.
T
2
(a) (1), (2) and (4)
(b) (2), (3) and (4)
(c) (1), (2) and (3)
(d) (1) and (4)
A particle undergoing simple harmonic motion has time
pt
. The
90
ratio of kinetic to potential energy of this particle at
t = 210s will be :
[11 Jan 2019, I]
23. A pendulum is executing simple harmonic motion and its
maximum kinetic energy is K1. If the length of the
pendulum is doubled and it performs simple harmonic
motion with the same amplitude as in the first case, its
maximum kinetic energy is K2.
[11 Jan 2019, II]
(a) K2 = 2K1
24.
(a)
(b) 1
(c) 2
(d)
1
3
K2 =
K1
2
K1
(d) K2 = K1
4
A particle is executing simple harmonic motion (SHM) of
amplitude A, along the x-axis, about x = 0. When its
potential Energy (PE) equals kinetic energy (KE), the
position of the particle will be:
[9 Jan 2019, II]
(c)
K2 =
(a)
A
2
(b)
A
2 2
A
(d) A
2
25. A particle is executing simple harmonic motion with a time
period T. At time t = 0, it is at its position of equilibrium.
The kinetic energy-time graph of the particle will look like:
[2017]
(c)
(a)
(b)
(c)
(d)
26. A block of mass 0.1 kg is connected to an elastic spring of
spring constant 640 Nm–1 and oscillates in a medium of
constant 10–2 kg s–1. The system dissipates its energy
gradually. The time taken for its mechanical energy of
vibration to drop to half of its initial value, is closest to :
[Online April 9, 2017]
(a) 2 s
(b) 3.5 s
(c) 5 s
(d) 7 s
27. For a simple pendulum, a graph is plotted between its
kinetic energy (KE) and potential energy (PE) against its
displacement d. Which one of the following represents these
correctly? (graphs are schematic and not drawn to scale)
[2015]
E
KE
E
PE
(a)
d
dependent displacement given by x ( t ) = A sin
1
9
(b)
(b)
KE
PE
E
(c)
E
KE
PE
(d)
d
PE
KE
d
P-202
28.
29.
30.
A pendulum with time period of 1s is losing energy. At
certain time its energy is 45 J. If after completing 15
oscillations, its energy has become 15 J, its damping
constant (in s–1) is :
[Online April 11, 2015]
1
1
1
ln3 (c) 2
ln3
(a)
(b)
(d)
30
15
2
This question has Statement 1 and Statement 2. Of the
four choices given after the Statements, choose the one
that best describes the two Statements.
If two springs S1 and S2 of force constants k1 and k 2
respectively, are stretched by the same force, it is found
that more work is done on spring S1 than on spring S2.
Statement 1 : If stretched by the same amount work done
on S1
Statement 2 : k1 < k2
[2012]
(a) Statement 1 is false, Statement 2 is true.
(b) Statement 1 is true, Statement 2 is false.
(c) Statement 1 is true, Statement 2 is true, Statement 2 is
the correct explanation for Statement 1
(d) Statement 1 is true, Statement 2 is true, Statement 2 is
not the correct explanation for Statement 1
A particle of mass m executes simple harmonic motion with
amplitude a and frequency n. The average kinetic energy
during its motion from the position of equilibrium to the
end is
[2007]
2
2 2
2
2
2
(a) 2p ma n
(b) p ma n
1
2 2
ma n
(d) 4p 2 ma 2 n2
4
Starting from the origin a body oscillates simple
harmonically with a period of 2 s. After what time will its
kinetic energy be 75% of the total energy?
[2006]
(c)
31.
1
1
1
1
s
s
s
s
(b)
(d)
(c)
6
3
4
12
The total energy of a particle, executing simple harmonic
motion is
[2004]
(a) independent of x
(b) µ x2
(c) µ x
(d) µ x1/2
where x is the displacement from the mean position, hence
total energy is independent of x.
A body executes simple harmonic motion. The potential
energy (P.E), the kinetic energy (K.E) and total energy (T.E)
are measured as a function of displacement x. Which of
the following statements is true ?
[2003]
(a) K.E. is maximum when x = 0
(b) T.E is zero when x = 0
(c) K.E is maximum when x is maximum
(d) P.E is maximum when x = 0
In a simple harmonic oscillator, at the mean position
[2002]
(a) kinetic energy is minimum, potential energy is maximum
(b) both kinetic and potential energies are maximum
(c) kinetic energy is maximum, potential energy is minimum
(d) both kinetic and potential energies are minimum
(a)
32.
33.
34.
Physics
Time Period, Frequency,
TOPIC 3 Simple Pendulum and Spring
Pendulum
35. An object of mass m is suspended at the end of a massless
wire of length L and area of cross-section, A. Young
modulus of the material of the wire is Y. If the mass is
pulled down slightly its frequency of oscillation along the
vertical direction is:
[Sep. 06, 2020 (I)]
(a)
f =
1 mL
2p YA
(b) f =
1 YA
2p mL
1 mA
1 YL
(d) f =
2p YL
2p mA
36. When a particle of mass m is attached to a vertical spring
of spring constant k and released, its motion is described
by y (t) = y0 sin2wt, where ‘y’ is measured from the lower
(c)
f =
end of unstretched spring. Then w is :
[Sep. 06, 2020 (II)]
(a)
1
2
g
y0
(b)
g
2 y0
(d)
g
y0
2g
y0
37. A block of mass m attached to a massless spring is
performing oscillatory motion of amplitude 'A' on a
frictionless horizontal plane. If half of the mass of the block
breaks off when it is passing through its equilibrium point,
the amplitude of oscillation for the remaining system
become fA. The value of f is :
[Sep. 03, 2020 (II)]
1
1
(a)
(b) 1
(c)
(d)
2
2
2
38. A person of mass M is, sitting on a swing of length L and
swinging with an angular amplitude q0. If the person stands
up when the swing passes through its lowest point, the
work done by him, assuming that his centre of mass
moves by a distance l (l<<L), is close to :
[12 April 2019, II]
(b) mgl (1+q02)
(a) mgl (1– q02)
(c)
æ q0 2
(d) Mgl çç 1 + 2
è
(c) mgl
ö
÷
÷
ø
39. A simple pendulum oscillating in air has period T. The bob
of the pendulum is completely immersed in a non-viscous
liquid. The density of the liquid is
1
th of the material of
16
the bob. If the bob is inside liquid all the time, its period of
oscillation in this liquid is :
[9 April 2019 I]
(a) 2T
1
10
(b) 2T
1
1
(c) 4T
14
15
(d) 4T
1
14
P-203
Oscillations
40.
Two light identical springs of spring constant k are attached
horizontally at the two ends of a uniform horizontal rod AB of
length l and mass m. The rod is pivoted at its centre ‘O’ and
can rotate frreely in horizontal plane. The other ends of two
springs are fixed to rigid supports as shown in figure. The rod
is gently pushed through a small angle and released. The
frequency of resulting oscillation is:
[12 Jan 2019, I]
A
2
3
s
s
(b)
3
2
3
s
(c)
(d) 2 3s
2
45. A particle executes simple harmonic motion with an
amplitude of 5 cm. When the particle is at 4 cm from the
mean position, the magnitude of its velocity in SI units
is equal to that of its acceleration. Then, its periodic time
in seconds is:
[10 Jan 2019, II]
(a)
y
(a)
O
4p
3
(b)
3
p
8
8p
7
(d)
p
3
3
A cylindrical plastic bottle of negligible mass is filled
with 310 ml of water and left floating in a pond with still
water. If pressed downward slightly and released, it
starts performing simple harmonic motion at angular
frequency w. If the radius of the bottle is 2.5 cm then w
is close to: (density of water = 103 kg/m3)
[10 Jan 2019, II]
–1
(a) 3.75 rad s
(b) 1.25 rad s–1
–1
(c) 2.50 rad s
(d) 5.00 rad s–1
A rod of mass 'M' and length '2L' is suspended at its middle
by a wire. It exhibits torsional oscillations; If two masses
each of 'm' are attached at distance 'L/2' from its centre on
both sides, it reduces the oscillation frequency by 20%.
The value of ratio m/M is close to :
[9 Jan 2019, II]
(a) 0.77
(b) 0.57
(c) 0.37
(d) 0.17
A silver atom in a solid oscillates in simple harmonic motion
in some direction with a frequency of 1012/sec. What is the
force constant of the bonds connecting one atom with the
other? (Mole wt. of silver = 108 and Avagadro number
= 6.02 ×1023 gm mole –1)
[2018]
(a) 6.4 N/m (b) 7.1 N/m (c) 2.2 N/m (d) 5.5 N/m
A particle executes simple harmonic motion and is located
at x = a, b and c at times t0, 2t0 and 3t0 respectively. The
frequency of the oscillation is [Online April 16, 2018]
(c)
x
46.
B
1 3k
1 2k
1 6k
1 k
(b)
(c)
(d)
2π m
2π m
2π m
2π m
A simple pendulum, made of a string of length l and a bob
of mass m, is released from a small angle q0. It strikes a
block of mass M, kept on a horizontal surface at its lowest
point of oscillations, elastically. It bounces back and goes
up to an angle q1. The M is given by : [12 Jan 2019, I]
(a)
41.
(a)
m æ θ0 + θ1 ö
ç
÷
2 è θ0 - θ1 ø
(b)
47.
æ θ -θ ö
mç 0 1 ÷
è θ0 + θ1 ø
æθ +θ ö
m æ θ0 - θ1 ö
mç 0 1 ÷
ç
÷
(d)
2 è θ0 + θ1 ø
è θ0 - θ1 ø
A simple harmonic motion is represented by :
(c)
42.
48.
y = 5(sin3pt + 3 cos3pt) cm
The amplitude and time period of the motion are :
[12 Jan 2019, II]
3
s
2
3
2
(c) 5 cm, s
(d) 5 cm, s
2
3
A simple pendulum of length 1 m is oscillating with an
angular frequency 10 rad/s. The support of the pendulum
starts oscillating up and down with a small angular
frequency of 1 rad/s and an amplitude of 10–2 m. The
relative change in the angular frequency of the pendulum
is best given by :
[11 Jan 2019, II]
(a) 10–3 rad/s
(b) 1 rad/s
(c) 10–1 rad/s
(d) 10–5 rad/s
The mass and the diameter of a planet are three times the
respective values for the Earth. The period of oscillation
of a simple pendulum on the Earth is 2s. The period of
oscillation of the same pendulum on the planet would be:
[11 Jan 2019, II]
(a) 10 cm,
43.
44.
2
s
3
(b) 10 cm,
49.
(a)
1
æa+bö
cos -1 ç
÷
2pt 0
è 2c ø
(b)
1
æa+bö
cos -1 ç
÷
2pt 0
è 3c ø
(c)
1
æ 2a + 3c ö
cos -1 ç
÷ (d)
2 pt 0
è b ø
1
æa +cö
cos-1 ç
÷
2 pt 0
è 2b ø
50. In an experiment to determine the period of a simple pendulum of length 1 m, it is attached to different spherical
bobs of radii r1 and r2. The two spherical bobs have uniform mass distribution. If the relative difference in the periods, is found to be 5 × 10–4 s, the difference in radii, |r1– r2|
is best given by:
[Online April 9, 2017]
(a) 1 cm
(b) 0.1 cm (c) 0.5 cm (d) 0.01 cm
P-204
52.
53.
54.
1
1
1
Hz (c)
Hz
Hz
(a)
(b)
(d) 2 Hz
2 2
4
2
A pendulum clock loses 12 s a day if the temperature is
40°C and gains 4 s a day if the temperature is 20° C. The
temperature at which the clock will show correct time, and
the co-efficient of linear expansion (a) of the metal of the
pendulum shaft are respectively :
[2016]
(a) 30°C; a = 1.85 × 10–3/°C
(b) 55°C; a = 1.85 × 10–2/°C
(c) 25°C; a = 1.85 × 10–5/°C
(d) 60°C; a = 1.85 × 10–4/°C
In an engine the piston undergoes vertical simple harmonic
motion with amplitude 7 cm. A washer rests on top of the
piston and moves with it. The motor speed is slowly
increased. The frequency of the piston at which the washer
no longer stays in contact with the piston, is close to :
[Online April 10, 2016]
(a) 0.7 Hz (b) 1.9 Hz (c) 1.2 Hz (d) 0.1 Hz
A pendulum made of a uniform wire of cross sectional area
A has time period T. When an additional mass M is added
to its bob, the time period changes to TM. If the Young's
1
modulus of the material of the wire is Y then
is equal
Y
2
2
A 1 kg block attached to a spring vibrates with a frequency
of 1 Hz on a frictionless horizontal table. Two springs
identical to the original spring are attached in parallel to an
8 kg block placed on the same table. So, the frequency of
vibration of the 8 kg block is :
[Online April 8, 2017]
T (in s )
51.
Physics
8.0
6.0
4.0
2.0
0.5 1.0 1.5 2.0 L (in m)
What is the value of g at the place?
(a) 9.81 m/s2
(b) 9.87 m/s2
2
(c) 9.91 m/s
(d) 10.0 m/s2
57. The amplitude of a simple pendulum, oscillating in air with
a small spherical bob, decreases from 10 cm to 8 cm in 40
seconds. Assuming that Stokes law is valid, and ratio of
the coefficient of viscosity of air to that of carbon dioxide
is 1.3. The time in which amplitude of this pendulum will
reduce from 10 cm to 5 cm in carbon dioxide will be close to
(In 5 = 1.601, In 2 = 0.693).
[Online April 9, 2014]
(a) 231 s
(b) 208 s
(c) 161 s
(d) 142 s
58. Two bodies of masses 1 kg and 4 kg are connected to a
vertical spring, as shown in the figure. The smaller mass
executes simple harmonic motion of angular frequency
25 rad/s, and amplitude 1.6 cm while the bigger mass
remains stationary on the ground. The maximum force
exerted by the system on the floor is (take g = 10 ms–2)
[Online April 9, 2014]
to:
1 kg
(g = gravitational acceleration)
(a)
é æ T ö2 ù A
ê1 - ç M ÷ ú
ë è T ø û Mg
éæ T ö2 ù A
êç M ÷ - 1ú
ëè T ø
û Mg
55.
56.
[2015]
é æ T ö2 ù A
(b) ê1 - ç T ÷ ú Mg
ëê è M ø úû
éæ T
êç M
ëè T
2
ù Mg
ö
(d)
(c)
÷ - 1ú
ø
û A
A particle moves with simple harmonic motion in a straight
line. In first t s, after starting from rest it travels a distance
a, and in next t s it travels 2a, in same direction, then:
(a) amplitude of motion is 3a
[2014]
(b) time period of oscillations is 8t
(c) amplitude of motion is 4a
(d) time period of oscillations is 6t
In an experiment for determining the gravitational
acceleration g of a place with the help of a simple
pendulum, the measured time period square is plotted
against the string length of the pendulum in the figure.
[Online April 19, 2014]
4 kg
(a) 20 N
(b) 10 N
(c) 60 N
(d) 40 N
59. An ideal gas enclosed in a vertical cylindrical container
supports a freely moving piston of mass M. The piston
and the cylinder have equal cross sectional area A. When
the piston is in equilibrium, the volume of the gas is V0
and its pressure is P0. The piston is slightly displaced
from the equilibrium position and released. Assuming that
the system is completely isolated from its surrounding,
the piston executes a simple harmonic motion with
frequency
[2013]
V
MP
1 0 0
1 AgP0
(a)
(b)
2p A 2 g
2p V0 M
(c)
1
2p
A 2 gP0
MV0
(d)
1
2p
MV0
AgP0
P-205
Oscillations
60.
A mass m = 1.0 kg is put on a flat pan attached to a vertical
spring fixed on the ground. The mass of the spring and the
pan is negligible. When pressed slightly and released, the
mass executes simple harmonic motion. The spring
constant is 500 N/m. What is the amplitude A of the
motion, so that the mass m tends to get detached from
the pan ?
(Take g = 10 m/s2).
The spring is stiff enough so that it does not get distorted
during the motion.
[Online April 22, 2013]
m
61.
62.
63.
(b) b
(c)
1
b
(d)
2
b
A ring is suspended from a point S on its rim as shown in
the figure. When displaced from equilibrium, it oscillates
with time period of 1 second. The radius of the ring is
(take g =
p2)
[Online May 19, 2012]
S
(a) 0.15 m
64.
(b) 1.5 m
(b)
ld
(r - d ) g
(d) 2p
2p
lr
dg
lr
(r - d ) g
65. If x, v and a denote the displacement, the velocity and
the acceleration of a particle executing simple harmonic
motion of time period T, then, which of the following
does not change with time?
[2009]
(a) aT/x
(b) aT + 2pv
(c) aT/v
(d) a2T2 + 4p2v2
(a) 2 f
(d) 0.5 m
A wooden cube (density of wood ‘d’) of side ‘l’ floats in a
liquid of density ‘r’ with its upper and lower surfaces
horizontal. If the cube is pushed slightly down and
released, it performs simple harmonic motion of period
‘T’
[2011 RS]
(b) f /2
k2
(c) f /4
(d) 4 f
67. The displacement of an object attached to a spring and
executing simple harmonic motion is given by x = 2 × 10–2
cos pt metre.The time at which the maximum speed first
occurs is
[2007]
(a) 0.25 s
(b) 0.5 s
(c) 0.75 s
(d) 0.125 s
68. The bob of a simple pendulum is a spherical hollow ball
filled with water. A plugged hole near the bottom of the
oscillating bob gets suddenly unplugged. During
observation, till water is coming out, the time period of
oscillation would
[2005]
(a) first decrease and then increase to the original value
(b) first increase and then decrease to the original value
(c) increase towards a saturation value
(d) remain unchanged
69. If a simple harmonic motion is represented by
2
d x
dt 2
(a)
(c) 1.0 m
m
k1
If a simple pendulum has significant amplitude (up to a
factor of 1/e of original) only in the period between t =
0s to t = t s, then t may be called the average life of the
pendulum. When the spherical bob of the pendulum
suffers a retardation (due to viscous drag) proportional
to its velocity with b as the constant of proportionality,
the average life time of the pendulum in second is
(assuming damping is small)
[2012]
0.693
b
(c) 2p
ld
rg
(b) A = 2.0 cm
(c) A < 2.0 cm
(d) A = 1.5 cm
Two simple pendulums of length 1 m and 4 m respectively
are both given small displacement in the same direction
at the same instant. They will be again in phase after the
shorter pendulum has completed number of oscillations
equal to :
[Online April 9, 2013]
(a) 2
(b) 7
(c) 5
(d) 3
(a)
2p
66. Two springs, of force constants k1 and k2 are connected
to a mass m as shown. The frequency of oscillation of
the mass is f. If both k1 and k2 are made four times their
original values, the frequency of oscillation becomes
[2007]
k
(a) A > 2.0 cm
(a)
+ ax = 0 , its time period is
2p
a
(b)
2p
a
(c)
[2005]
2p a
(d)
2pa
70. The bob of a simple pendulum executes simple harmonic
motion in water with a period t, while the period of oscillation
of the bob is t0 in air. Neglecting frictional force of water and
given that the density of the bob is (4 / 3) ´ 1000 kg/m 3 .
Which relationship between t and t0 is true?
[2004]
(a) t = 2t 0
(b) t = t 0 / 2
(c) t = t0
(d) t = 4t 0
P-206
71.
72.
73.
Physics
A particle at the end of a spring executes S.H.M with a
period t1. while the corresponding period for another spring
is t2. If the period of oscillation with the two springs in
series is T then
[2004]
-1
-1
-1
= t1 + t 2
(a)
T
(c)
T = t1 + t2
(b) T 2 = t12 + t 22
(d) T -2 = t1-2 + t 2-2
A mass M is suspended from a spring of negligible mass.
The spring is pulled a little and then released so that the
mass executes SHM of time period T. If the mass is
5T
increased by m, the time period becomes
. Then the
3
m
ratio of
is
[2003]
M
3
25
16
5
(b)
(c)
(d)
(a)
5
9
9
3
The length of a simple pendulum executing simple harmonic
motion is increased by 21%. The percentage increase in
the time period of the pendulum of increased length is
[2003]
(a) 11%
74.
(c) 42%
(d) 10%
If a spring has time period T, and is cut into n equal parts,
then the time period of each part will be
[2002]
T n
(b) T / n (c) nT
(d) T
A child swinging on a swing in sitting position, stands up,
then the time period if the swing will
[2002]
(a)
75.
(b) 21%
(a) increase
(b) decrease
(c) remains same
(d) increases if the child is long and decreases if the
child is short
TOPIC 4
Damped, Forced
Oscillations and Resonance
76. A damped harmonic oscillator has a frequency of 5
oscillations per second. The amplitude drops to half its
value for every 10 oscillations. The time it will take to drop
to
1
of the original amplitude is close to :
1000
(a) 50s
(b) 100s
(c) 20s
[8 April 2019, II]
(d) 10s
77. The displacement of a damped harmonic oscillator is given
by x(t) = e–0.1t. cos(10pt + j). Here t is in seconds.
The time taken for its amplitude of vibration to drop to half
of its initial value is close to :
[9 Jan 2019, II]
(a) 4s
(b) 7s
(c) 13s
(d) 27s
78. An oscillator of mass M is at rest in its equilibrium position
1
2
in a potential V = k(x - X) . A particle of mass m comes
2
from right with speed u and collides completely inelastically
with M and sticks to it. This process repeats every time
the oscillator crosses its equilibrium position. The
amplitude of oscillations after 13 collisions is:
(M = 10, m = 5, u = 1, k = 1).
[Online April 16, 2018]
(a)
1
2
(b)
1
3
(c)
2
3
(d)
3
5
79. The angular frequency of the damped oscillator is given
æk
r2 ö
÷ where k is the spring constant, m
by, w = çç m 4m 2 ÷ø
è
is the mass of the oscillator and r is the damping constant.
r2
is 8%, the change in time period
mk
compared to the undamped oscillator is approximately
as follows:
[Online April 11, 2014]
If the ratio
(a) increases by 1%
(b) increases by 8%
(c) decreases by 1%
(d) decreases by 8%
80. The amplitude of a damped oscillator decreases to 0.9 times
its original magnitude in 5s. In another 10s it will decrease
to a times its original magnitude, where a equals [2013]
(a) 0.7
(b) 0.81
(c) 0.729
(d) 0.6
81. A uniform cylinder of length L and mass M having crosssectional area A is suspended, with its length vertical, from
a fixed point by a massless spring, such that it is half
submerged in a liquid of density s at equilibrium position.
When the cylinder is given a downward push and released,
it starts oscillating vertically with a small amplitude. The
time period T of the oscillations of the cylinder will be :
[Online April 25, 2013]
é
ù
M
(a) Smaller than 2p ê
ú
ë (k + Asg ) û
(b) 2p
M
k
é
ù
M
(c) Larger than 2p ê
ú
ë (k + Asg ) û
(d)
1
é
ù
M
2p ê
ú
(
k
+
A
s
g
)
ë
û
1
2
1
2
2
P-207
Oscillations
82. Bob of a simple pendulum of length l is made of iron.
The pendulum is oscillating over a horizontal coil
carrying direct current. If the time period of the pendulum
is T then :
[Online April 23, 2013]
(a) T < 2p
l
and damping is smaller than in air alone.
g
(b) T = 2p
l
and damping is larger than in air alone.
g
(c) T > 2p
l
and damping is smaller than in air alone.
g
(d) T < 2p
l
and damping is larger than in air alone.
g
83. In forced oscillation of a particle the amplitude is
maximum for a frequency w1 of the force while the energy
is maximum for a frequency w2 of the force; then
[2004]
(a) w1 < w2 when damping is small and w1 > w2 when
damping is large
(b) w1 > w2
(c) w1 = w2
(d) w1 < w2
84. A particle of mass m is attached to a spring (of spring
constant k) and has a natural angular frequency w0. An
external force F(t) proportional to cos wt (w ¹ w 0 ) is
applied to the oscillator. The displacement of the oscillator
will be proportional to
[2004]
1
1
(a)
(b) m (w 2 - w 2 )
2
2
m (w 0 + w )
0
m
m
(c)
2
2
(d)
2
2
w0 - w
(w 0 + w )
P-208
1.
Physics
(a) Here, vx = – a w sin wt, vy = a w cos wt and vz = aw
Þ v = v 2x + v 2y + v z2
Þv=
2.
( –aw sin wt ) + ( aw cos wt )
2
2
+ ( aw )
5.
12 0°
y2
2 xy
+ 2cos d = sin 2 d
2
AB
A
B
x = A sin (at + d)
y = B sin (bt + r)
3.
60°
Clearly A ¹ B hence ellipse.
(b) Maximum velocity in SHM, vmax = aw
Maximum acceleration in SHM, Amax = aw2
where a and w are maximum amplitude and angular
frequency.
A
Given that, max = 10
v max
i.e., w = 10 s–1
Displacement is given by
x = a sin (wt + p/4)
at t = 0, x = 5
5 = a sin p/4
p
rad.
3
If they cross each other at time t then
Angle covered to meet q = 60° =
t<
6.
2
2 æ 2A ö
Finally 3V = w A' - ç
÷
è 3 ø
7.
2
trebled)
On dividing we get
æ 2A ö
A2 - ç
÷
è 3 ø
0
0
x
t
0
0
1
sin 2t
2
(c) At t = 0, x (t) = 0 ; y (t) = 0
x (t) is a sinusoidal function
x = sin t –
Where A'= final amplitude (Given at x =
3
=
1
t
ò dV µ ò sin t dt
ò dx = ò (- cos t + 1) dt
2
2
(b) We know that V = w A - x
æ 2A ö
A' 2 - ç
÷
è 3 ø
0
V µ – cos t + 1
Maximum acceleration Amax = aw2 = 500 2 m/s2
2 æ 2A ö
Initially V = w A - ç
÷
è 3 ø
q
p
T
<
T<
2p 3≥2p
6
(c) As we know,
F = ma Þ a µ F
or, a µ sin t
dv
Þ
µ sin t
dt
Þ
5 = a sin 45° Þ a = 5 2
4.
A
O
A
2
(at time t = 0)
2
v = 2aw
(c) From the two mutually perpendicular S.H.M.’s, the
general equation of Lissajous figure,
x2
Equilibrium point
(d)
2A
, velocity to
3
2
p
; x (t) = a and y (t) = 0
2w
Hence trajectory of particle will look like as (c).
(c) Given: Time period, T = 0.5 sec
Amplitude, A = 1 cm
Average velocity in the interval in which body moves
from equilibrium to half of its amplitude, v = ?
S
At t =
8.
2
é 2 4A 2 ù
4A 2
9 ê A - 9 ú = A'2 –
9
ëê
ûú
\ A' =
7A
3
O
P-209
Oscillations
Time taken to a displacement A/2 where A is the
9.
T
amplitude of oscillation from the mean position ‘O’ is
12
0.5
Therefore, time, t =
sec
12
A 1
Displacement, s = = cm
2 2
A
1
\ Average velocity, v = 2 = 2 = 12 cm / s
0.5
t
12
(d) In linear S.H.M., the restoring force acting on particle
should always be proportional to the displacement of the
particle and directed towards the equilibrium position.
i.e., F µ x
æ
Þ A1 M = A2 M + m
A
1
Þ A =
2
m+M
M
14. (a) Given,
Displacement, x = x0 cos (wt – p / 4 )
\ Velocity, v =
dx
æ
= - x0 w sin ç wt è
dt
pö
÷
4ø
Acceleration,
a=
dv
pö
æ
2
= - x0w cos ç wt - ÷
è
dt
4ø
é
pö ù
æ
2
= x0w cos ê p + ç wt - ÷ ú
è
4øû
ë
x
= cos wt
a1
3p ö
æ
= x0 w 2 cos ç wt + ÷
è
4ø
and from eqn (2)
...(1)
Acceleration, a = A cos (wt + d) ...(2)
Comparing the two equations, we get
y
= 2cos wt
a2
æ
fö
15.
N
A
x
mg
mean
position
mg – N = mw2x
where x is the distance from mean position
For block to leave contact N = 0
f
x2 – x1 = 2A cos çè wt + 2 ÷ø sin 2
The above equation is SHM with amplitude 2A sin
3p
.
4
(b) For block A to move in SHM.
A = x0w2 and d =
a
\ y= 2 2 x
a1
11. (a) Displacement y (t) = A sin (wt + f)[Given]
2p
For f =
3
2p
at t = 0; y = A sin f = A sin
3
= A sin 120° = 0.87 A [Q sin 120° ; 0.866]
Graph (a) depicts y = 0.87A at t = 0
12. (a) Let, x1 = A sin w t and x2 = A sin (w t + f)
f
2
f
2
\ 2 A sin = A
13.
k ö
\ çV = A M ÷
è
ø
or F = - bx where b is a positive constant.
10. (b) Two perpendicular S.H.Ms are
x = a1 cos wt
....(1)
and y = a2 2 cos wt
....(2)
From eqn (1)
Þ sin
k
k
= ( M + m) A2
M
m+M
MA1
f 1
p
= Þf=
2 2
3
(c) At mean position, F net = 0
Therefore, by principal of conseruation of linear
momentum.
\ Mv1 = (M + m)v2
M w, a, = (M + m) w2 a2
Þ mg = mw 2 x Þ x =
g
w2
16. (a) Maximum velocity,
vmax = a w
Here, a = amplitude of SHM
w = angular velocity of SHM
vmax = a ´
ÞT =
2p
\
T
2p ö
æ
çèQ w = ÷ø
T
2pa 2 ´ 3.14 ´ 7 ´ 10
=
vmax
4.4
-3
» 0.01 s
P-210
17.
Physics
(a) Clearly sin 2wt is a periodic function with period
p
w
x = 4 2 sin(p t + 45°)
On comparing it with standard equation x = A sin(wt + f)
we get A = 4 2
21. (c) From graph equation of SHM
0
2
For SHM
d y
dt
2
y = sin2 wt =
=
p/w
2p/w
X = A cos wt
3p/w
3T
particle is at mean position.
4
\ Acceleration = 0, Force = 0
(2) At T particle again at extreme position so acceleration
is maximum.
(1) At
µ -y
1 – cos 2wt
2
1 1
– cos 2wt
2 2
(3) At t =
dy 1
= ´ 2w sin 2wt = 2w sin wt cos wt
dt 2
= w sin 2wt
maximum.
Acceleration = 0
(4) When KE = PE
v=
2
d y
2
= 2w cos 2wt which is not
dt 2
proportional to –y. Hence, it is not in SHM.
18. (b) Velocity of particle 1,
Acceleration, a =
dy1
pö
æ
= 0.1 ´ 100p cos ç100pt + ÷
è
dt
3ø
Velocity of particle 2,
v1 =
p p 2 p - 3p
p
=–
= f1 - f 2 = - =
3 2
6
6
(c) Maximum velocity during SHM, Vmax = Aw
But k = mw2
\ w=
k
m
\ Maximum velocity of the body in SHM
=A
k
m
20.
(c)
A
1
22. (d) Kinetic energy, k = mw 2 A 2 cos 2 wt
2
1
Potential energy, U = mw 2 A 2 sin 2 wt
2
k
1
2
2 p
= cot wt = cot
(210) =
U
90
3
1
2 2
23. (a) K = mw x
2
w=
k2
Þ
2
1
mw 2 A 2
2
A = Lq
k1
k
= A2 2
m
m
Þ A1 k 1 = A 2
+A
T
= A cos wt Þ t =
2
2
\ x = – A which is not possible
\ 1, 2 and 3 are correct.
Þ
Þ K max =
As maximum velocities are equal
\ A1
1
1
k ( A2 - x 2 ) = kx 2
2
2
Here, A = amplitude of SHM
x = displacement from mean position
Þ
Þ A2 = 2 x 2 Þ x =
dy
pö
æ
v2 = 2 = - 0.1p sin pt = 0.1p cos ç pt + ÷
è
dt
2ø
\ Phase difference of velocity of particle 1 with respect to
the velocity of particle 2 is
19.
T
, particle is at mean position so velocity is
4
A1
=
A2
Displacement, x = 4(cos pt + sin pt )
æ sin pt cos pt ö
= 2 ´ 4ç
+
÷
è 2
2 ø
= 4 2(sin p t cos 45° + cos p t sin 45°)
k2
k1
g
L
Þ K=
1 g 2 2
m. .L q
2 L
1
mgLq 2
2
K
L 1
\ 1 =
=
Þ K 2 = 2K1
K 2 2L 2
=
P-211
Oscillations
\
K1
L 1
=
=
Þ K 2 = 2K1
K 2 2L 2
1 2
kx
2
24. (c) Potential energy (U) =
1 2 1 2
kA - kx
2
2
According to the question, U = k
25.
Kinetic energy (K) =
Work done by spring S1, w1 =
1
k1x 2
2
1
1
1
\ kx 2 = kA2 - kx 2
2
2
2
A
Þ x2 = A2 or, x =±
2
(b) For a particle executing SHM
At mean position; t = 0, wt = 0, y = 0, V = Vmax = aw
1
\ K.E. = KEmax = mw2a2
2
T
p
At extreme position : t = , wt = , y = A, V = Vmin = 0
4
2
\ K.E. = KEmin = 0
Work done by spring S2, w2 =
1
k2 x 2
2
1
Kinetic energy in SHM, KE = mw2(a2 – y2)
2
1
= mw2a2cos2wt
2
Hence graph (b) correctly depicts kinetic energy time graph.
26. (b) Since system dissipates its energy gradually, and
hence amplitude will also decreases with time according to
a = a0 e–bt/m
....... (i)
Q Energy of vibration drop to half of its initial value
(E0), as E µ a2 Þ a µ E
-2
a
a = 0 Þ bt = 10 t = t
2
m
0.1
10
From eqn (i),
a0
= a 0e - t 10
2
t
27.
28.
Taking log on both sides
b
1
=
ln3
m 15
1 2
29. (b) Work done, w = kx
2
1
= e - t 10 or 2 = e10
2
t
ln 2 =
\ t = 3.5 seconds
10
1
2
2
(d) K.E = k ( A - d )
2
1 2
and P.E. = kd
2
At mean position d = 0. At extreme positions d = A
(d) As we know, E = E0 e
–
–
bt
m
b15
15 = 45e m
[As no. of oscillations = 15 so t = 15sec]
–
1
=e
3
b15
m
Since w1 > w2 Thus (k1 > k2)
30. (b) The kinetic energy of a particle executing S.H.M. at
any instant t is given by
1 2 2 2
ma w sin wt
2
where, m = mass of particle
a = amplitude
w = angular frequency
t = time
K=
The average value of sin 2wt over a cycle is
1
.
2
1ö
æ 1ö æ
1
2
\KE = mw2a2 çè ÷ø çè Q < sin q > = ÷ø
2
2
2
=
1
1
2 2
mw a = ma2 (2pn)2
4
4
2
(Q w = 2pn)
2 2
or, < K > = p ma n
31. (a) K.E. of a body undergoing SHM is given by,
1
2 2
2
ma w cos wt
2
Here, a = amplitude of SHM
w = angular velocity of SHM
K .E. =
1 2 2
Total energy in S.H.M = ma w
2
Given K.E. = 75% T.E.
1
75 1
ma 2 w 2 cos 2 w t =
´ ma 2 w 2
2
100 2
p
2
Þ 0.75 = cos wt Þ wt =
6
p
p´2
1
Þt=
Þt=
Þt= s
6´w
6 ´ 2p
6
32. (a) At any instant the total energy in SHM is
1 2
kA = constant,
2 0
where A0 = amplitude
k = spring constant
hence total energy is independent of x.
33. (a) K.E. of simple harmonic motion
1
= mw2 (a 2 - x 2 )
2
P-212
Physics
34.
(c) The kinetic energy (K. E.) of particle in SHM is given by,
1
K.E = k ( A2 - x 2 ) ;
2
1 2
Potential energy of particle in SHM is U = kx
2
Where A = amplitude and k = mw2
x = displacement from the mean position
At the mean position x = 0
1 2
\ K.E. = kA = Maximum
2
and U = 0
35. (b) An elastic wire can be treated as a spring and its spring
constant.
é
F
YA
k=
êQ Y = A
L
ë
Frequency of oscillation,
1
2p
f =
36.
Dl ù
l0 úû
l
g
When immersed non viscous liquid
39. (c) T = 2p
g ö 15 g
æ
amt = ç g - ÷ =
16 ø 16
è
l
T ¢ = 2p
0net
Now
l
t = -2Kx cos q
2
x=
k
1 YA
=
m 2p mL
y0
(1 - cos 2wt )
2
1 - cos 2wt ö
æ
2
çèQ sin wt =
÷ø
2
y0 - y0
=
cos 2wt
2
2
Þ y = A cos 2wt
y0
2
Angular velocity = 2w
37.
f=
2
2g
1 2g
Þw=
=
y0
2 y0
g
2 y0
l
2
1 2
kx
2
Here, x = distance of block from mean position,
k = spring constant
1 C
1
=
2p I 2p
Kl2
2 = 1 6K
Ml 2 2p M
12
41. (b)
q0 l
(a) Potential energy of spring =
1
At mean position, potential energy = kA2
2
At equilibrium position, half of the mass of block breaks
off, so its potential energy becomes half.
Remaining energy =
1 æ 1 2ö 1
2
ç kA ÷ø = kA '
2è2
2
Here, A' = New distance of block from mean position
Þ A' =
38.
T
q
ky0
k 2g
= mg Þ =
m y0
2
Also, spring constant k = m(2w)
k
=
m
15
Kx
æ Kl 2 ö
é
Kl 2 ù
Þ t=ç
q = -Cq êlet C =
ú
÷
2 ûú
è 2 ø
ëê
Þ So, frequency of resulting oscillations
\ Amplitude =
Þ 2w =
4
q
Þ y-
For equilibrium of mass,
15 g
16
=
40. (c) Net torque due to spring force:
(c) y = y0 sin 2 wt
Þy=
l
= 2p
(b)
mM
Velocity before colision
Velocity after colision
v1 = 2gl(1 - cos q1)
Using momentum conservation
mv = MVm – mV1
m 2gl(1 - cos q0 ) = MVm - m 2gl(1 - cos q)
Þ m 2gl
{
A
2
v = 2gl(1 - cos q0 )
and e = 1 =
}
1 - cos q0 + 1 - cos q1 = MVm
Vm + 2gl(1 - cos q1)
2gl(1 - cos q0 )
P-213
Oscillations
(
2gl (
2gl
m
... (i)
)
1 - cos q1 ) = MVM ... (ii)
1 - cos q0 - 1 - cos q1 = Vm
1 - cos q 0 +
Dividing (ii) by (i) we get
(
(
)=M
1 - cos q1 ) m
Þ w A 2 – x2 = w2 x
Þ A 2 – x 2 = w2 x 2
Þ 5 2 – 4 2 = w2 ( 4 2 )
1 - cos q0 + 1 - cos q1
3 = w × 4 Þ w=
1 - cos q0 -
\ T = 2p/w =
By componendo and dividendo rule
46. (Bonus)
æq ö
sin ç 1 ÷
1 - cos q1
è 2ø
m-M
=
=
m+M
æ q0 ö
1 - cos q0
sin ç ÷
è 2ø
2 p 8p
=
3/ 4 3
B0– B )
x
q -q
M q0 - q1
Þ
=
ÞM=m 0 1
m q0 + q1
q0 + q1
42.
pö
mg
Extra boyant force = rAxg
B0 + B = mg + ma
\ B = ma = rAxg = (pr2rg)x
ø
\ Amplitude = 10 cm
Time period, T =
2p
2p
2
=
= s
w
3p
3
( pr rg ) x
2
(a) Angular frequency of pendulum w =
\ relative change in angular frequency
g
l
Dw 1 Dg [as length remains constant]
=
w 2 g
D g = 2Aws2 [ws = angular frequency of support and, A =
amplitude]
a=
m
using, a = w2x
pr 2rg
m
W ;7.95 rads–1
47. (c)
Þw=
L/2
Dw 1 2Aws2
= ´
w 2
g
Mp æ R e
=
ç
g e M e çè R p
L– X–L
Tp
1
g
Þ = e= 3
gp
g Te
(c) Velocity, v = w A 2 – x 2
acceleration, a = –w2x
and according to question,
| v | =| a |
f1 =
1 C
2p 1
=
1 3C
2 ML2
2
2
ö
æ1ö 1
÷÷ = 3 ç ÷ =
è3ø 3
ø
C
æM M ö
L2 ç + ÷
2 ø
è 3
As frequency reduces by 80%
f2
\ f2 = 0.8 f1 Þ f = 0.8
1
Solving equations (i), (ii) & (iii)
f2 =
Þ Tp = 2 3 s
45.
m
M
GM
(d) Acceleration due to gravity g = 2
R
gp
L/2
m
1 2 ´12 ´10 –2
Dw = ´
= 10–3 rad/sec.
2
10
Also T µ
at
equilibriu
B0 = mg
y = 5 éësin(3pt) + 3 cos(3pt) ùû
(a) Given :
æ
è
44.
a
x0
Þ y = 10sin ç 3pt + 3 ÷
43.
3
4
...(i)
...(ii)
1
2p
Ratio
m
= 0.37
M
...(i)
...(ii)
...(iii)
P-214
Physics
As frequency reduces by 80%
f2
\ f2 = 0.8 f1 Þ f = 0.8
1
Solving equations (i), (ii) & (iii)
F=
...(iii)
52. (c) Time lost/gained per day =
m
= 0.37
M
48. (b) As we know, frequency in SHM
Ratio
1 k
= 1012
2p m
where m = mass of one atom
f=
108
´10-3 kg
Mass of one atom of silver, = 6.02 ´ 1023
(
49.
)
1
k
´ 6.02 ´ 1023 = 1012
2p 108 ´ 10-3
Solving we get, spring constant,
K = 7.1N/m
(d) Using y = A sin wt
a = A sin wt0
b = A sin 2wt0
c = A sin 3wt0
a + c = A[sin wt0 + A sin 3wt0] = 2A sin 2wt0 cos wt0
f<
TM
=
T
2
2
Mg
æ TM ö
= 1+
or, ç
÷
è T ø
AY
5 ´ 10-4 =
Þ
4p 2 =
k
m
l + Dl
l
l + Dl
æ TM ö
=
Þç
÷
è T ø
l
DT 1 Dl
=
T
2 l
Q change in length Dl = r1 – r2
Mgl ù
é
êëQ Dl = AY úû
2
ù A
1 éæ TM ö
ú
= êç
1
÷
Y êè T ø
úû Mg
ë
(d) In simple harmonic motion, starting from rest,
At t = 0 , x = A
x = Acoswt
..... (i)
When t = t , x = A – a
When t = 2 t , x = A –3a
From equation (i)
A – a = Acosw t
......(ii)
A – 3a = A cos2w t
....(iii)
As cos2w t = 2 cos2 w t – 1...(iv)
From equation (ii), (iii) and (iv)
\
55.
1 k
= 1 Hz
2p m
m = 1 kg
If block of mass m = 1 kg is attached then,
k = 4p2
Now, identical springs are attached in parallel with mass m
= 8 kg. Hence,
keq = 2k
k
8 kg
l
g
l + Dl
g
TM = 2p
differentiating both side,
(c) Frequency of spring (f ) =
....(ii)
When additional mass M is added then
Tµ l
51.
.... (i)
ω
g 1
1000 1
<
<
< 1.9 Hz.
2π
A 2π
7 2p
54. (c) As we know, time period, T = 2p
(b) As we know, Time-period of simple pendulum,
1 r1 - r2
2 1
r1 – r2 = 10 × 10–4
10–3 m = 10–1 cm = 0.1 cm
1
µ Dq ´ 86400 second
2
1
12 = a (40 – q) ´ 86400
2
1
4 = a (q – 20) ´ 86400
2
40 – q
On dividing we get, 3 =
q – 20
3q – 60 = 40 – q
4q = 100 Þ q = 25°C
53. (b) Washer contact with piston Þ N = 0
Given Amplitude A = 7 cm = 0.07 m.
amax = g = w2A
The frequency of piston
a+c
= 2cos wt 0
b
1
1
æ a + cö
æ a + cö
Þ w = cos -1 ç
Þf =
cos -1 ç
è 2b ÷ø
è 2b ÷ø
t0
2 pt 0
50.
1 k´2 1
= Hz
2p
g
2
2
A - 3a
æ A-a ö
= 2ç
÷ -1
A
è A ø
Þ
A - 3a 2 A2 + 2a 2 - 4 Aa - A2
=
A
A2
P-215
Oscillations
Þ A2 – 3aA = A2 + 2a2 – 4Aa
Þ 2a2 = aA
Þ A = 2a
=
=
=
a 1
=
A 2
Now, A – a = A coswt
Þ
Þ
cos wt =
59.
4 × 10 + 625 × 1.6 × 10–2 + 1 × 10
40 + 10 + 10
60 N
Mg
= P0
(c)
A
P0V0 g = PV g
Mg = P0A
… (1)
Let piston is displaced by distance x
A- a
A
P0 Ax0 g = PA( x0 - x ) g
2p
p
1
t=
Þ cos wt =
or
T
3
2
Þ T = 6t
56. (b) From graph it is clear that when
L = 1m, T2 = 4s2
As we know,
g
P=
P0 x0
( x0 - x )g
Piston
x
L
T = 2p
g
Þ
g=
x0
4 p2 L
æ P xg ö
Mg - ç 0 0 g ÷ A = Frestoring
ç ( x - x) ÷
è 0
ø
T2
2
2
æ 22 ö 1 æ 22 ö
= 4´ç ÷ ´ = ç ÷
è 7 ø 4 è 7 ø
484
g=
= 9.87m / s 2
\
49
57. (d) As we know,
x = x0 e–bt/2m
From question,
8 = 10e
-
40b
2m
Similarly, 5 = 10e
æ
ö
x0g
÷ = Frestoring [ x - x » x ]
P0 A ç1 g
0
0
ç ( x - x) ÷
0
è
ø
gP Ax
F=- 0
x0
\ Frequency with which piston executes SHM.
....(i)
-
bt
2m
....(ii)
Solving eqns (i) and (ii) we get
t @ 142 s
58. (c) Mass of bigger body M = 4 kg
Mass of smaller body m = 1 kg
Smaller mass (m = 1 kg) executes S.H.M of angular
frequency w = 25 rad/s
Amplitude x = 1.6 cm = 1.6 × 10–2
As we know,
T = 2p
or,
m
K
2p
m
= 2p
w
K
1
1
=
[Q m = 1kg; w = 25 rad / s ]
25
K
or, K = 625 Nm–1.
The maximum force exerted by the system on the floor
=
Mg + Kx + mg
or,
Cylinder
containing
ideal gas
f =
1
2p
gP0 A
1
=
x0 M
2p
gP0 A2
MV0
60. (c) As F = -kx
61. (a) Let T1 and T2 be the time period of the two pendulums
T1 = 2p
4
1
and T2 = 2p
g
g
As l1 < l 2 therefore T1 < T2
Let at t = 0 they start swinging together. Since their time
periods are different, the swinging will not be in unison
always. Only when number of completed oscillations
differ by an integer, the two pendulums will again begin
to swing together
Let longer length pendulum complete n oscillation and
shorter length pendulum complete (n + 1) oscillation. For
unison swinging
(n + 1)T1 = nT2
(n + 1) ´ 2p
l
4
= (n) ´ 2p
g
g
Þn=1
\n+1=1+1=2
P-216
62.
Physics
(d) The equation of motion for the pendulum, for
damped harmonic motion
F = – kx - bv
Þ ma + kx + bv = 0
d 2x
dx
m 2 + kx + b
=0
dt
dt
Þ
k
b dx
+ x+
=0
m dt
dt 2 m
d2 x
Þ
dt
2
+
b dx k
+ x =0
m dt m
… (1)
Let x = elt is the solution of the equation (1)
dx
d2 x
= lelt Þ
= l 2 elt
dt
dt 2
Substituting in the equation (1)
b
k
l 2 elt + l elt + elt = 0
m
m
b
k
l2 + l + = 0
m
m
b
b2
k
±
-4
2
m
m
-b ±
m
l=
=
2
Solving the equation (1) for x,
-
x=
b2 - 4km
2m
we have
-b
t
2
e m
k
+b
w = w0 2 - l 2 where w0 = , l =
m
2
1 2
The average life = =
l b
63. (a)
64. (a) Let the cube be at a depth x from the equilibrium
position.
Force acting on the cube = up thrust on the portion of
length x.
F = – rl 2 xg [\ mass density X volume ] ....(i)
Clearly F µ – x, Hence it is a SHM.
Equation of SHM is F = –kx
....(ii)
Comparing equation (i) and (ii) we have
k = rl2g
Now, Time period, T = 2p
Þ T = 2p
= 2p
a = -w2 x where w 2 is a constant.
a=
d2 x
Þ
rg
ld
Þ T = 2p
dl
rg
65. (a) For an SHM, the acceleration
m
k
w=
\
–4p 2 x
T2
Þ
aT –4p2
=
x
T
aT
The time period T is also constant. Therefore,
is a
x
constant.
66. (a) The two springs are in parallel.
\ Effective spring constant,
k = k1 + k2
Initial frequency of oscillation is given by
1 k1 + k 2
....(i)
m
2p
When both k1 and k2 are made four times their original
values, the new frequency is given by
v =
v' =
=
1
2p
1
2p
4 k1 + 4 k 2
m
æ 1
4(k1 + 4k 2 )
= 2ç
ç 2p
m
è
dx
= 2 × 10–2 p sin p t
dt
For the first time, the when velocity becomes maximum,
sin p t = 1
Þ sin p t = sin p
2
v=
1
p
or,, t = = 0.5 sec.
2
2
68. (b) When plugged hole near the bottom of the oscillating
bob gets suddenly unplugged, centre of mass of
combination of liquid and hollow portion (at position l ),
Þ pt =
first goes down ( to l + D l) and when total water is drained
out, centre of mass regain its original position (to l ),
Time period, T = 2p
l
g
\ ‘T ’ first increases and then decreases to original value.
l d
rl 2 g
Comparing the above equation with
a = –w2x, we get
ö
÷÷ = 2v
ø
67. (b) Here, Displacement, x = 2 × 10–2 cos p t
Velocity is given by
3
ld
rg
k1 + k2
m
c
P-217
Oscillations
70.
(a) Time period, t = 2p
l
g eff
Þ 1+
;
l
g
In air, t 0 = 2p
Þ
Buoyant
force
4
´ 1000 Vg
3
1000
æ4 ö
Vg
Net force = ç - 1 ÷ ´ 1000 Vg =
3
è3 ø
1000 Vg
g
geff =
=
4
3 ´ ´1000V 4
3
\ t = 2p
l
l
= 2 ´ 2p
g /4
g
m
(b) Time period for first spring, t1 = 2p
,
k1
m
Time period for second spring, t 2 = 2p
k2
k1k 2
kl + k 2
\ Time period of oscillation for series combination
Force constant of the series combination keff =
m( k l + k 2 )
k1k 2
T = 2p
\ T = 2p
2
t1
m m
+
= 2p
+
2
2
k 2 k1
(2p)
(2p)
Þ T 2 = t12 + t 22
72.
where x is the displacement from the mean position
(c) With mass M, the time period of the spring.
M
k
With mass M + m, the time period becomes,
T = 2p
T ' = 2p
\ 2p
Þ
l' = l + 0.21 l
Þ l' = 1.21 l
T ' = 2p
M + m 5T
=
k
3
1.21l
g
% increase in length =
1.21l - l
l
´ 100 =
T '- T
´100
T
(
)
1.21 - 1 ´ 100
= (1.1 - 1) ´ 100 = 10%
74. (b) Let k be the spring constant of the original spring.
Time period T = 2p
m
where m is the mas s of oscillating
k
body.
When the spring is cut into n equal parts, the spring
constant of one part becomes nk. Therefore the new time
period,
T ' = 2p
2
t2
l
g
New length, l ' = l + 2 1 % o f l
=
t = 2t0
71.
16
m 25
=
-1 =
M
9
9
73. (d) Time period, T = 2p
1000 Vg
Weight
m 25
=
M
9
m
T
=
nk
n
l
where l = distance
g
between the point of suspension and the centre of mass of
the child.
As the child stands up, her centre of mass is raised. The
distance between point of suspension and centre of mass
decreases ie length l decreases.
75. (b) The time period T = 2p
\ l¢ < l
\ T ¢ < T i.e., the period decreases.
point of suspension
l'
l
M +m 5
M
= ´ 2p
3
k
k
M +m =
25
´M
9
Case (ii) child standing
Case (i) child sitting
P-218
Physics
Putting value of M, m, u and K we get amplitude
point of suspension
1 75
1
=
15 1
3
79. (b) The change in time period compared to the undamped
oscillator increases by 8%.
A=
l'
l
Case (ii) child standing
80.
Case (i) child sitting
0.9A 0 = A 0 e
...(i)
A0
= Ae e- kt
and
1000
...(ii)
Dividing (i) by (ii) and solving, we get
t = 20 s
77. (b) Amplitude of vibration at time t = 0 is given by
A = A0e –0.1× 0 = 1 × A0 = A0
also at t = t, if A =
Þ
78.
A0
2
-
b(5)
2m
… (i)
After 10 more second,
A = A0 e
-
b(15)
2m
…(ii)
From eqns (i) and (ii)
A = 0.729 A0
\ a = 0.729
81. (a)
82. (d) When the pendulum is oscillating over a current
carrying coil, and when the direction of oscillating
pendulum bob is opposite to the direction of current. Its
instantaneous acceleration increases.
l
g
and damping is larger than in air alone due energy dissipation.
83. (c) As energy µ ( Amplitude)2, the maximum for both
of them occurs at the same frequency and this is only
possible in case of resonance.
In resonance state w1 = w 2
84. (b) Equation of displacement in forced oscillation is given
by
Hence time period T < 2p
1
= e –0.1t
2
t = 10 ln 2 ;7s
(b) In first collision mu momentum will be imparted to
system, in second collision when momentum of (M + m) is in
opposite direction mu momentum of particle will make its
momentum zero.
On 13th collision, m ® M + 12 ; M + 13m ® V
mu = (M + 13m)v Þ v =
bt
2m
(where, A0 = maximum amplitude)
According to the questions, after 5 second,
76. (c) Time of half the amplitude is = 2s
Using, A = A0e–kt
A0
= Ae e - k ´2
2
(c) Q A = A 0e
-
mu
u
=
M + 13m 15
u
K
´A
v = wA Þ =
15
M - 13m
y=
F0
2
2 2
m (w 0 - w )
=
F0
m (w 0 2 - w 2 )
Here damping effect is considered to be zero
\x µ
1
2
2
m (w 0 - w )
14
Waves
Basic of Mechanical Waves,
TOPIC 1 Progressive and Stationary
Waves
1.
2.
Assume that the displacement (s) of air is proportional
to the pressure difference (Dp) created by a sound wave.
Displacement (s) further depends on the speed of sound
(v), density of air (r) and the frequency ( f ). If Dp ~
10Pa, v ~ 300 m/s, r ~ 1 kg/m3 and f ~ 1000 Hz, then s
will be of the order of (take the multiplicative constant
to be 1)
[Sep. 05, 2020 (I)]
3
(a)
mm
(b) 10 mm
100
1
(c)
mm
(d) 1 mm
10
For a transverse wave travelling along a straight line, the
distance between two peaks (crests) is 5 m, while the
distance between one crest and one trough is 1.5 m. The
possible wavelengths (in m) of the waves are :
[Sep. 04, 2020 (I)]
1 1 1
, , , .......
1 3 5
1 1 1
, , , .......
(c) 1, 2, 3, .....
(d)
2 4 6
A progressive wave travelling along the positive x-direction
is represented by y(x,t) = Asin (kx – wt + f). Its snapshot at
t = 0 is given in the figure.
[12 April 2019 I]
(a) 1, 3, 5, .....
3.
(b)
4.
5.
6.
7.
8.
For this wave, the phase f is :
p
(a) (b) p
(c) 0
2
(d)
p
2
A small speaker delivers 2 W of audio output. At what
distance from the speaker will one detect 120 dB intensity
sound ? [Given reference intensity of sound as 10–12W/m2]
[12 April 2019 II]
(a) 40 cm (b) 20 cm
(c) 10 cm (d) 30 cm
The pressure wave,
P = 0.01 sin[1000t – 3x] Nm–2, corresponds to the sound
produced by a vibrating blade on a day when atmospheric
temperature is 0°C. On some other day when temperature
is T, the speed of sound produced by the same blade and
at the same frequency is found to be 336 ms–1. Approximate
value of T is :
[9 April 2019 I]
(a) 4°C
(b) 11°C
(c) 12°C (d)
15°C
A travelling harmonic wave is represented by the equation y(x, t)=10–3 sin(50t+2x), where x and y are in meter and
t is in seconds. Which of the following is a correct statement about the wave?
[12 Jan. 2019 I]
(a) The wave is propagating along the negative x-axis
with speed 25 ms–1.
(b) The wave is propagating along the positive x-axis with
speed 100 ms–1.
(c) The wave is propagating along the positive x-axis with
speed 25 ms–1.
(d) The wave is propagating along the negative x-axis
with speed 100 ms–1.
A transverse wave is represented by
10 æ 2p 2p ö
y = sin ç t x÷
p
l ø
è T
For what value of the wavelength the wave velocity is
twice the maximum particle velocity?
[Online April 9, 2014]
(a) 40 cm (b) 20 cm
(c) 10 cm (d) 60 cm
In a transverse wave the distance between a crest and
neighbouring trough at the same instant is 4.0 cm and the
distance between a crest and trough at the same place is
1.0 cm. The next crest appears at the same place after a
time interval of 0.4s. The maximum speed of the vibrating
particles in the medium is :
[Online April 25, 2013]
3p
5p
cm/s
cm/s
(a)
(b)
2
2
p
cm/s
(c)
(d) 2p cm/s
2
P-220
Physics
9.
When two sound waves travel in the same direction in a
medium, the displacements of a particle located at 'x' at
time ‘t’ is given by :
y1 = 0.05 cos (0.50 px – 100 pt)
y2 = 0.05 cos (0.46 px – 92 pt)
where y1, y2 and x are in meters and t in seconds. The
speed of sound in the medium is : [Online April 9, 2013]
(a) 92 m/s (b) 200 m/s (c) 100 m/s (d) 332 m/s
10. The disturbance y (x, t) of a wave propagating in the
1
positive x-direction is given by y =
at time t = 0
1 + x2
1
and by y =
at t = 2 s, where x and y are in
é1 + ( x - 1) 2 ù
ë
û
meters. The shape of the wave disturbance does not change
during the propagation. The velocity of wave in m/s is
[Online May 26, 2012]
(a) 2.0
(b) 4.0
(c) 0.5
(d) 1.0
11. The transverse displacement y (x, t) of a wave is given by
y( x, t ) = e
(
- ax 2 + bt 2 + 2 ab ) xt
This represents a:
).
[2011]
(a) wave moving in – x direction with speed
b
a
(b) standing wave of frequency
b
1
(c) standing wave of frequency
b
(d) wave moving in + x direction speed
a
b
12.
A wave travelling along the x-axis is described by the
equation y(x, t) = 0.005 cos (a x – bt). If the wavelength
and the time period of the wave are 0.08 m and 2.0s,
respectively, then a and b in appropriate units are [2008]
0.08
2.0
,b =
(a) a = 25.00 p , b = p (b) a =
p
p
0.04
1.0
p
,b =
(c) a =
(d) a = 12.50p, b =
p
p
2.0
13. A sound absorber attenuates the sound level by 20 dB.
The intensity decreases by a factor of
[2007]
(a) 100
(b) 1000
(c) 10000 (d) 10
14. The displacement y of a particle in a medium can be
pö
æ
-6
expressed as, y = 10 sin ç100t + 20 x + ÷ m where t is
è
4ø
in second and x in meter. The speed of the wave is [2004]
(a) 20 m/s
(b) 5 m/s
(c) 2000 m/s
(d) 5p m/s
15. The displacement y of a wave travelling in the x -direction
pö
æ
is given by y = 10 - 4 sin ç 600 t - 2 x + ÷ metres
3ø
è
where x is expressed in metres and t in seconds. The speed
of the wave - motion, in ms–1 , is
[2003]
(a) 300
(b) 600
(c) 1200 (d) 200
16. When temperature increases, the frequency of a tuning
fork
[2002]
(a) increases
(b) decreases
(c) remains same
(d) increases or decreases depending on the material
TOPIC 2 Vibration of String and Organ
Pipe
17. In a resonance tube experiment when the tube is filled
with water up to a height of 17.0 cm from bottom, it resonates with a given tuning fork. When the water level is
raised the next resonance with the same tuning fork occurs at a height of 24.5 cm. If the velocity of sound in air
is 330 m/s, the tuning fork frequency is :
[Sep. 05, 2020 (I)]
(a) 2200 Hz
(b) 550 Hz
(c) 1100 Hz
(d) 3300 Hz
18. A uniform thin rope of length 12 m and mass 6 kg hangs
vertically from a rigid support and a block of mass 2 kg
is attached to its free end. A transverse short wave-train
of wavelength 6 cm is produced at the lower end of the
rope. What is the wavelength of the wavetrain (in cm)
when it reaches the top of the rope ?[Sep. 03, 2020 (I)]
(a) 3
(b) 6
(c) 12
(d) 9
19. Two identical strings X and Z made of same material have
tension T X an d T Z in them. If their fundamental
frequencies are 450 Hz and 300 Hz, respectively, then
the ratio TX/TZ is:
[Sep. 02, 2020 (I)]
(a) 2.25
(b) 0.44
(c) 1.25
(d) 1.5
20. A wire of density 9 × 10–3 kg cm–3 is stretched between
two clamps 1 m apart. The resulting strain in the wire is
4.9 × 10–4. The lowest frequency of the transverse
vibrations in the wire is
(Young’s modulus of wire Y = 9 × 1010 Nm–2), (to the nearest
integer), ___________.
[Sep. 02, 2020 (II)]
21. A one metre long (both ends open) organ pipe is kept in a
gas that has double the density of air at STP. Assuming
the speed of sound in air at STP is 300 m/s, the frequency
difference between the fundamental and second harmonic
of this pipe is ______ Hz.
[NA 8 Jan. 2020 (I)]
22. A transverse wave travels on a taut steel wire with a
velocity of v when tension in it is 2.06 ´ l04 N. When
the tension is changed to T, the velocity changed to v/2.
The value of T is close to:
[8 Jan. 2020 (II)]
(a) 2.50 ´ l04 N
(b) 5.15 ´ l03 N
(c) 30.5 ´ l04 N
(d) 10.2 ´ l02 N
23. Speed of a transverse wave on a straight wire (mass 6.0
g, length 60 cm and area of cross-section 1.0 mm2) is 90
ms–1. If the Young’s modulus of wire is 16 ´ l011 Nm–2 the
extension of wire over its natural length is:
[7 Jan. 2020 (I)]
P-221
Waves
(a) 0.03 mm
(c) 0.04 mm
(b) 0.02 mm
(d) 0.01 mm
24. Equation of travelling wave on a stretched string of linear
density 5 g/m is y = 0.03 sin (450 t – 9x) where distance and
time are measured in SI units. The tension in the string is:
[11 Jan 2019 (I)]
(a) 10 N
(b) 7.5 N
(c) 12.5 N (d) 5 N
25. A heavy ball of mass M is suspended from the ceiling of
a car by a light string of mass m (m<<M). When the car is
at rest, the speeed of transverse waves in the string is 60
ms–1. when the car has acceleration a, the wave-speed
increases to 60.5 ms–1. The value of a, in terms of
gravitational acceleration g, is closest to: [9 Jan. 2019 (I)]
g
g
g
g
(b)
(c)
(d)
30
5
10
20
A wire of length L and mass per unit length 6.0 × 10–3 kgm–1
is put under tension of 540 N. Two consecutive
frequencies that it resonates at are: 420 Hz and 490 Hz.
Then L in meters is:
[9 Jan. 2020 (II)]
(a) 2.1 m
(b) 1.1 m
(c) 8.1 m
(d) 5.1 m
A tuning fork of frequency 480 Hz is used in an experiment
for measuring speed of sound (v) in air by resonance
tube method. Resonance is observed to occur at two
successive lengths of the air column, l1 = 30 cm and l2 = 70
cm. Then, v is equal to :
[12 April 2019 (II)]
(a) 332 ms–1
(b) 384 ms–1
(c) 338 ms–1
(d) 379 ms–1
A string 2.0 m long and fixed at its ends is driven by a 240
Hz vibrator. The string vibrates in its third harmonic mode.
The speed of the wave and its fundamental frequency is:
[9 April 2019 (II)]
(a) 180 m/s, 80 Hz
(b) 320 m/s, 80 Hz
(c) 320 m/s, 120 Hz
(d) 180 m/s, 120 Hz
A string is clamped at both the ends and it is vibrating in
its 4th harmonic. The equation of the stationary wave is Y
= 0.3 sin(0.157x) cos(200At). The length of the string is:
(All quantities are in SI units.)
[9 April 2019 (I)]
(a) 20 m
(b) 80 m
(c) 40 m (d)
60 m
A wire of length 2L, is made by joining two wires A and B
of same length but different radii r and 2r and made of the
same material. It is vibrating at a frequency such that the
joint of the two wires forms a node. If the number of
antinodes in wire A is p and that in B is q then the ratio
p : q is :
[8 April 2019 (I)]
(a)
26.
27.
28.
29.
30.
(a) 3 : 5
(b) 4 : 9
(c) 1 : 2
(d)
1: 4
31. A closed organ pipe has a fundamental frequency of 1.5
kHz. The number of overtones that can be distinctly
heard by a person with this organ pipe will be: (Assume
that the highest frequency a person can hear is 20,000 Hz)
[10 Jan. 2019 (I)]
(a) 6
(b) 4
(c) 7
(d) 5
32. A string of length 1 m and mass 5 g is fixed at both
ends. The tension in the string is 8.0 N. The string is
set into vibration using an external vibrator of frequency
100 Hz. The separation between successive nodes on
the string is close to:
[10 Jan. 2019 (I)]
(a) 10.0 cm (b) 33.3 cm (c) 16.6 cm (d) 20.0 cm
33. A granite rod of 60 cm length is clamped at its middle point
and is set into longitudinal vibrations. The density of
granite is 2.7 × 103 kg/m3 and its Young's modulus is
9.27×1010 Pa.
What will be the fundamental frequency of the longitudinal
vibrations?
[2018]
(a) 5 kHz (b) 2.5 kHz (c) 10 kHz (d) 7.5 kHz
34. The end correction of a resonance column is 1cm. If the
shortest length resonating with the tuning fork is 10cm,
the next resonating length should be
[Online April 16, 2018]
(a) 32cm
(b) 40cm
(c) 28cm (d) 36cm
35. Two wires W1 and W2 have the same radius r and
respective densities r1 and r2 such that r2 = 4r1. They
are joined together at the point O, as shown in the figure.
The combination is used as a sonometer wire and kept
under tension T. The point O is midway between the two
bridges. When a stationary waves is set up in the
composite wire, the joint is found to be a node. The ratio
of the number of antinodes formed in W1 to W2 is :
[Online April 8, 2017]
r1
r2
O
W
W1
2
(a) 1 : 1
(b) 1 : 2
(c) 1 : 3
(d) 4 : 1
36. A uniform string of length 20 m is suspended from a rigid
support. A short wave pulse is introduced at its lowest
end. It starts moving up the string. The time taken to reach
the supports is :
[2016]
(take g = 10 ms–2)
(a) 2 2s (b) 2 s
(d)2 s
(c) 2p 2 s
37. A pipe open at both ends has a fundamental frequency f
in air. The pipe is dipped vertically in water so that half of
it is in water. The fundamental frequency of the air column
is now :
[2016]
f
3f
(a) 2f
(b) f
(c)
(d)
2
4
38. A pipe of length 85 cm is closed from one end. Find the
number of possible natural oscillations of air column in
the pipe whose frequencies lie below 1250 Hz. The velocity
of sound in air is 340 m/s.
[2014]
(a) 12
(b) 8
(c) 6
(d) 4
39. The total length of a sonometer wire between fixed ends is
110 cm. Two bridges are placed to divide the length of wire
in ratio 6 : 3 : 2. The tension in the wire is 400 N and the
mass per unit length is 0.01 kg/m. What is the minimum
common frequency with which three parts can vibrate?
[Online April 19, 2014]
P-222
40.
41.
42.
43.
44.
45.
46.
47.
(a) 1100 Hz
(b) 1000 Hz
(c) 166 Hz
(d) 100 Hz
A sonometer wire of length 1.5 m is made of steel. The
tension in it produces an elastic strain of 1%. What is the
fundamental frequency of steel if density and elasticity of
steel are 7.7 × 103 kg/m3 and 2.2 × 1011 N/m2 respectively?
(a) 188.5 Hz
(b) 178.2 Hz
[2013]
(c) 200.5 Hz
(d) 770 Hz
A sonometer wire of length 114 cm is fixed at both the ends.
Where should the two bridges be placed so as to divide the
wire into three segments whose fundamental frequencies
are in the ratio 1 : 3 : 4 ?
[Online April 23, 2013]
(a) At 36 cm and 84 cm from one end
(b) At 24 cm and 72 cm from one end
(c) At 48 cm and 96 cm from one end
(d) At 72 cm and 96 cm from one end
A cylindrical tube, open at both ends, has a fundamental
frequency f in air. The tube is dipped vertically in water so
that half of it is in water. The fundamental frequency of the
air-column is now :
[2012]
(a) f
(b) f / 2
(c) 3 f /4 (d) 2 f
An air column in a pipe, which is closed at one end, will
be in resonance wtih a vibrating tuning fork of frequency
264 Hz if the length of the column in cm is (velocity of
sound = 330 m/s)
[Online May 26, 2012]
(a) 125.00 (b) 93.75
(c) 62.50 (d) 187.50
A uniform tube of length 60.5 cm is held vertically with
its lower end dipped in water. A sound source of frequency
500 Hz sends sound waves into the tube. When the length
of tube above water is 16 cm and again when it is 50 cm,
the tube resonates with the source of sound. Two lowest
frequencies (in Hz), to which tube will resonate when it
is taken out of water, are (approximately).
[Online May 19, 2012]
(a) 281, 562
(b)
281, 843
(c) 276, 552
(d)
272, 544
The equation of a wave on a string of linear mass density
0.04 kg m–1 is given by
é æ t
x öù
y = 0.02(m) sin ê2p ç
÷ú .
ë è 0.04(s ) 0.50(m) ø û
The tension in the string is
[2010]
(a) 4.0 N (b) 12.5 N (c) 0.5 N (d) 6.25 N
While measuring the speed of sound by performing a
resonance column experiment, a student gets the first
resonance condition at a column length of 18 cm during
winter. Repeating the same experiment during summer,
she measures the column length to be x cm for the second
resonance. Then
[2008]
(a) 18 > x
(b) x > 54
(c) 54 > x > 36
(d) 36 > x > 18
A string is stretched between fixed points separated by
75.0 cm. It is observed to have resonant frequencies of
420 Hz and 315 Hz. There are no other resonant
frequencies between these two. Then, the lowest resonant
frequency for this string is
[2006]
Physics
(a) 105 Hz
(b) 1.05 Hz
(c) 1050 Hz
(d) 10.5 Hz
48. Tube A has both ends open while tube B has one end
closed, otherwise they are identical. The ratio of
fundamental frequency of tube A and B is
[2002]
(a) 1 : 2
(b) 1 : 4
(c) 2 : 1
(d) 4 : 1
49. A wave y = a sin(wt–kx) on a string meets with another
wave producing a node at x = 0. Then the equation of the
unknown wave is
[2002]
(a) y = a sin( w t + kx)
(b) y = –a sin( w t + kx)
(c) y = a sin( w t – kx)
(d) y = –a sin( w t – kx)
TOPIC 3
Beats, Interference and
Superposition of Waves
50. There harmonic waves having equal frequency n and same
p
p
and - respectively..
4
4
When they are superimposed the intensity of the resultant
wave is close to:
[9 Jan. 2020 I]
(a) 5.8 I0
(b) 0.2 I0
(c) 3 I0
(d) I0
51. The correct figure that shows, schematically, the wave
pattern produced by superposition of two waves of
frequencies 9 Hz and 11 Hz is :
[10 April 2019 II]
intensity I0, have phase angles 0,
(a)
(b)
(c)
(d)
52. A resonance tube is old and has jagged end. It is still used
in the laboratory to determine velocity of sound in air. A
tuning fork of frequency 512 Hz produces first resonance
when the tube is filled with water to a mark 11 cm below
a reference mark, near the open end of the tube. The
experiment is repeated with another fork of frequency
256 Hz which produces first resonance when water
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Waves
reaches a mark 27 cm below the reference mark. The
velocity of sound in air, obtained in the experiment, is
close to:
[12 Jan. 2019 II]
(a) 322 ms–1
(b) 341 ms–1
(c) 335 ms–1
(d) 328 ms–1
53. A tuning fork vibrates with frequency 256 Hz and gives
one beat per second with the third normal mode of
vibration of an open pipe. What is the length of the pipe?
(Speed of sound of air is 340 ms–1)
[Online April 15, 2018]
(a) 190 cm
(b) 180 cm
(c) 220 cm
(d) 200 cm
54. 5 beats/ second are heard when a turning fork is sounded
with a sonometer wire under tension, when the length of
the sonometer wire is either 0.95m or 1m . The frequency
of the fork will be:
[Online April 15, 2018]
(a) 195Hz (b) 251Hz (c) 150Hz (d) 300Hz
55. A standing wave is formed by the superposition of two
waves travelling in opposite directions. The transverse
displacement is given by
æ 5p ö
y(x, t) = 0.5 sin ç x ÷ cos(200 pt).
è 4 ø
What is the speed of the travelling wave moving in the
positive x direction ?
(x and t are in meter and second, respectively.)
[Online April 9, 2017]
(a) 160 m/s (b) 90 m/s (c) 180 m/s (d) 120 m/s
56. A wave represented by the equation y1 = acos (kx – wt) is
superimposed with another wave to form a stationary wave
such that the point x – 0 is node. The equation for the
other wave is
[Online May 12, 2012]
(a) a cos (kx – wt + p)
(b) a cos (kx + wt + p)
pö
pö
æ
æ
a cos ç kx + wt + ÷
a cos ç kx - wt + ÷
(d)
è
è
2ø
2ø
57. Following are expressions for four plane simple harmonic
waves
[Online May 7, 2012]
æ
ö
x
(i) y1 = A cos 2p ç n1t + ÷
l1 ø
è
58. A travelling wave represented by
y = A sin (wt – kx) is superimposed on another wave
represented by y = A sin (wt + kx). The resultant is
(a) A wave travelling along + x direction [2011 RS]
(b) A wave travelling along – x direction
(c) A standing wave having nodes at
59.
60.
61.
62.
(c)
æ
ö
x
(ii) y2 = A cos 2p ç n1t + + p÷
l1
è
ø
æ
xö
(iii) y3 = A cos 2p ç n2t + ÷
l2 ø
è
æ
xö
(iv) y4 = A cos 2p ç n2t - ÷
l2 ø
è
The pairs of waves which will produce destructive
interference and stationary waves respectively in a
medium, are
(a) (iii, iv), (i, ii)
(b) (i, iii), (ii, iv)
(c) (i, iv), (ii, iii)
(d) (i, ii), (iii, iv)
nl
, n = 0,1, 2....
2
(d) A standing wave having nodes at
1ö l
æ
x = ç n + ÷ ; n = 0,1, 2....
è
2ø 2
Statement - 1 : Two longitudinal waves given by
equations : y1(x, t) = 2a sin (wt – kx) and y2(x, t) = a
sin (2wt - 2kx) will have equal intensity..
[2011 RS]
Statement - 2 : Intensity of waves of given frequency in
same medium is proportional to square of amplitude only.
(a) Statement-1 is true, statement-2 is false.
(b) Statement-1 is true, statement-2 is true, statement2 is the correct explanation of statement-1
(c) Statement-1 is true, statement-2 is true, statement2 is not the correct explanation of statement-1
(d) Statement-1 is false, statement-2 is true.
Three sound waves of equal amplitudes have frequencies
(n –1), n, (n + 1). They superpose to give beats. The number
of beats produced per second will be :
[2009]
(a) 3
(b) 2
(c) 1
(d) 4
When two tuning forks (fork 1 and fork 2) are sounded
simultaneously, 4 beats per second are heard. Now, some
tape is attached on the prong of the fork 2. When the
tuning forks are sounded again, 6 beats per second are
heard. If the frequency of fork 1 is 200 Hz, then what was
the original frequency of fork 2?
[2005]
(a) 202 Hz (b) 200 Hz (c) 204 Hz (d) 196 Hz
A tuning fork of known frequency 256 Hz makes 5 beats
per second with the vibrating string of a piano. The beat
frequency decreases to 2 beats per second when the
tension in the piano string is slightly increased. The
frequency of the piano string before increasing the tension
was
[2003]
(a) (256 + 2) Hz
(b) (256 – 2) Hz
(c) (256 – 5) Hz
(d) (256 + 5) Hz
A tuning fork arrangement (pair) produces 4 beats/sec with
one fork of frequency 288 cps. A little wax is placed on the
unknown fork and it then produces 2 beats/sec. The
frequency of the unknown fork is
[2002]
(a) 286 cps (b) 292 cps (c) 294 cps (d) 288 cps
x=
63.
TOPIC 4
Musical Sound and Doppler's
Effect
64. A sound source S is moving along a straight track with
speed v, and is emitting sound of frequency vo (see
figure). An observer is standing at a finite distance, at the
point O, from the track. The time variation of frequency
heard by the observer is best represented by:
[Sep. 06, 2020 (I)]
P-224
Physics
(t0 represents the instant when the distance between the
source and observer is minimum)
v
v
(a) vo
v
(b)
t0 t
vo
71.
t0
v
t
vo
vo
(c)
72.
(d)
t0
t
t0
t
65. A driver in a car, approaching a vertical wall notices that
the frequency of his car horn, has changed from 440 Hz to
480 Hz, when it gets reflected from the wall. If the speed of
sound in air is 345 m/s, then the speed of the car is :
[Sep. 05, 2020 (II)]
(a) 54 km/hr
(b) 36 km/hr
(c) 18 km/hr
(d) 24 km/hr
66. The driver of a bus approaching a big wall notices that the
frequency of his bus’s horn changes from 420 Hz to 490 Hz
when he hears it after it gets reflected from the wall. Find
the speed of the bus if speed of the sound is 330 ms–1.
[Sep. 04, 2020 (II)]
(a) 91 kmh–1
(b) 81 kmh–1
(c) 61 kmh–1
(d) 71 kmh–1
67. Magnetic materials used for making permanent magnets
(P) and magnets in a transformer (T) have different
properties of the following, which property best matches
for the type of magnet required?
[Sep. 02, 2020 (I)]
(a) T : Large retentivity, small coercivity
(b) P : Small retentivity, large coercivity
(c) T : Large retentivity, large coercivity
(d) P : Large retentivity, large coercivity
68. A stationary observer receives sound from two identical
tuning forks, one of which approaches and the other one
recedes with the same speed (much less than the speed
of sound). The observer hears 2 beats/sec. The oscillation
frequency of each tuning fork is v0 = 1400 Hz and the
velocity of sound in air is 350 m/s. The speed of each
tuning fork is close to:
[7 Jan. 2020 I]
1
1
1
m/s (b) 1m/s
m/s
(a)
(c)
(d) 8 m/s
2
4
69. A submarine (A) travelling at 18 km/hr is being chased
along the line of its velocity by another submarine (B)
travelling at 27 km/hr. B sends a sonar signal of 500 Hz to
detect A and receives a reflected sound of frequency v.
The value of v is close to :
[12 April 2019 I]
(Speed of sound in water = 1500 ms–1)
(a) 504 Hz
(b) 507 Hz
(c) 499 Hz
(d) 502 Hz
70. Two sources of sound S1 and S2 produce sound waves of
same frequency 660 Hz. A listener is moving from source
S1 towards S2 with a constant speed u m/s and he hears
73.
74.
75.
76.
77.
78.
10 beats/s. The velocity of sound is 330 m/s. Then u
equals:
[12 April 2019 II]
(a) 5.5 m/s
(b) 15.0 m/s
(c) 2.5 m/s
(d) 10.0 m/s
A stationary source emits sounds waves of frequency
500 Hz. Two observers moving along a line passing
through the source detect sound to be of frequencies
4801 Hz and 530 Hz. Their respective speeds are, in
ms–1,
(Given speed of sound = 300 m/s)
[10 April 2019 I]
(a) 12, 16 (b) 12, 18 (c) 16, 14 (d) 8, 18
A source of sound S is moving with a velocity of 50 m/s
towards a stationary observer. The observer measures
the frequency of the source as 1000 Hz. What will be the
apparent frequency of the source when it is moving away
from the observer after crossing him? (Take velocity of
sound in air 350 m/s)
[10 April 2019 II]
(a) 750 Hz (b) 857 Hz (c) 1143 Hz (d) 807 Hz
Two cars A and B are moving away from each other in
opposite directions. Both the cars are moving with a speed
of 20 ms–1 with respect to the ground. If an observer in car A
detects a frequency 2000 Hz of the sound coming from car B,
what is the natural frequency of the sound source in car B?
(speed of sound in air = 340 ms–1)
[9 April 2019 II]
(a) 2250 Hz
(b) 2060 Hz
(c) 2300 Hz
(d) 2150 Hz
A train moves towards a stationary observer with speed
34 m/s. The train sounds a whistle and its frequency
registered by the observer is f1 If the speed of the train
is reduced to 17 m/s, the frequency registered is f2 If
speed of sound is 340 m/s, then the ratio f1/f2 is:
[10 Jan. 2019 I]
(a) 18/17
(b) 19/18
(c) 20/19 (d) 21/20
A musician using an open flute of length 50 cm
produces second harmonic sound waves. A person runs
towards the musician from another end of a hall at a
speed of 10 km/h. If the wave speed is 330 m/s, the
frequency heard by the running person shall be close to:
[9 Jan. 2019 II]
(a) 666 Hz
(b) 753 Hz
(c) 500 Hz
(d) 333 Hz
Two sitar strings, A and B, playing the note 'Dha' are slightly
out of tune and produce beats and frequency 5 Hz. The
tension of the string B is slightly increased and the beat
frequency is found to decrease by 3 Hz . If the frequency
of A is 425 Hz, the original frequency of B is
[Online April 16, 2018]
(a) 430 Hz (b) 428 Hz (c) 422 Hz (d) 420 Hz
A toy–car, blowing its horm, is moving with a steady speed
of 5 m/s, away from a wall. An observer, towards whom the
toy car is moving, is able to hear 5 beats per second. If the
velocity of sound in air is 340 m/s, the frequency of the
horn of the toy car is close to : [Online April 10, 2016]
(a) 680 Hz (b) 510 Hz (c) 340 Hz (d) 170 Hz
Two engines pass each other moving in opposite
directions with uniform speed of 30 m/s. One of them is
blowing a whistle of frequency 540 Hz. Calculate the
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Waves
frequency heard by driver of second engine before they
pass each other. Speed of sound is 330 m/sec:
[Online April 9, 2016]
(a) 450 Hz (b) 540 Hz (c) 270 Hz (d) 648 Hz
79. A train is moving on a straight track with speed 20 ms–1.
It is blowing its whistle at the frequency of 1000 Hz. The
percentage change in the frequency heard by a person
standing near the track as the train passes him is (speed
of sound = 320 ms–1) close to :
[2015]
(a) 18%
(b) 24%
(c) 6%
(d) 12%
80. A source of sound emits sound waves at frequency f0. It
is moving towards an observer with fixed speed
vs (vs < v, where v is the speed of sound in air). If the
observer were to move towards the source with speed
v0, one of the following two graphs (A and B) will given
the correct variation of the frequency f heard by the
observer as v0 is changed.
(B)
(A)
f
f
v0
1/v0
The variation of f with v0 is given correctly by :
[Online April 11, 2015]
f0
(a) graph A with slope =
(v + vs )
(b) graph B with slope =
f0
(v – vs )
(c) graph A with slope =
f0
(v – vs )
f0
(v + vs )
81. A bat moving at 10 ms–1 towards a wall sends a sound
signal of 8000 Hz towards it. On reflection it hears a
sound of frequency f. The value of f in Hz is close to
(speed of sound = 320 ms–1)
[Online April 10, 2015]
(a) 8516
(b) 8258
(c) 8424 (d) 8000
82. A source of sound A emitting waves of frequency 1800
Hz is falling towards ground with a terminal speed v. The
observer B on the ground directly beneath the source
receives waves of frequency 2150 Hz. The source A
receives waves, reflected from ground of frequency
nearly: (Speed of sound = 343 m/s)
[Online April 12, 2014]
(a) 2150 Hz
(b) 2500 Hz
(c) 1800 Hz
(d) 2400 Hz
83. Two factories are sounding their sirens at 800 Hz. A man
goes from one factory to other at a speed of 2m/s. The
velocity of sound is 320 m/s. The number of beats heard
by the person in one second will be:
[Online April 11, 2014]
(d) graph B with slope =
(a) 2
(b) 4
(c) 8
(d) 10
84. A and B are two sources generating sound waves. A
listener is situated at C. The frequency of the source at A
is 500 Hz. A, now, moves towards C with a speed 4 m/s.
The number of beats heard at C is 6. When A moves away
from C with speed 4 m/s, the number of beats heard at C
is 18. The speed of sound is 340 m/s. The frequency of
the source at B is :
[Online April 22, 2013]
A
C
B
(a) 500 Hz (b) 506 Hz (c) 512 Hz (d) 494 Hz
85. An engine approaches a hill with a constant speed. When
it is at a distance of 0.9 km, it blows a whistle whose echo
is heard by the driver after 5 seconds. If the speed of
sound in air is 330 m/s, then the speed of the engine is :
[Online April 9, 2013]
(a) 32 m/s (b) 27.5 m/s (c) 60 m/s (d) 30 m/s
86. This question has Statement 1 and Statement 2. Of the
four choices given after the Statements, choose the one
that best describes the two Statements.
Statement 1: Bats emitting ultrasonic waves can detect
the location of a prey by hearing the waves reflected from it.
Statement 2: When the source and the detector are
moving, the frequency of reflected waves is changed.
[Online May 12, 2012]
(a) Statement 1 is false, Statement 2 is true.
(b) Statement 1 is true, Statement 2 is false.
(c) Statement 1 is true, Statement 2 is true, Statement
2 is not the correct explanation of Statement 1.
(d) Statement 1 is true, Statement 2 is true, Statement
2 is the correct explanation of Statement 1.
87. A motor cycle starts from rest and accelerates along a
straight path at 2m/s2. At the starting point of the motor
cycle there is a stationary electric siren. How far has the
motor cycle gone when the driver hears the frequency of
the siren at 94% of its value when the motor cycle was at
rest? (Speed of sound = 330 ms–1)
[2009]
(a) 98 m
(b) 147 m (c) 196 m (d) 49 m
88. A whistle producing sound waves of frequencies 9500
HZ and above is approaching a stationary person with
speed v ms–1. The velocity of sound in air is 300 ms–1. If
the person can hear frequencies upto a maximum of
10,000 HZ, the maximum value of v upto which he can
hear whistle is
[2006]
(a) 15 2 ms -1
(b)
15
ms -1
2
(c) 15
(d) 30 ms–1
89. An observer moves towards a stationary source of sound,
with a velocity one-fifth of the velocity of sound. What is
the percentage increase in the apparent frequency ?[2005]
(a) 0.5%
(b) zero
(c) 20 % (d) 5 %
ms–1
P-226
1.
Physics
6.
(a) As we know,
w
´ rV 2
V
é
w
Bù
êQ K = , V =
ú
V
rû
ë
Pressure amplitude, DP0 = aKB = S 0 KB = S 0 ´
DP0
10
1
3
»
m=
mm »
mm
rV w 1 ´ 300 ´ 1000
30
100
(b) Given : Distance between one crest and one trough
= 1.5 m
Þ S0 =
2.
= (2n1 + 1)
l
2
Distance between two crests = 5 m = n2 l
1.5 (2n1 + 1)
=
Þ 3n2 = 10n1 + 5
5
2n2
Here n1 and n2 are integer.
If
3.
4.
n1 = 1, n2 = 5
\l = 1
n1 = 4, n2 = 15
\l = 1/ 3
n1 = 7, n2 = 25
\l = 1/ 5
1 1 1
Hence possible wavelengths , , metre.
1 3 5
(b) At t = 0, x = 0, y = 0
f = p rad
(a) Using, b = 10
æ I ö
or 120 = 10 log10 çè -12 ÷ø
10
Also I =
5.
P
=
2
...(i)
...(ii)
4pr
4pr 2
On solving above equations, we get
r = 40 cm.
(a) On comparing with P = P0 sin (wt – kx), we have
w = 1000 rad/s, K = 3 m–1
\ v0 =
2
w 1000
=
= 333.3m/s
k
3
v1
T
= 1
v2
T2
333.3
=
336
\ t = 4°C
or
273 + 0
273 + t
(a) Comparing the given equation
y = 10–3sin(50t + 2x) with standard equation,
y = a sin(wt – kx)
Þ wave is moving along –ve x-axis with speed
w
50
v= Þv=
= 25 m/sec.
K
2
7. (a) Given, amplitude a = 10 cm
wave velocity = 2 × maximum particle velocity
i.e, wl = 2 aw
2p
p
or, l = 4a = 4 × 10 = 40 cm
8. (b)
9. (b) Standard equation
æw
ö
y(x, t) = A cos ç x - wt ÷
èV
ø
From any of the displacement equation
Say y1
w
= 0.50 p and w = 100 p
V
100p
= 0.5p
\
V
100p
\ V=
= 200 m/s
0.5p
10. (c) The equation of wave at any time is obtained by
putting X = x – vt
1
1
y=
...(i)
2 = 1 + ( x - vt ) 2
1+ x
We know at t = 2 sec,
1
...(ii)
y=
1 + ( x - 1) 2
On comparing (i) and (ii) we get
vt = 1
1
V=
t
As t = 2 sec
1
\V=
=0.5 m/s.
2
11. (a) Given
y (x, t) = e
(-ax2 +bt2 +2
ab xt
)
-[( ax )2 +( b t )2 + 2 a x . b t ]
= e
- ( a x + bt )2
= e
æ
ö
b ÷
-ç x +
t
è
a ø
e
2
=
It is a function of type y = f (x + vt)
\ y (x, t) represents wave travelling along –ve x direction
w
b
=
Þ Speed of wave =
a
k
P-227
Waves
12. (a) Given,
Wavelength, l = 0.08m
Time period, T = 2.05
y(x, t) = 0.005 cos (ax - bt) (Given)
Comparing it with the standard equation of wave
y(x, t) = a cos (kx - wt) we get
2p
2p
and w = b =
l
T
2p
2p
=p
\ a=
= 25p and b =
2
0.08
k=a =
l = 2(l2 - l1 ) = 2 ´ (24.5 - 17) = 15 cm
Now, from v = f l Þ 330 = l ´ 15 ´ 10-2
\l =
330
1100 ´ 100
´ 100 =
= 2200 Hz
15
5
18. (c) Using, V = f l
V1 V2
V
=
Þ l 2 = 2 l1
l1 l 2
V1
æI ö
13. (a) Loudness of sound. L1 = 10 log ç 1 ÷ ;
è I0 ø
T2
12 m, 6 kg
æI ö
L2 = 10log ç 2 ÷
è I0 ø
æI
\ L1 – L2 = 10 log ç 1
è I0
l = 6 cm
ö
æ I2 ö
÷ - 10 log ç ÷
ø
è I0 ø
Again using,
æI I ö
or, DL = 10 log ç 1 ´ 0 ÷
è I0 I 2 ø
æI ö
or, DL = 10 log ç 1 ÷
è I2 ø
The sound level attenuated by 20 dB ie
L1 – L2 = 20 dB
æI ö
æI ö
or, 20 = 10 log ç 1 ÷ or, 2 = log ç 1 ÷
è I2 ø
è I2 ø
I1
I
= 102 or, I2 = 1 .
I2
100
Þ Intensity decreases by a factor 100.
pö
æ
14. (b) Given, y = 10–6 sin çè100t + 20 x + ÷ø m
4
Comparing it with standard equation, we get
w = 100 and k = 20
w 100
v= =
= 5m / s
k
20
pö
æ
-4
15. (a) y = 10 sin çè 600t - 2x + ÷ø
3
On
comparing
with
standard
equation
y= A sin ( wt - kx + f)
we get w = 600; k=2
Velocity of wave
w 600
= 300 ms -1
v= =
k
2
16. (b) The frequency of a tuning fork is given by
n=
=
m2 k
Y
r
4 3 pl
As temperature increases, the length or dimension of the
prongs increases and also young's modulus increases
therefore f decreases.
17. (a) Here, l1 = 17 cm and l2 = 24.5 cm, V = 330 m/s,
f=?
2
V
=
l
T
T
l 2 = 2 l1
M
T1
1 T
,
2l m
where, T = tension and m =
fx =
T2 = 8g (Top)
8g
l1 = 2l1 = 12 cmT = 2g (Bottom)
1
2g
19. (a) Using f =
or,
f=
T1
2 kg
mass
length
1 Tx
1 Tz
and f z =
2l m
2l m
f x 450
T
=
= x
f z 300
Tz
\
Tx 9
= = 2.25.
Tz 4
20. (35.00)
Given,
Denisty of wire, s = 9 ´ 10 -3 kg cm–3
Young's modulus of wire, Y = 9 × 1010 Nm–2
Strain = 4.9 × 10–4
Y=
Stress T / A
=
Strain Strain
T
= Y ´ Strain = 9 × 109 × 4.9 × 10–4
A
Also, mass of wire, m = Al s
\
Mass per unit length, m =
m
= As
J
P-228
Physics
Fundamental frequency in the string
f =
=
1 T
1 T
=
2l m 2l sA
1
9 ´ 109 ´ 4.9 ´ 10 -4
2 ´1
9 ´ 103
DL =
1
1
49 ´ 109 - 4 -3 = ´ 70 = 35 Hz
2
2
21. (106) Given : Vair = 300 m/s, rgas = 2 r air
=
B
r
Using, V =
Vgas
Vair
=
Þ Vgas =
B
2rair
B
rair
Vair
2
=
300
2
And fnth harmonic =
=
Tl mv 2 ´ l
=
YA l (YA)
6 ´ 10 –3 ´ 902
11
–6
= 3 ´ 10 –4 m
16 ´ 10 ´ 10
= 0.03 mm
24. (c) We have given,
y = 0.03 sin(450 t – 9x)
Comparing it with standard equation of wave, we get w =
450 k = 9
w 450
= 50m/s
\ v= =
k
9
Velocity of travelling wave on a stretched string is given
by
T
T
Þ = 2500
m
m
m = linear mass density
Þ T = 2500 × 5 × 10–3
Þ 12.5 N
v=
= 150 2m/s
nv
(in open organ pipe)
2L
(L = 1 metre given)
\ f2nd harmonic – ffundamental =
2v
v
v
–
=
2 ´1 2 ´ 1 2
\ f2n harmonic – ffundamental = 150 2 = 150 » 106 Hz
2
2
22. (b) The velocity of a transverse wave in a stretched wire
is given by
v=
T
mv 2
´l ÞT =
m
I
T
Again from, Y = DL / L0
A
Using, v =
T
m
25. (b) Wave speed V =
T
m
when car is at rest a = 0
\ 60 =
Mg
m
Similarly when the car is moving with acceleration a,
(
M g2 + a 2
60.5 =
)
12
m
Where,
T = Tension in the wire
m = linear density of wire
60.5
=
60
(Q V µ T )
g2 + a2
2
æ 0.5 ö
=1+
çè1+
÷ø =
2
60
60
g
2
Þ g2 + a 2 = g 2 + g 2 ×
60
g2
4
v
T
\ 1 = 1
v2
T2
2.06 ´ 10 4
v
Þ ´2=
v
T2
4
2.06 ´ 10
= 0.515 ´ 104 N
4
Þ T2 = 5.15 × 103 N
23. (a) Given, l = 60 cm, m = 6 g, A = 1 mm2, v = 90 m/s and Y
= 16 × 1011 Nm–2
Þ T2 =
g2 + a2
a =g
2
g
=
60
30
[which is closest to g/5]
26. (a) Fundamental frequency, f = 70 Hz.
The fundamental frequency of wire vibrating under
tension T is given by
f =
1 T
2L m
P-229
Waves
Here, µ = mass per unit length of the wire
L = length of wire
Since rod is clamped at middle fundamental wave shape
is as follow
1
540
2 L 6 ´ 10-3
Þ L » 2.14 m
(b) V = f l = f × 2 (l 2 - l1 )
= 480 × 2(0.70 – 0.30)
= 384 m/s
3l
4
= 2 or l = m
(b)
2
3
4
Velocity, v = f l = 240 ´ = 320 m/sec
3
240
= 80 Hz
Also f1 =
3
(b) Given, y = 0.3 sin (0.157 x) cos (200 pt)
So k = 0.157 and w = 200p
w 200p
= 4000m/s
or f = 100 Hz, v = =
k 0.157
nv 4v 2v
=
=
Now, using f =
2l 2l
l
2v 2 ´ 4000
\l =
=
= 80m
f
100
(c) As there must be node at both ends and at the joint
of the wire A and B so
l
A
N
= L Þ l = 2L
2
l/2
l = 1.2m (Q L = 60 cm = 0.6m (given)
Using v = fl
70 =
27.
28.
29.
30.
v 5.85 ´103
=
l
1.2
= 4.88 × 103 Hz ; 5 KHz
Þ
l
= l1 + e = 11 cm
4
(Q end correction e = 1 cm given)
3l
For second resonance,
= l2 + e
4
Þ l 2 = 3 ´ 11 - 1 = 32 cm
35. (b) n1 = n2
T ® Same
r ® Same
l ® Same
Frequency of vibration
T
8
=
´ 1000 = 40m/s
m
5
Here, T = tension and µ = mass/length
v 40
m
Wavelength of wave l = =
n 100
Separation b/w successive nodes,
l
40
20
=
=
m = 20 cm
2 2 ´ 100 100
33. (a) In solids, Velocity of wave
n=
p
2l
p1
=
T
pr 2 r
As T, r, and l are same for both the wires
n1 = n2
p2
r1
r2
p
1
Þ 1 =
p2 2
Þ l A = 2l B
P 1
=
q 2
31. (a) If a closed pipe vibration in Nth mode then frequency
( 2N - 1) v = 2N - 1 n
of vibration n =
(
) 1
4l
(where n1 = fundamental frequency of vibration)
Hence 20,000 = (2N – 1) × 1500
Þ N = 7.1 » 7
\ Number of over tones = (No. of mode of vibration) – 1
=7–1=6
32. (d) Velocity of wave on string
f=
34. (a) For first resonance,
VA
uB rB
l
=
=
=2= A
VB
u A rA
lB
Þ
36.
Y
9.27 ´ 1010
=
r
2.7 ´ 103
3
v = 5.85 × 10 m/sec
Q r2 = 4 r1
(a ) We know that velocity in string is given by
v=
T
m
where m =
...(i)
m mass of string
=
l length of string
The tension T =
From (1) and (2)
m
´ x´g
l
..(ii)
l
V=
V=
A
T
x
dx
= gx
dt
x -1/2 dx = g dt
l
l
0
0
\ ò x -1/2 dx - g ò dt
Þ2 l
P-230
Physics
= g´t \ t = 2
37.
41. (d) Total length of the wire, L = 114 cm
n1 : n2 : n3 = 1 : 3 : 4
Let L1, L2 and L3 be the lengths of the three parts
1
As n µ
L
1 1 1
\ L1 : L2 : L3 = : : = 12: 4 : 3
1 3 4
æ 12
ö
\ L1 = ç
´ 114÷ = 72cm
è 12 + 4 + 3
ø
20
l
=2
=2 2
g
10
f
(b)
l
l
(a)
(b)
The fundamental frequency in case (a) is f =
The fundamental frequency in case (b) is
v
2l
v
v
=
=f
4(l / 2) 2l
38. (c) Length of pipe = 85 cm = 0.85m
Frequency of oscillations of air column in closed organ
pipe is given by,
(2n - 1)u
f =
4L
(2n - 1)u
f =
£ 1250
4L
(2n - 1) ´ 340
£ 1250
Þ
0.85 ´ 4
Þ 2n – 1 < 12.5 » 6
39. (b) Total length of sonometer wire, l = 110 cm = 1.1 m
Length of wire is in ratio, 6 : 3 : 2 i.e. 60 cm, 30 cm, 20 cm.
Tension in the wire, T = 400 N
Mass per unit length, m = 0.01 kg
Minimum common frequency = ?
As we know,
f'=
1 T 1000
Hz
=
2l m
11
1000
Hz
Similarly, n1 =
6
1000
n2 =
Hz
3
1000
n3 =
Hz
2
Hence common frequency = 1000 Hz
40. (b) Fundamental frequency,
é
T
mù
v
1 T
1 T
and m = ú
f =
=
=
êQ v =
m
lû
2l 2l m 2l Ar ë
Tl
T Y Dl
Also, Y =
Þ
=
ADl
A
l
Frequency, n =
Þ f =
1
2l
y Dl
lr
....(i)
Dl
= 0.01,
l
r = 7.7 × 103 kg/m3 (given)
y = 2.2 × 1011 N/m2 (given)
Dl
Putting the value of l,
, r and y in eqn. (i) we get,
l
2 103
f =
´
7 3 or f » 178.2 Hz
l = 1.5 m,
æ 4
ö
L2 = ç ´ 114÷ = 24 cm
è 19
ø
æ 3
ö
and L3 = ç ´ 114÷ = 18 cm
è 19
ø
Hence the bridges should be placed at 72 cm and 72 + 24
= 96 cm from one end.
42. (a) Initially for open organ pipe, fundamental frequency
v
n0 =
… (i)
2l0
where l0 is the length of the tube
v = speed of sound
But when it is half dipped in water, it becomes closed organ
l
pipe of length 0 .
2
Fundamental frequency of closed organ pipe
v
nc =
… (ii)
4lc
l0
New length, lc =
2
v
v
Þ nc =
Thus n c =
… (iii)
4l0 / 2
2l
From equations (i) and (iii)
n 0 = nc
Thus, nc = f ( Q n0 = f is given)
43. (b) Given : Frequency of tuning fork, n = 264 Hz
Length of column L = ?
For closed organ pipe
v
n=
4l
v
330
Þl=
=
= 0.3125
4n
4 ´ 264
or, l = 0.3125 × 100 = 31.25 cm
In case of closed organ pipe only odd harmonics are
possible.
Therefore value of l will be (2n – 1) l
Hence option (b) i.e. 3 × 31.25 = 93.75 cm is correct.
44. (d) Two lowest frequencies to which tube will resonates
are 272 Hz and 544 Hz.
é æ t ö
x ù
45. (d) y = 0.02(m)sin ê2p ç
ú
÷0.04(
s
)
0.50(
m) û
ø
ë è
Comparing it with the standard wave equation
y = a sin(wt - kx )
we get
2p
w=
rad s–1
0.04
P-231
Waves
and k =
2p
0.50
Wave velocity, v =
w
k
2p / 0.04
= 12.5 m / s
2p / 0.5
Velocity on a string is given by
49.
Þv=
v=
T
m
\ T = v2 ´ m = (12.5)2 × 0.04 = 6.25 N
46. (b) Fundamental frequency for first resonant length
v
v
n=
=
(in winter)
4l1 4 ´ 18
Fundamental frequency for second resonant length
3v ' 3v '
(in summer)
n' =
=
4l 2 4x
According to questions,
v
3v'
\
=
4 ×18 4× x
v'
\ x = 3 ´ 18 ´
v
v'
\ x = 54 ´ cm
v
v' > v because velocity of light is greater in summer as
compared to winter (v µ T )
\ x > 54cm
47. (a) It is given that 315 Hz and 420 Hz are two resonant
frequencies, let these be n th and (n + 1)th harmonies, then
we have nv = 315
2l
v
= 420
and (n + 1)
2l
n + 1 420
Þ
=
n
315
Þn=3
v
v
= 105 Hz
Hence 3 ´
= 315 Þ
2
l
2l
The lowest resonant frequency is when
n=1
Therefore lowest resonant frequency
= 105 Hz.
48. (c) The fundamental frequency for tube B closed at one
end is given by
lù
v
é
Ql = ú
uB =
ê
4û
4l
ë
Where l = length of the tube and v is the velocity of
sound in air.
The fundamental frequency for tube A open with both ends
is given by
lù
é
v
Ql = ú
uA =
ê
2û
2l
ë
50.
51.
52.
53.
54.
55.
uA
v 4l 2
\ u = 2l ´ v = 1
B
(b) To form a node there should be superposition of this
wave with the reflected wave. The reflected wave should
travel in opposite direction with a phase change of p. The
equation of the reflected wave will be
y = a sin (wt + kx + p)
Þ y = – a sin (wt + kx)
(a)
(c) Beat frequency
= difference in frequencies of two waves
= 11 – 9 = 2 Hz
(d)
(d) According to question, tuning fork gives 1 beat/second
with (N) 3rd normal mode. Therefore, organ pipe will have
frequency (256 ± 1) Hz. In open organ pipe, frequency
NV
n=
2l
3 ´ 340
Þ l = 2 m = 200 cm
or, 255 =
2´ l
(a) Probable frequencies of tuning fork be n ± 5
1
Frequency of sonometer wire, n µ
l
n + 5 100
\
=
Þ 95(n + 5) = 100( n - 5)
n - 5 95
or, 95 n + 475 = 100 n – 500
or, 5 n = 975
975
or, n =
= 195 Hz
5
æ 5p ö
(a) Given, y (x, t) = 0.5 sin ç x ÷ cos (200 pt),
è 4 ø
comparing with equation – y (x, t) = 2 a sin kx cos wt
5p
w = 200 p, k =
4
w 200p
speed of travelling wave v = =
= 160 m/s
k 5p 4
56. (b) Since the point x = 0 is a node and reflection is taking
place from point x = 0. This means that reflection must be
taking place from the fixed end and hence the reflected ray
must suffer an additional phase change of p or a path
l
change of .
2
So, if yincident = a cos ( kx – wt)
Þ yincident = a cos (– kx – wt + p)
= – a cos (wt + kx)
Hence equation for the other wave
y = a cos(kx + wt + p)
57. (d) In case of destructive interference
Phase difference f = 180° or p
So wave pair (i) and (ii) will produce destructive
interference.
Stationary or standing waves will produce by equations
(iii) & (iv) as two waves travelling along the same line but
in opposite direction.
n¢ = n + x
58. (d) y = A sin (wt – kx) + A sin (wt + kx)
y = 2A sin wt cos kx
P-232
Physics
This is an equation of standing wave. For position of
nodes
cos kx = 0
2p
p
.x = (2n + 1)
Þ
l
2
2 n + 1) l
(
Þ x=
, n = 0,1, 2,3,...........
4
59. (a) Intensity of a wave
1
I = pw2 A2 v
2
Since, I µ A2w 2
\ I1 µ (2a)2 w 2
60.
61.
62.
63.
and I 2 µ a 2 (2w ) 2
I1 = I 2
In the same medium, p and v are same.
Intensity depends on amplitude and frequency.
(b) Maximum number of beats
= Maximum frequency – Minimum frequency
= ( n + 1) – ( n – 1) = 2 Beats per second
(d) Frequency of fork 1, no = 200 Hz
No. of beats heard when fork 2 is sounded with fork 1 = Dn =
4
Now on loading (attaching tape) on unknown fork, the
mass of tuning fork increases, So the beat frequency
increases (from 4 to 6 in this case) then the frequency
of the unknown fork 2 is given by,
n = n 0 – Dn = 200 – 4 = 196 Hz
(c) It is given that tuning fork of frequency 256 Hz makes
5 beats/second with the vibrating string of a piano.
Therefore, possible frequency of the piano are (256 ± 5)
Hz. i.e., either 261Hz or 251 Hz. When the tension in the
piano string increases, its frequency will increases. As
the original frequency was 261Hz, the beat frequency
should decreases, we can conclude that the frequency of
piano string is 251Hz
(b) Frequency of unknown fork = known frequency ± Beat
frequency = 288 + 4 cps or 288 – 4 cps i.e. 292 cps or 284
cps. When a little wax is placed on the unknown fork, it
produces 2 beats/sec. When a little wax is placed on the
unknown fork, its frequency decreases and simultaneously
the beat frequency decreases confirming that the frequency
of the unknown fork is 292 cps.
Note : Had the frequency of unknown fork been 284 cps,
then on placing wax its frequency would have decreased
thereby increasing the gap between its frequency and the
frequency of known fork. This would produce high beat
frequency.
64. (b) Frequency heard by the observer
æ
ö
vsound
vobserved = ç
÷ v0
è vsound - v cos q ø
Observer
O
D
q
Source V
Initially q will be less so cos q more.
\ vobserved more, then it will decrease.
65. (a) Let f 1 be the frequency heard by wall,
æ v ö
f1 = ç
f0
è v - vc ÷ø
Here, v = Velocity of sound,
vc = Velocity of Car,
f0 = actual frequency of car horn
Let f2 be the frequency heard by driver after reflection
from wall.
æ v + vc ö
æ v + vc ö
f2 = ç
f =
f
è v ÷ø 1 çè v - v ÷ø 0
c
é 345 + vc ù
12 345 + vc
Þ 480 = ê
=
ú 440 Þ
11 345 - vc
ë 345 - vc û
Þ vc = 54 km/hr
66. (a) From the Doppler's effect of sound, frequency
appeared at wall
330
fw =
×f
...(i)
330 - v
Here, v = speed of bus,
f = actual frequency of source
Frequency heard after reflection from wall (f') is
330 + v
330 + v
f '=
× fw =
×f
330
330 - v
330 + v
Þ 490 =
× 420
330 - v
330 ´ 7
Þv=
» 25.38 m/s = 91 km/s
91
67. (d) Permanent magnets (P) are made of materials with large
retentivity and large coercivity. Transformer cores (T) are
made of materials with low retentivity and low coercivity.
68. (c) From Doppler’s effect, frequency of sound heard (f1)
when source is approaching
c
f1 = f 0
c–v
Here, c = velocity of sound
v = velocity of source
Frequency of sound heard (f2) when source is receding
c
c+v
Beat frequency = f1 – f2
1 ù
é 1
Þ 2 = f1 – f 2 = f 0c ê
–
–
c
v
c
+
v úû
ë
2v
= f0c
é v2 ù
c 2 ê1 – 2 ú
êë c úû
For c>> v
c
2c
350 1
Þ v=
=
=
= m/s
f 0 1400 4
2 f0
f2 = f0
æ v - vo ö
æ 1500 - 5 ö
69. (d) f1 = f ç v - v ÷ = f çè 1500 - 7.5 ÷ø
è
ø
s
No reflected signal,
P-233
Waves
æ v + vo ö
æ 1500 + 7.5 ö
f2 = f1 ç v + v ÷ = f1 çè 1500 + 5 ÷ø
è
ø
s
æ 1500 - 5 ö æ 1500 + 7.5 ö
f2 = 500 çè
֍
÷
1500 - 7.5 ø è 1500 + 5 ø
502 Hz
70. (c) f1 = f
S1
é V ù
é 340 ù
fapp = f0 ê
ú Þ fl = f0 ê
ú
ë V – Vs û
ë 340 – 34 û
v - v0
v + v0
and f2 = f
v
v
L
u
S2
But frequency,
2v0
f2 – f1 = f ´
v
2u
330
\ u = 2.5 m/s.
71. (b) Frequency of sound source (f0) = 500 Hz
When observer is moving away from the source
æ n - n'0 ö
Apparent frequency f1 = 480 = f0 ç n ÷ ....(i)
è
ø
And when observer is moving towards the source
or 10 = 660 ´
æ n - n"0 ö
f 2 = 530 = f0 ç
÷
è n ø
From equation (i)
æ 300 - v '0
480 = 500 ç
ç 300
è
v’0 = 12 m/s
From equation (ii)
....(ii)
ö
÷
÷
ø
æ v'' ö
530 = 500 ç1 + 0 ÷
vø
è
\ V”0 = 18 m/s
72. (a) When source is moving towards a stationary
observer,
æ V -0 ö
fapp = f source ç
è V - 50 ÷ø
æ 350 ö
1000 = fsource ç
è 300 ÷ø
When source is moving away from observer
æ 350 ö
f ' = fsource çè
÷
350 + 50 ø
1000 ´ 300 350
´
350
400
f ' » 750 Hz
v , v0
73. (a) f ' = f
v ∗ vs
f'=
340 , 20
340 ∗ 20
\ f = 2250 Hz.
74. (b) According to Doppler’s effect, when source is moving
but observer at rest
or 2000 < f
é 340 ù
and, f 2 = f 0 ê
ú
ë 340 –17 û
f 340 –17 323
f 19
\ l =
=
or, 1 =
f 2 340 – 34 306 f 2 18
75. (a) Frequency of the sound produced by open flute.
æ v ö 2 ´ 330
= 660 Hz
f = 2ç ÷ =
è 2l ø 2 ´ 0.5
5 25
Velocity of observer, v0 = 10 ´ = m / s
18 9
As the source is moving towards the observer therefore,
according to Doppler's effect.
\ Frequency detected by observer,
é 25
ù
ê 9 + 330 ú
é v + v0 ù
ê
ú 660
f'= ê
ú f = ê 330 ú
ë v û
ë
û
2995
´ 660 or, f ' = 665.55 ; 666 Hz
=
9 ´ 330
76. (d) nA = 425 Hz, nB = ?
Beat frequency x = 5 Hz which is decreasing (5 ® 3) after
increasing the tension of the string B.
Also tension of string B increasing so
nB­ (Q n µ T)
Hence nA – nB­= x¯ ¾¾
® correct
nB­ – nA= x¯ ¾¾
® incorrect
\ nB = nA – x = 425 – 5 = 420 Hz
77. (d) From Doppler's effect
æ 340 ö
f (direct) = f ç
÷ = f1
è 340 - 5 ø
æ 340 ö
f (by wall)=f ç
÷ =f
è 340+5 ø 2
Beats = (f1 – f2)
340 ö
æ 340
5=fç
÷
è 340 - 5 340 + 5 ø
Þ f < 170 Hz.
78. (d) We know that the apperent frequency
æ v - v0 ö
f'=ç
÷ f from Doppler's effect
è v - vs ø
where v0 = vs = 30 m/s, velocity of observer and source
Speed of sound v = 330 m/s
330 ∗ 30
\ f '<
≥ 540 = 648 Hz.
330 , 30
Q Frequency of whistle (f) = 540 Hz.
é
79.
v ù
320
Hz
ú= f´
v
v
300
sû
ë
(d ) f1 = f ê
é v ù
320
f2 = f ê
Hz
ú= f ´
340
ë v + vs û
P-234
Physics
æ f2 ö
æ 300 ö
çè f - 1÷ø ´ 100 = çè 340 - 1÷ø ´ 100 ; 12%
1
80. (c) According to Doppler’s effect,
æ V +V ö
0
Apparent, frequency f = ç V – V ÷ f 0
S ø
è
æ
f
ö
V f
0
0
Now, f = ç V – V ÷ V0 + V – V
è
S ø
s
f0
So, slope =
V – VS
Hence, option (c) is the correct answer.
81. (a) Reflected frequency of sound reaching bat
é V - (-V0 ) ù
é V + V0 ù
V + 10
= ê
f
úf= ê
úf=
V - 10
ë V - Vs û
ëV - Vs û
æ 320 + 10 ö
= ç
÷ ´ 8000 = 8516 Hz
è 320 - 10 ø
82. (b) Given fA = 1800 Hz
vt = v
fB = 2150 Hz
Reflected wave frequency received by A, fA¢ = ?
Applying doppler’s effect of sound,
vs f
f¢ =
vs - v t
æ f ö
here, v t = vs ç1 - A ÷
è fB ø
æ 1800 ö
= 343 ç 1 ÷
è 2150 ø
vt = 55.8372 m/s
Now, for the reflected wave,
æ vs + v t ö
¢
\ fA = ç v - v ÷ fA
t ø
è s
æ 343 + 55.83 ö
=ç
÷ ´ 1800
è 343 - 55.83 ø
= 2499.44 » 2500Hz
83. (d) Given: Frequency of sound produced by siren, f =
800 Hz
Speed of observer, u = 2 m/s
Velocity of sound, v = 320 m/s
No. of beats heard per second = ?
No. of extra waves received by the observer per second = + 4l
\ No. of beats/ sec
2 æ 2ö 4
= - ç- ÷ =
l è lø l
æ
2´ 2
Vö
= 320
çQ l = ÷
f ø
è
800
2 ´ 2 ´ 800
= 10
=
320
84. (c) f = 500 Hz
A
4 m/s C
B
Listener
Case 1 : When source is moving towards stationary
listener
æ v ö
æ 340 ö = 506
apparent frequency h ' = h ç v - v ÷ = 500 ç
÷
s ø
è
è 336 ø
Hz
Case 2 : When source is moving away from the stationary
listener
æ v ö
æ 340 ö
h" = h ç
÷ = 500 ç
÷ = 494 Hz
è 344 ø
è v + vs ø
In case 1 number of beats heard is 6 and in case 2 number of
beats heard is 18 therefore frequency of the source at B = 512 Hz
85. (d)
ENGINE
A
0.9 km
H
B I
L
L
C
Let after 5 sec engine at point C
AB BC
0.9 ´ 1000 BC
t=
+
+
5=
330 330
330
330
\ BC = 750 m
Distance travelled by engine in 5 sec
= 900 m – 750 m = 150 m
Therefore velocity of engine
150 m
=
= 30 m/s
5sec
86. (c) Bats catch the prey by hearing reflected ultrasonic
waves.
When the source and the detector (observer) are moving,
frequency of reflected waves change. This is according to
Doppler’s effect.
2
u=0
vm
a = 2m/s
87. (a)
Electric
s
Motor
siren
cycle
Let the motorcycle has travelled a distances, its velocity
at that point
2
vm
- u 2 = 2as \ v2m = 2 ´ 2 ´ s
\ vm = 2 s
The observed frequency will be
é v - vm ù
n'= nê
ú
ë v û
é 330 - 2 s ù
0.94n = n ê
ú Þ s = 98.01 m
ë 330 û
é v ù
88. (c) Apparent frequency n' = n ê
ú
ë v - vs û
é 300 ù
Þ 10000 = 9500 ê
ú Þ 300 - v = 300 ´ 0.95
ë 300 - v û
Þ v = 300 - 285 = 15 ms -1
89. (c) Apparent frequency
vù
é
v+ ú
ê
+
v
v
é
5 = n é6ù n' = 6
0ù = n
n' = nê
ê
ú
ê5ú n 5
ú
ë û
ë v û
ê v ú
ë
û
The percentage increase in apparent
n'- n 6 - 5
=
´ 100 = 20%
frequency
n
5
15
P-235
Electric Charges and Fields
Electric Charges
and Fields
TOPIC 1
1.
2.
Electric Charges and Coulomb's
Law
+Q
Three charges + Q, q, + Q are placed respectively, at
distance, d/2 and d from the origin, on the x-axis. If the net
force experienced by + Q, placed at x = 0, is zero, then
value of q is:
[9 Jan. 2019 I]
(a) – Q/4
(b) + Q/2 (c) + Q/4
(d)
– Q/2
Charge is distributed within a sphere of radius R with a
A -2r a
where A and a
e
r2
are constants. If Q is the total charge of this charge
distribution, the radius R is:
[9 Jan. 2019, II]
–Q
volume charge density p(r) =
Q ö
æ
(a) a log ç 1 ÷
2paA ø
è
3.
4.
æ
ö
ç
÷
a
1
log ç
(b)
÷
2
çç 1 - Q ÷÷
2paA ø
è
æ
ö
ç
÷
1
a
Q ö
æ
(c) a log ç
log ç 1 ÷ (d)
÷
2
2paA ø
è
çç 1 - Q ÷÷
2paA ø
è
Two identical conducting spheres A and B, carry equal
charge. They are separated by a distance much larger than
their diameter, and the force between them is F. A third
identical conducting sphere, C, is uncharged. Sphere C is
first touched to A, then to B, and then removed. As a
result, the force between A and B would be equal to
[Online April 16, 2018]
3F
F
3F
(b)
(c) F
(d)
4
2
8
Shown in the figure are two point charges +Q and –Q
inside the cavity of a spherical shell. The charges are kept
near the surface of the cavity on opposite sides of the
centre of the shell. If s1 is the surface charge on the inner
surface and Q1 net charge on it and s2 the surface charge
on the outer surface and Q2 net charge on it then :
[Online April 10, 2015]
(a)
5.
(a) s1 ¹ 0, Q1 = 0
(b) s1 ¹ 0, Q1 = 0
s2 = 0, Q2 = 0
s2 ¹ 0, Q2 = 0
(c) s1 = 0, Q1 = 0
(d) s1 ¹ 0, Q1 ¹ 0
s2 = 0, Q2 = 0
s2 ¹ 0, Q2 ¹ 0
Two charges, each equal to q, are kept at x = – a and x = a
q
on the x-axis. A particle of mass m and charge q 0 =
is
2
placed at the origin. If charge q0 is given a small
displacement (y <<a) along the y-axis, the net force acting
on the particle is proportional to
[2013]
1
1
(d) –
y
y
Two balls of same mass and carrying equal charge are
hung from a fixed support of length l. At electrostatic
equilibrium, assuming that angles made by each thread is
small, the separation, x between the balls is proportional
to :
[Online April 9, 2013]
(a) l
(b) l 2
(c) l 2/3
(d) l 1/3
Two identical charged spheres suspended from a common
point by two massless strings of length l are initially a
distance d(d << l) apart because of their mutual repulsion.
The charge begins to leak from both the spheres at a
constant rate. As a result charges approach each other
with a velocity v. Then as a function of distance x between
them,
[2011]
(a) v µ x–1 (b) v µ x½ (c) v µ x
(d) v µ x–½
A charge Q is placed at each of the opposite corners of a
square. A charge q is placed at each of the other two
corners. If the net electrical force on Q is zero, then Q/q
equals:
[2009]
1
(a) –1
(b) 1
(c) (d) -2 2
2
If gE and gM are the accelerations due to gravity on the
surfaces of the earth and the moon respectively and if
Millikan’s oil drop experiment could be performed on the
(a) y
6.
7.
8.
9.
(b) –y
(c)
P-236
Physics
[2007]
two surfaces, one will find the ratio
electronic charge on the moon
to be
electronic charge on the earth
10.
11.
(a) gM / g E (b) 1
(c) 0
(d) g E / g M
Two spherical conductors B and C having equal radii and
carrying equal charges on them repel each other with a
force F when kept apart at some distance. A third spherical
conductor having same radius as that B but uncharged is
brought in contact with B, then brought in contact with C
and finally removed away from both. The new force of
repulsion between B and C is
[2004]
(a) F/8
(b) 3 F/4 (c) F/4
(d) 3 F/8
Three charges –q1 , +q2 and –q3 are place as shown in the
figure. The x - component of the force on –q 1 is
proportional to
[2003]
Y
q3
a
b
+q 2 X
q2 q3
q 2 q3
(a) 2 - 2 cos q
(b) 2 +
sin q
b
b
a
a2
q
q
q
q
(c) 2 + 3 cos q
(d) 2 - 3 sin q
2
2
2
b
b
a
a2
If a charge q is placed at the centre of the line joining two
equal charges Q such that the system is in equilibrium
then the value of q is
[2002]
(a) Q/2
(b) –Q/2 (c) Q/4
(d) –Q/4
q1
12.
TOPIC 2
13.
x1
14.
x13
x23
(b)
Q2
B
x2
O
x2
x1
(c)
x1
x2
(d)
x2 2
x12
Consider the force F on a charge ‘q’ due to a uniformly
charged spherical shell of radius R carrying charge Q distributed uniformly over it. Which one of the following statements is true for F, if ‘q’ is placed at distance r from the
centre of the shell?
[Sep. 06, 2020 (II)]
1 Qq
1 Qq
for r < R (b)
> F > 0 for r < R
4pe 0 R 2
4pe0 R 2
1 Qq
1 Qq
for r > R (d) F =
for all r
4pe0 R 2
4pe0 R 2
15. Two charged thin infinite plane sheets of uniform surface
(c) F =
charge density s + and s – , where | s + | > | s – |, intersect
at right angle. Which of the following best represents the
electric field lines for this system ? [Sep. 04, 2020 (I)]
s–
s+
(a)
s–
s+
(b)
s–
s+
(c)
s–
Electric Field and Electric Field
Lines
Charges Q1 and Q2 are at points A and B of a right angle
triangle OAB (see figure). The resultant electric field at
point O is perpendicular to the hypotenuse, then Q1/Q2 is
proportional to :
[Sep. 06, 2020 (I)]
A
Q1
(a)
(a) F =
s+
(d)
16. A particle of charge q and mass m is subjected to an electric
field E = E0 (1 – ax2) in the x-direction, where a and E0 are
constants. Initially the particle was at rest at x = 0. Other
than the initial position the kinetic energy of the particle
becomes zero when the distance of the particle from the
origin is :
[Sep. 04, 2020 (II)]
3
1
2
(c)
(d)
a
a
a
17. A charged particle (mass m and charge q) moves along X
axis with velocity V0. When it passes through the origin it
r
enters a region having uniform electric field E = - Ejˆ which
(a) a
(b)
extends upto x = d. Equation of path of electron in the
region x > d is :
[Sep. 02, 2020 (I)]
P-237
ur
20. An electric dipole of moment p = (iˆ - 3 ˆj + 2kˆ) ´ 10-29 C.m
is at the origin (0, 0, 0). The electric field due to this dipole
r
at r = +iˆ + 3 ˆj + 5kˆ
r ur
(note that r . p = 0) is parallel to:
[9 Jan. 2020, I]
Electric Charges and Fields
Y
E
O
(a) y =
(c) y =
18.
qEd
mV02
qEd
V0
X
d
(x - d )
(b) y =
x
(d) y =
qEd æ d
ö
- x÷
2 ç
mV0 è 2
ø
qEd 2
x
mV02
A small point mass carrying some positive charge on it, is
released from the edge of a table. There is a uniform electric
field in this region in the horizontal direction. Which of the
following options then correctly describe the trajectory of
the mass ? (Curves are drawn schematically and are not to
scale).
[Sep. 02, 2020 (II)]
E
x
mV02
(a) (+iˆ - 3 ˆj - 2kˆ)
(c) (+iˆ + 3 ˆj - 2kˆ)
(d) (-iˆ - 3 ˆj + 2kˆ)
21. A charged particle of mass ‘m’ and charge ‘q’ moving under
the influence of uniform electric field Eiˆ and a uniform
r
magnetic field Bk follows a trajectory from point P to Q as
shown in figure. The velocities at P and Q are respectively,
r
r
vi and -2vj . Then which of the following statements
(A, B, C, D) are the correct? (Trajectory shown is
schematic and not to scale)
[9 Jan. 2020, I]
Y
E
P
y
y
a
y
(a)
O
(b)
x
x
y
y
(c)
19.
(d)
x
x
Consider a sphere of radius R which carries a uniform
R
charge density r. If a sphere of radius
is carved out of
2
ur
EA
it, as shown, the ratio ur
of magnitude of electric
EB
ur
ur
field E A and E B , respectively, at points A and B due to
the remaining portion is:
[9 Jan. 2020, I]
(b) (-iˆ + 3 ˆj - 2kˆ)
B
v
Q
2v
2a
3 æ mv 2 ö
(A) E = 4 çç qa ÷÷
è
ø
X
3 æ mv 2 ö
(B) Rate of work done by the electric field at P is 4 çç a ÷÷
è
ø
(C) Rate of work done by both the fields at Q is zero
(D) The difference between the magnitude of angular
momentum of the particle at P and Q is 2 mav.
(a) (A), (C), (D)
(b) (B), (C), (D)
(c) (A), (B), (C)
(d) (A), (B), (C) , (D)
22. Three charged particles
y
2q
B d
150°
O
d
30°
30°
d
–4q
A
x
C
–2q
A, B and C with charges – 4q, 2q and –2q are present on
the circumference of a circle of radius d. The charged
particles A, C and centre O of the circle formed an
equilateral triangle as shown in figure. Electric field at O
along x-direction is:
[8 Jan. 2020, I]
(a)
21
34
(b)
18
34
(c)
17
54
(d)
18
54
(a)
3q
p Î0 d
2
(b)
2 3q
p Î0 d
2
(c)
3q
4p Î0 d
2
(d)
3 3q
4p Î0 d 2
P-238
23.
Physics
A particle of mass m and charge q is released from rest in
a uniform electric field. If there is no other force on the
particle, the dependence of its speed v on the distance x
travelled by it is correctly given by (graphs are schematic
and not drawn to scale)
[8 Jan. 2020, II]
v
v
(a)
x
v
v
(c)
(d)
x
x
Two infinite planes each with uniform surface charge
density +s are kept in such a way that the angle between
them is 30°. The electric field in the region shown between
them is given by:
[7 Jan. 2020, I]
30°
x
é
æ
s
3ö
xˆ ù
s é
xˆ ù
yˆ + ú
(a)
(1 + 3) yˆ - ú (b)
êç1 +
÷
ê
Î0 ëêè
2 ø
2 ûú
2 Î0 ë
2û
s éæ
3ö
xˆ ù
xˆ ù
s é
yˆ - ú
1 + 3 yˆ + ú (d)
êç1 ÷
ê
2 Î0 ëêè
2 ø
2 ûú
2 Î0 ë
2û
A particle of mass m and charge q has an initial velocity
r
r
r
v = v0 $j . If an electric field E = E0 i and magnetic field
r
B = B0iˆ act on the particle, its speed will double after a
(
25.
)
time:
26.
[7 Jan 2020, II]
2mv0
2mv0
3mv0
3mv0
(a) qE
(b) qE
(c)
(d) qE0
qE0
0
0
A simple pendulum of length L is placed between the
plates of a parallel plate capacitor having electric field E,
as shown in figure. Its bob has mass m and charge q. The
time period of the pendulum is given by :
[10 April 2019, II]
L
2p
qE ö
æ
(a)
çg+
÷
m ø
è
2p
(c)
L
qE ö
æ
çg÷
m ø
è
2p
(b)
(d)
1
R
(b) E µ
R
(c) R
(d)
R 2
5
2
30. Two point charges q1 ( 10 mC) and q2 (– 25 mC) are
placed on the x-axis at x = 1 m and x = 4 m respectively.
The electric field (in V/m) at a point y = 3 m on y-axis
is,
[9 Jan 2019, II]
é
ù
1
= 9 ´ 109 Nm 2 C -2 ú
ê take
4p Î0
ë
û
(b)
(a) (63 î – 27 ĵ ) × 102
(b) (– 63 î + 27 ĵ ) × 102
(c) (81 î – 81 ĵ ) × 102 (d) (–81 î + 81 ĵ ) × 102
31. A body of mass M and charge q is connected to a spring
of spring constant k. It is oscillating along x-direction about
its equilibrium position, taken to be at x = 0, with an
amplitude A. An electric field E is applied along the
x-direction. Which of the following statements is correct?
[Online April 15, 2018]
1
1 q2 E2
mw2 A2 +
2
2 k
2qE
(b) The new equilibrium position is at a distance:
k
from x = 0
qE
(c) The new equilibrium position is at a distance:
2k
from x = 0
2 2
1
2 2 1q E
(d) The total energy of the system is mw A –
2
2 k
32. A solid ball of radius R has a charge density r given by
(a) The total energy of the system is
q2 E2
m2
the ball is:
L
(a)
æ qE ö
g2 + ç
÷
è m ø
3
rö
æ
r = r0 ç1 - ÷ for 0 £ r £ R. The electric field outside
è Rø
L
g2 -
2p
29.
(a)
y
(c)
1
1
1
(c) E µ 4 (d) E µ 2
D
D
D
D
The bob of a simple pendulum has mass 2 g and a charge
of 5.0 ¼C. It is at rest in a uniform horizontal electric field
of intensity 2000 V/m. At equilibrium, the angle that the
pendulum makes with the vertical is :
[8 April 2019 I]
(take g = 10 m/s2)
(a) tan–1 (2.0) (b) tan –1 (0.2)
(c) tan–1 (5.0) (d) tan –1 (0.5)
For a uniformly charged ring of radius R, the electric field
on its axis has the largest magnitude at a distance h from
its centre. Then value of h is:
[9 Jan. 2019 I]
(a) E µ
28.
(b)
x
24.
27. Four point charges –q, +q, + q and –q are placed on y-axis
at y = –2d, y = –d, y = +d and y = +2d, respectively. The
magnitude of the electric field E at a point on the x-axis at
x = D, with D>> d, will behave as:
[9 April 2019, II]
2
r0 R 3
e0 r 2
[Online April 15, 2018]
(b)
4r0 R 3
3e 0 r 2
(c)
3r 0 R 3
4e 0 r 2
(d)
r0 R 3
12e0 r 2
P-239
Electric Charges and Fields
33.
34.
A long cylindrical shell carries positive surface charge s in
the upper half and negative surface charge - s in the lower
half. The electric field lines around the cylinder will look
like figure given in : (figures are schematic and not drawn
to scale)
[2015]
(a)
(b)
(c)
(d)
C ield Lines
A wire of length L (=20 cm), is bent into a semicircular
arc. If the two equal halves of the arc were each to be
uniformly charged with charges ± Q, [|Q| = 103e0
Coulomb where e0 is the permittivity (in SI units) of free
space] the net electric field at the centre O of the
semicircular arc would be : [Online April 11, 2015]
Y
X
O
(a) (50 × 10 N/C) $j
3
35.
36.
O
(b) (50 × 103 N/C) $i
(c) (25 × 103 N/C) $j
(d) (25 × 103 N/C) $i
A thin disc of radius b = 2a has a concentric hole of radius
‘a’ in it (see figure). It carries uniform surface charge ‘s’
on it. If the electric field on its axis at height ‘h’ (h << a)
from its centre is given as ‘Ch’ then value of ‘C’ is :
[Online April 10, 2015]
s
(a)
4aÎ0
s
(b)
8aÎ0
s
(c)
aÎ0
s
(d)
2aÎ0
A spherically symmetric charge distribution is characterised
by a charge density having the following variations:
rö
æ
r ( r ) = ro ç1 - ÷ for r < R
è Rø
r(r) = 0 for r ³ R
Where r is the distance from the centre of the charge
distribution ro is a constant. The electric field at an internal
point (r < R) is:
[Online April 12, 2014]
(a)
ro æ r r 2 ö
ç ÷
4eo çè 3 4R ÷ø
ro æ r r 2 ö
(c)
ç ÷
3eo çè 3 4R ÷ø
(b)
37. The magnitude of the average electric field normally
present in the atmosphere just above the surface of the
Earth is about 150 N/C, directed inward towards the center
of the Earth. This gives the total net surface charge carried
by the Earth to be:
[Online April 9, 2014]
[Given eo = 8.85 × 10–12 C2/N-m2, RE = 6.37 × 106 m]
(a) + 670 kC
(b) – 670 kC
(c) – 680 kC
(d) + 680 kC
38. The surface charge density of a thin charged disc of radius
R is s. The value of the electric field at the centre of the
disc is
s
. With respect to the field at the centre, the
2 Î0
electric field along the axis at a distance R from the centre
of the disc :
[Online April 25, 2013]
(a) reduces by 70.7%
(b) reduces by 29.3%
(c) reduces by 9.7%
(d) reduces by 14.6%
39. A liquid drop having 6 excess electrons is kept stationary
under a uniform electric field of 25.5 kVm–1. The density of
liquid is 1.26 × 103 kg m–3. The radius of the drop is (neglect
buoyancy).
[Online April 23, 2013]
(a) 4.3 × 10–7 m
(b) 7.8 × 10–7 m
(c) 0.078 × 10–7 m
(d) 3.4 × 10–7 m
40. In a uniformly charged sphere of total charge Q and radius
R, the electric field E is plotted as function of distance
from the centre, The graph which would correspond to the
above will be:
[2012]
E(r)
E(r)
(a)
(b)
r
r
E(r)
E(r)
(c)
(d)
r
r
41. Three positive charges of equal value q are placed at
vertices of an equilateral triangle. The resulting lines of
force should be sketched as in [Online May 26, 2012]
(a)
(b)
(c)
(d)
ro æ r r 2 ö
ç ÷
eo çè 3 4R ÷ø
ro æ r r 2 ö
(d)
ç ÷
12eo çè 3 4R ÷ø
P-240
42.
Physics
A thin semi-circular ring of radius r has a positive charge q
ur
distributed uniformly over it. The net field E at the centre
O is
[2010]
(a)
j
(b)
O
43.
q
ˆj
2
2
(b) -
2
æ5 rö
with charge density varying as r(r ) = r0 çè - ÷ø upto r
4 R
= R , and r(r ) = 0 for r > R , where r is the distance from
the origin. The electric field at a distance r(r < R) from the
origin is given by
[2010]
r0 r æ 5 r ö
4pr0 r æ 5 r ö
(a)
ç - ÷
(b) 3e çè 3 - R ÷ø
4e 0 è 3 R ø
0
r0 r æ 5 r ö
r0 r æ 5 r ö
ç - ÷
(d) 3ε çè 4 - ÷ø
R
4ε0 è 4 R ø
0
This question contains Statement-1 and Statement-2. Of
the four choices given after the statements, choose the
one that best describes the two statements.
Statement-1 : For a charged particle moving from point P
to point Q, the net work done by an electrostatic field on
the particle is independent of the path connecting point P
to point Q.
Statement-2 : The net work done by a conservative force
on an object moving along a closed loop is zero. [2009]
(a) Statement-1 is true, Statement-2 is true; Statement-2
is the correct explanation of Statement-1.
(b) Statement-1 is true, Statement-2 is true; Statement-2
is not the correct explanation of Statement-1.
(c) Statement-1 is false, Statement-2 is true.
(d) Statement-1 is true, Statement-2 is false.
Q
Let r (r ) =
r be the charge density distribution for
p R4
a solid sphere of radius R and total charge Q. For a point
‘P’ inside the sphere at distance r1 from the centre of the
sphere, the magnitude of electric field is :
[2009]
2
Q
Qr1
(a)
(b)
2
4p Î0 r1
4p Î0 R4
(c)
45.
(c)
46.
Qr12
r
E(r)
(d) 0
3p Î0 R4
A thin spherical shell of radus R has charge Q spread
uniformly over its surface. Which of the following graphs
most closely represents the electric field E(r) produced by
the shell in the range 0 £ r < ¥, where r is the distance from
the centre of the shell?
[2008]
O
r
R
(d)
O
R
r
47. Two spherical conductors A and B of radii 1 mm and 2 mm
are separated by a distance of 5 cm and are uniformly
charged. If the spheres are connected by a conducting
wire then in equilibrium condition, the ratio of the
magnitude of the electric fields at the surfaces of spheres
A and B is
[2006]
(a) 4 : 1
(b) 1 : 2
(c) 2 : 1
(d) 1 : 4
48. Two point charges + 8q and – 2q are located at
x = 0 and x = L respectively. The location of a point on the
x axis at which the net electric field due to these two point
charges is zero is
[2005]
(c)
44.
R
R
q
ˆj
4p e 0 r
4p e 0 r 2
q
q
ˆj
ˆ
(c) - 2 2 j
(d)
2
2p e 0 r
2p e 0 r 2
Let there be a spherically symmetric charge distribution
(a)
O
r
E(r)
i
O
E(r)
E(r)
L
(b) 2 L
(c) 4 L
(d) 8 L
4
A charged ball B hangs from a silk thread S, which makes
an angle q with a large charged conducting sheet P, as
shown in the figure. The surface charge density s of the
sheet is proportional to
[2005]
(a)
49.
P
q
S
B
(a) cot q
(b) cos q (c) tan q
(d) sin q
50. Four charges equal to -Q are placed at the four corners of
a square and a charge q is at its centre. If the system is in
equilibrium the value of q is
[2004]
(a) -
(b)
Q
(1 + 2 2)
4
Q
Q
(1 + 2 2)
(1 + 2 2)
(d)
4
2
A charged oil drop is suspended in a uniform field of 3×104
v/m so that it neither falls nor rises. The charge on the
drop will be (Take the mass of the charge = 9.9×10–15 kg
and g = 10 m/s2)
[2004]
(a) 1.6×10–18 C
(b) 3.2×10–18 C
(c) 3.3×10–18 C
(d) 4.8×10–18 C
(c) -
51.
Q
(1 + 2 2)
2
P-241
Electric Charges and Fields
(a) surface change density on the inner surface is uniform
Electric Dipole, Electric Flux
TOPIC 3 and Gauss's Law
52.
Two identical electric point dipoles have dipole moments
®
®
P1 = P$i and P2 = - P$i and are held on the x axis at distance
‘a’ from each other. When released, they move along xaxis with the direction of their dipole moments remaining
unchanged. If the mass of each dipole is ‘m’, their speed
when they are infinitely far apart is : [Sep. 06, 2020 (II)]
P
1
(a)
a pe 0 ma
(c)
53.
P
1
(b)
a 2pe 0 ma
P
2
a pe 0 ma
(d)
P
2
a 2 pe 0 ma
ur
An electric field E = 4 xiˆ - ( y 2 + 1) ˆj N/C passes through
the box shown in figure. The flux of the electric field
through surfaces ABCD and BCGF are marked as f1 and
f11 respectively. The difference between (f1 – f11) is (in
Nm2/C) _______.
[9 Jan 2020, II]
z
A (0, 0, 2)
(0, 2, 2)
54.
y
D
B
(3, 0, 2)
C
(3, 2, 2)
E
F
x
(0, 0, 0)
(3, 0, 0)
H
G
(0, 2, 0)
(3, 2, 0)
In finding the electric field using Gauss law the formula
r
q
| E | = enc is applicable. In the formula Î is
0
Î | A|
0
55.
and equal to
permittivity of free space, A is the area of Gaussian surface
and qenc is charge enclosed by the Gaussian surface. This
equation can be used in which of the following situation?
[8 Jan 2020, I]
(a) Only when the Gaussian surface is an equipotential
surface.
Only when the Gaussian surface is an
r
(b) equipotential surface and | E | is constant on the surface.
r
(c) Only when | E | = constant on the surface.
(d) For any choice of Gaussian surface.
Shown in the figure is a shell made of a conductor. It has
inner radius a and outer radius b, and carries charge Q. At
ur
its centre is a dipole p as shown. In this case :
[12 April 2019, I]
56.
Q/2
4 pa 2
(b) electric field outside the shell is the same as that of a
point charge at the centre of the shell.
(c) surface charge density on the outer surface depends
r
on P
(d) surface charge density on the inner surface of the
shell is zero everywhere.Let a total charge 2 Q be distributed in a sphere of radius
R, with the charge density given by r(r) = kr, where r is
the distance from the centre. Two charges A and B, of – Q
each, are placed on diametrically opposite points, at equal
distance, a, from the centre. If A and B do not experience
any force, then.
[12 April 2019, II]
(a) a = 8–1/4 R
(b) a =
(c) a = 2–1/4 R
(d)
3R
21/ 4
a = R/ 3
57. An electric dipole is formed by two equal and opposite
charges q with separation d. The charges have same mass
m. It is kept in a uniform electric field E. If it is slightly
rotated from its equilibrium orientation, then its angular
frequency w is :
[8 April 2019, II]
qE
qE
qE
2qE
(b)
(c) 2
(d)
md
md
2md
md
58. An electric field of 1000 V/m is applied to an electric dipole
at angle of 45°. The value of electric dipole moment is
10–29 C.m. What is the potential energy of the electric
dipole?
[11 Jan 2019, II]
(a) –20 × 10–18 J
(b) –7 × 10 –27 J
(c) –10 × 10–29 J
(d) – 9 × 10–20 J
59. Charges – q and + q located at A and B, respectively,
constitute an electric dipole. Distance AB = 2a, O is the
mid point of the dipole and OP is perpendicular to AB.
A charge Q is placed at P where OP = y and y >> 2a. The
charge Q experiences an electrostatic force F. If Q is now
moved along the equatorial line to P¢ such that OP¢
(a)
æ yö
æy
ö
= ç ÷ , the force on Q will be close to: ç >> 2a ÷
3
3
è ø
è
ø
[10 Jan 2019, II]
P
Q P¢
A
(a) 3 F
O
–q
(b)
+q
F
3
(c) 9 F
B
(d) 27 F
P-242
60.
Physics
A charge Q is placed at a distance a/2 above the centre of the
square surface of edge a as shown in the figure. The electric
flux through the square surface is:
[Online April 15, 2018]
Q
(a) 3e
0
P
a/2
Q
(b) 6e
0
Q
(c) 2e
0
61.
62.
a
Q
(d) e
0
ur
An electric dipole has a fixed dipole moment p , which
makes angle q with respect to x-axis. When subjected to
uur
ur
an electric field E1 = Eiˆ , it experiences a torque T1 = t iˆ .
uur
When subjected to another electric field E2 = 3E1 ˆj it
uur
ur
experiences torque T2 = -T1 . The angle q is :
[2017]
(a) 60°
(b) 90°
(c) 30°
(d) 45°
Four closed surfaces and corresponding charge distributions are shown below.
[Online April 9, 2017]
5q
q
2q
q
S1
8q
–2q
–4q
–q
q
q
q
S3
S2
Q
b
(c)
64.
(
2
p a -b
Q
2 pa 2
2
)
(b)
(d)
(a) pa2 E
S4
a
2Q
45°
2Q
pa 2
(
®
E
3q
Let the respective electric fluxes through the surfaces be
F 1, F 2, F 3, and F 4. Then :
(a) F 1< F2 = F 3 > F4
(b) F 1> F2 > F 3 > F4
(c) F 1= F2 = F 3 = F4
(d) F 1> F 3 ; F 2 < F4
63. The region between two concentric spheres of radii 'a' and
'b', respectively (see figure), have volume charge density
A
r= , where A is a constant and r is the distance from
r
the centre. At the centre of the spheres is a point charge
Q. The value of A such that the electric field in the region
between the spheres will be constant, is:
[2016]
(a)
through a circular surface of radius 0.02 m parallel to the YZ plane is nearly:
[Online April 19, 2014]
(a) 0.125 Nm2/C
(b) 0.02 Nm2/C
2
(c) 0.005 Nm /C
(d) 3.14 Nm2/C
ur
ur
65. Two point dipoles of dipole moment p1 and p 2 are at a
ur ur
distance x from each other and p1 || p 2 . The force between
the dipoles is :
[Online April 9, 2013]
1 4 p1 p2
1 3 p1 p2
(a)
(b)
4pe0 x 4
4pe0 x3
1 8 p1 p2
1 6 p1 p2
(c)
(d)
4pe0 x 4
4pe0 x 4
66. The flat base of a hemisphere of radius a with no charge
inside it lies in a horizontal plane. A uniform electric field
®
p
E is applied at an angle
with the vertical direction. The
4
electric flux through the curved surface of the hemisphere
is
[Online May 19, 2012]
(
)
The electric field in a region of space is given by,
r
E = Eoˆi + 2Eoˆj where Eo = 100 N/C. The flux of the field
)
2
(d)
2 2
2 2
67. An electric dipole is placed at an angle of 30° to a nonuniform electric field. The dipole will experience [2006]
(a) a translational force only in the direction of the field
(b) a translational force only in a direction normal to
the direction of the field
(c) a torque as well as a translational force
(d) a torque only
68. If the electric flux entering and leaving an enclosed surface
respectively is f1 and f2, the electric charge inside the surface
will be
[2003]
(a) (f2 – f1)eo
(b) (f1 – f2)/eo
(c) (f2 – f1)/eo
(d) (f1 – f2)eo
69. A charged particle q is placed at the centre O of cube of
length L (A B C D E F G H). Another same charge q is
placed at a distance L from O. Then the electric flux
through ABCD is
[2002]
E
F
(c)
c
O
2 p b2 - a 2
2
( p + 2) pa 2 E
pa2 E
D
Q
pa2 E
(b)
H
A
(a) q /4 p Î0 L
(c) q/2 p Î0 L
q
q
G
B
L
(b) zero
(d) q/3 p Î0 L
P-243
Electric Charges and Fields
y
d
1.
(a)
Fa
Fb
+Q
q
d/2
5.
Force due to charge + Q,
KQQ
Fa = 2
d
Force due to charge q,
KQq
Fb =
2
ædö
ç ÷
è2ø
For equilibrium,
r r
Fa + Fb = 0
Þ
kQQ
d2
+
kQq
( d / 2)
2
F
+Q
d/2
(b)
Fnet =
= 0 \q = -
Q
4
A –2r/a
e
( 4pr 2dr )
r2
0
æ
ö
R
ç e –2r/a ÷
–2r/a
= 4 pA ò e dr = 4 p A ç
÷
0
ç –2 ÷
è a ø0
æ aö
= 4pA ç – ÷ e –2R/a –1
è 2ø
dr
Q = 2paA(1–e–2R/a)
æ
ö
r
ç
÷
a
1
R = log ç
÷
2
ç 1– Q ÷
2paA ø
è
(d) Spheres A and B carry equal charge say 'q'
kqq
\ Force between them, F = 2
r
q
When A and C are touched, charge on both q A = q C =
2
Then when B and C are touched, charge on B
q
+q
3q
qB = 2
=
2
4
Now, the force between charge qA and qB
4.
Þ
k´
F sin q
æqö
2kq ç ÷
è2ø
æ y2 + a2 ö
ç
÷
è
ø
6.
kq2 y
2
×
y
2
y + a2
(Q y << a)
So, F µ y
a3
(d)
)
q 3q
´
2
2 4 = 3 kq = 3 F
F' =
=
8 r2
8
r2
r2
(c) Inside the cavity net charge is zero.
\ Q1 = 0 and s1 = 0
There is no effect of point charges +Q, –Q and induced
charge on inner surface on the outer surface.
\ Q2 = 0 and s2 = 0
kq A q B
x
q
a
æqö
2kq ç ÷ y
2
Fnet = 2 è 2 ø 3/2
(y +a )
R
3.
a
2F cos q
Þ Fnet = 2F cosq
Q = ò rdv = ò
(
q
F
Þ F sin q
R
2.
(a)
q
l
q
Tcos q
Tsin q
q
q
x
mg
In equilibrium, Fe = T sin q
mg = T cos q
tan q =
Fe
q2
=
mg 4p Î0 x 2 ´ mg
also tan q » sin =
x/2
l
x
q2
Hence, 2l =
4p Î0 x 2 ´ mg
Þ x3 =
2q 2 l
4p Î0 mg
1/3
æ q2l ö
÷÷
\ x = çç
è 2p Î0 mg ø
Therefore x µ l1/3
Fe
P-244
7.
Physics
(d) From figure
T cos q = mg
T sin q = Fe
....(i)
....(ii)
Dividing equation (ii) by (i), we get
sin q
Fe
Þ cos q = mg
Þ
kq 2
Þ Fe = mg tan q
x 2mg tan q
Þ q2 =
k
= mg tan q
x2
Since q is small
\ q2 =
x 3mg
2 kl
Fe
Þ
l
Tcosq
T
q Tsinq
q
l
x
dq 3
dx
3
a
x
xV
=
dt 2
dt
2
Since
8.
dq
= const.
dt
Þ v µ x–1/2
[Q q2 µ x3]
(d) Let F be the force between Q and Q. The force
between q and Q should be attractive for net force on Q to
be zero. Let F¢ be the force between Q and q. The resultant
of F¢ and F¢ is R. For equilibrium
p(q)
A(Q)
l
D(q)
R
F¢
C
Þ
2´k
l
2
= -k
x2
æ 3Q ö æ Q ö
ç
÷ç ÷
3 Q2
4 øè 2 ø
=kè
=
k
8 x2
x2
3
F
8
11. (b) Force applied by charge q2 on q1
qq
F12 = k 1 22
b
Force applied by charge q3 on q1
qq
F13 = k 1 23
a
The X-component of net
q
force (Fx) on
q1 is F12 + F13 sin q
qq
qq
\ Fx = k 1 22 + k 1 22 sin q
b
a
F13cos q
q
q
\ Fx µ 22 + 32 sin q
b
a
12. (d) At equilibrium net force is zero,
Q ´Q
Qq
+k 2 =0
\k
2
(2 x )
x
x
x
Q
Q
Q
Þ q=4
F
q
13. (c) Electric field due charge Q2, E2 =
2
Q
= -2 2
q
9.
(b) It is obvious that by charge conservaiton law,
electronic charge does not depend on acceleration due to
gravity as it is a universal constant.
So, electronic charge on earth
= electronic charge on moon
\ Required ratio = 1.
C
B
10. (d)
×
r
r
F12
F13 sin q
F13
Q
Electric field due charge Q1, E1 =
Q2
( 2 l)
QC¢ QB¢
F¢
Net force on Q at C is zero.
r r
\ R+ F = 0 Þ
2 F¢ = -F
Qq
Q ö 1 3Q
æ
.
çè Q + ÷ø =
2 2
4
or Fnew =
q2 µ x3/2
mg
Þ
Q
Q
and charge on third sphere becomes . Now it is
2
2
touched to C, charge then equally distributes themselves
to make potential same, hence charge on C becomes
i.e.,
\ Fnew = k
x
2l
\ tan q » sin q =
x is distance between the spheres. When third spherical
conductor comes in contact with B charge on B is halved
kQ2
x22
kQ1
x12
Q1 A
Þ
Q Q
Initial force, F = K B 2 C
x
x1
q
E2
O
x2
q
Enet
q
90–q
E1
From figure,
E
x
tan q = 2 = 1 Þ
E1 x2
kQ2
x
= 1
kQ
x2
x22 ´ 21
x1
B
Q2
P-245
Electric Charges and Fields
14.
Q2 x12
x1
Q
x
Q
x
Þ 2 = 2 or, 1 = 1 .
2
x
Q
x
Q
x
Q1 x2
2
1
1
2
2
(c) For spherical shell
Þ
=
y
v0
t=0
1 Q
(if r ³ R )
4pe 0 r 2
=0
(if r < R)
Force on charge in electried field, F = qE
(For r < R)
\F = 0
1 Qq
F=
(For r > R)
4pe 0 r 2
15. (c) The electric field produced due to uniformly charged
infinite plane is uniform. So option (b) and (d) are wrong.
And +ve charge density s+ is bigger in magnitude so its
field along Y direction will be bigger than field of –ve charge
density s– in X direction. Hence option (c) is correct.
–s
E1
ER
ER
–s
E1
E
E
E=
E2 E2
1
+s
+s
E
16.
E
(c) Given,
Also, F = ma = mv
dv ö
æ
çèQ a = v ÷ø
dx
dv
dx
qEd ü
ì
ïm =
ï
mv02 ý
y = mx + c, í
ï(d , - y ) ï
0 þ
î
- y0 =
y=
-qEd
mv02
- qEd
mv02
, d + c Þ c = - y0 +
x - y0 +
qEd 2
mv02
qEd 2
mv02
- qEdx
mv02
- qEd
1 qEd 2 qEd 2
+
2 mv02
mv02
+
Net acceleration of particle is constant, initial velocity is
zero therefore path is straight line.
x
qE0 (1 - x 2 )dx
m
0
ax =
Þ ò v dv = ò
v 2 qE0 æ
9 x3 ö
=
çç x ÷=0
m è
2
3 ÷ø
2E
m
2
ay = g
3
a
(b) Fx = 0, ax = 0, (v)x = constant
Þx=
Time taken to reach at ' P ' =
-
1 qEd 2
qEd æ d
ö
Þy=
- x÷
2
2
2 ç
ø
2 mv0
mv0 è 2
mv0
18. (d) Net force acting on the particle,
r
F = qEiˆ + mgjˆ
qE0 (1 - x 2 )dx
m
Integrating both sides we get,
17.
qEt0 æ
d ö
, çt = ÷
m × v0 è
v0 ø
qEd , Slope = - qEd
mv02
m × v02
r
No electric field Þ Fnet = 0, v = const.
y=
Þ v dv =
Þ
vx
=
tan q =
y=
dv
= qE0 (1 - x 2 )
dx
0
vy
2
\ Force, F = qE = qE0 (1 - x2 )
v
tan q =
(d, –y0)
q vx
vnet
1 qE æ d ö
1 qEd 2
y0 = ×
=
2 m çè v0 ÷ø
2 mv02
Electric field, E = E0 (1 - x 2 )
\ mv
P
vy
d
= t0 (let)
v0
1 qE 2
× t0
(Along – y), y0 = 0 + ×
2 m
...(i)
...(ii)
a=
æ 2E ö
2
çè
÷ +g
mø
Rö
æ
19. (b) Electric field at A ç R ' = ÷
2ø
è
q
E A .ds =
e0
B
3
4 æRö
r´ p ç ÷
r
3 è2ø
Þ EA =
2
æRö
e 0 × 4p ç ÷
è2ø
A
3R
2
R/2
P-246
Physics
r
s ( R / 2 ) æ sR ö
Þ EA =
=ç
÷
3e0
è 6e 0 ø
Electric fields at ‘B’
Electric field due to charge +2q at centre O
r
E1 =
1
2q é 3iˆ – ˆj ù
´ 2ê
ú
4pe0 d ë 2 û
Electric field due to charge –2q at centre O
r
1
2q é 3iˆ – ˆj ù
E2 =
´ 2ê
ú
4pe 0 d ë 2 û
Electric field due to charge –4q at centre O
r
1
4q é 3iˆ + ˆj ù
´ 2ê
E3 =
ú
4pe0 d ë 2 û
\ Net electric field at point O
3
4 æRö
4
k ´ r´ pR 3 k ´ r´ p ç ÷
r
3
è2ø
3
EB =
2
2
R
æ 3R ö
ç
÷
è 2 ø
3
r
sR æ 1 ö ( s ) 4p æ R ö
Þ EB =
-ç
÷
ç
÷
3e 0 è 4pe0 ø æ 3R ö2 3 è 2 ø
ç
÷
è 2 ø
r
sR
sR
Þ EB =
3e0 54e 0
Þ EB =
23. (b)
17 æ sR ö
ç
÷
54 è e0 ø
E A 1´ 54 æ 9 ö 9 2 18
=
=ç ÷= ´ =
EB 6 ´17 è 17 ø 17 2 34
r r
20. (c) Since r × p = 0
r
r
E must be antiparallel to p
(
\ Ê is parallel to iˆ + 3 ˆj - 2kˆ
3 mv2
3
0 + qE0 2a = mv 2 Þ E0 =
4 qa
2
(B) Rate of work done at P = power of electric force
3
3 mv
4 a
dw
= 0 for both the fields
(C) At, Q,
dt
(D) The difference of magnitude of angular momentum
of the particle at P and Q,
r
DL = - m2v 2akˆ - -mvakˆ
r
DL = 3mva
22. (a)
) (
)
y-axis
–4q
A
+ 2q
B
150°
O
d
30°
30°
x-axis
C
–2q
v
x
Using
v2 – u2 = 2aS ...(i)
Here, u = 0, s = x
Also, Felectric = ma
Þ qE = ma Þ a =
1
2 1
2
Wmg + Wele = m ( 2v ) - m ( v )
2
2
(
3q ˆ
i
pe 0 d 2
qE
qE
Þ a=
m
m
Substituting the values in (i) we get
)
21. (c) (A) By work energy theorem
= qE0V =
r
r r
r
E0 = E1 + E2 + E3 =
v2 =
24. (d)
2qE
.x
m
E1
60°
P
E2
+s
30°
+s
y
x
From figure,
r
r
s
s
E1 =
yˆ
(– cos 60° xˆ – sin 60° yˆ )
and E2 =
2e 0
2e0
s æ 1
3 ö
=
yˆ ÷
çç – xˆ –
2e 0 è 2
2 ÷ø
Electric field in the region shown in figure (P)
r
r r
æ
s é 1
3ö ù
EP = E1 + E2 =
ê – xˆ + çç 1–
÷ yˆ ú
2e0 ëê 2
2 ÷ø ûú
è
r
s éæ
3ö
xˆ ù
or, E P =
êçç1 –
÷÷ yˆ – ú
2e 0 êëè
2 ø
2 úû
25. (c) In the x direction
Fx = qE
Þ max = qE
E q
Þ ax = 0
m
For speed to be double,
P-247
Electric Charges and Fields
29. (b) Electric field on the axis of a ring of radius R at a
distance h from the centre,
v02 + v x2 = (2v0 )2
Þ vx = 3 v0 = ax t
E=
3v0 m
qE0t
Þ t=
m
E0 q
26. (d) Time period of the pendulum (T) is given by
Þ 3v0 = 0 +
T = 2p
L
geff
geff =
(mg )2 + (qE )2
m
æ gE ö
Þ geff = g 2 + ç
÷
è m ø
®
®
®
(h
kQh
2
+ R2
)
3/2
Condition: for maximum electric field
dE
=0
dh
é
ù
d ê
kQh
ú
Þ
ú =0
dh ê 2
2 3/2
ê R +h
ú
ë
û
By using the concept of maxima and minima we get,
(
2
®
Þ T = 2p
L
æ qE ö
g2 + ç
÷
è m ø
2
h=
)
R
2
30. (a)
®
27. (d) E = ( E + E ) + ( E + E )
1
2
3
4
or E = 2E cos a – 2E cos b
y=3
–q
d
q
d
E3
a
D
d
E1
q
d
–q
2kq
D
2 kq
D
´
´
2
2
= ( D2 + d 2 )
2
2
2
( D + (2 d )
D +d
D + (2d )2
=
2 kqD
( D 2 + d 2 ) 3/2
For d < < D
Eµ
D
3
µ
-
2kqD
[ D 2 + (2d ) 2 ]3/2
1
D
D2
28. (d) At equilibrium resultant force on bob must be zero, so
T cos q = mg
..... (i)
T sin q = qE
..... (ii) Y
Solving (i) and (ii) we get
qE
tan q =
q
mg
q
T
qE
X
5 ´10-6 ´ 2000 1
tan q =
=
q
2
2 ´ 10-3 ´ 10
mg
[Here, q = 5 × 10–6 C,
E = 2000 v/m, m = 2 × 10–3 kg]
æ1ö
Þ tan -1 ç ÷
è2ø
x
(0, 0)
r
r
Let E1 and E2 are the vaues of electric field due to charge,
q1 and q2 respectively
magnitude of E = 1 q1
1
4p Î0 r12
=
1
10 ´ 10 -6
4p Î0 12 + 32
(
)
10
3
q1
= ( 9 ´ 109 ) ´ 10 ´ 10-7
= 9 10 ´ 10 2
r
r
r
\ E1 = 9 10 ´ 10 2 éëcos q1 ( - i ) + sin q1 j ùû
3 ˆù
é 1
Þ E1 = 9 ´ 10 ´ 10 2 ê
-iˆ +
j
10 úû
ë 10
Þ E1 = 9 ´102 éë -iˆ + 3 ˆj ùû = éë –9iˆ + 27 ˆj ùû102
1 q2
Similarly, E2 =
4 p Î0 r 2
( )
E2 =
9 ´109 ´ ( 25 ) ´ 10 -6
( 4r + 3 )
2
2
E2 = 9 × 103 V/m
(
)
\ E2 = 9 ´ 103 cos q2iˆ - sin q2 ˆj Q tan q2 =
r
æ4 3 ö
\ E2 = 9 ´103 ç iˆ - ˆj ÷ = 72iˆ - 54 ˆj ´102
è5 5 ø
r r
r
\ E = E1 + E 2 = 63iˆ - 27 ˆj ´ 10 2 V/m
(
(
)
)
3
4
P-248
31.
Physics
(a) Equilibrium position will shift to point where resultant
force = 0
kxeq = qE Þ xeq =
qE
k
35. (a) Electric field due to complete disc (R = 2a) at a
distance x and on its axis
1
2 2 1
2
Total energy = mw A + kx eq
2
2
=
2 2
1
1q E
Total energy = mw 2 A 2 +
2
2 k
32.
s é
ê1 –
2e 0 ê
ë
E1 =
ù
h
s é
ù
ú
ú E1 = 2e ê1 –
2
2 ú
0 ëê
2
2 ú
4
a
+
h
û
R +x û
x
s é hù
1–
2e0 êë 2a úû
é here x =h ù
êë and, R = 2a úû
2a
a
rö
æ
(d) Charge density, r = r0 ç1 - ÷
è Rø
dq = rdv
o
ò
qin = dq = rdv
Similarly, electric field due to disc (R = a)
rö
æ
= r0 ç 1 - ÷ 4pr 2 dr (Q dv = 4pr2dr)
è Rø
= 4pr0
= 4pr0
Ræ
rö
ò0 çè1 - R ÷ø r 2dr
R
ò0
r 2 dr -
s æ hö
ç 1– ÷
2e0 è a ø
Electric field due to given disc
E = E1 – E2
E2 =
s
2e 0
r2
dr
R
éé 3 ù R é 4 ù R ù
é R3 R 4 ù
r
r
= 4pr0 ê ê ú - ê ú ú = 4pr0 ê
ú
êê 3 ú
ê 4R ú ú
êë 3 4 R úû
ëë û0 ë û0 û
hù s
é
ê1– 2a ú – 2e
ë
û
0
é hù
sh
ê1– a ú =
ë
û
4e0 a
s
4ae 0
36. (b) Let us consider a spherical shell of radius x and
thickness dx.
Hence, c =
é R3 R 3 ù
é R3 ù
= 4pr0 ê
- ú = 4pr0 ê ú
4 úû
ëê 3
ëê 12 ûú
pr R3
q= 0
3
O
dx
x
Shell
æ pr R3 ö
E.4pr 2 = ç 0 ÷
è 3Î0 ø
\
Electric field outside the ball, E =
r0 R3
12 Î0 r 2
33. (c) Field lines originate perpendicular from positive
charge and terminate perpendicular at negative charge.
Further this system can be treated as an electric dipole.
34. (d) Given: Length of wire L = 20 cm
charge Q = 103e0
We know, electric field at the centre of the semicircular arc
2K l
E=
r
or,
=
æ 2Q ö
2K ç
÷
è pr ø é Asl = 2Q ù
E=
ê
r
pr úû
ë
4 KQ
pr
2
=
4 KQp
2
pL
2
=
4pKQ
2
L
= 25 ´ 103 N / Ci$
Charge on this shell
xö
æ
dq = r.4px2dx = r0 ç1 - ÷ .4px 2dx
è Rø
\ Total charge in the spherical region from centre
to r (r < R) is
r
xö
æ
q = ò dq = 4pr0 ò ç1 - ÷ x 2dx
è Rø
0
r
é r3 r 4 ù
é x3 x 4 ù
r ù
3 é1
= 4pr0 ê ú = 4pr0 ê - ú = 4pr0r ê - ú
ë 3 4R û
êë 3 4R úû
êë 3 4R úû 0
1 .q
\ Electric field at r, E =
4pe 0 r 2
=
1 4 pr0 r3 é 1 r ù r0 é r r 2 ù
.
ê ú
ê ú =
4pe 0
r 2 ë 3 4R û e0 êë 3 4R úû
P-249
Electric Charges and Fields
37. (c) Given,
Electric field E = 150 N/C
Total surface charge carried by earth q = ?
or, q = Î0 E A
dx
x
= Î0 E p r2.
= 8.85 × 10–12 × 150 × (6.37 × 106)2.
; 680 Kc
As electric field directed inward hence
q = – 680 Kc
38. (a) Electric field intensity at the centre of the disc.
E=
s
(given)
2 Î0
\
Electric field along the axis at any distance x from the
centre of the disc
æ
ö
x
ç1 ÷
ç
2
2 ÷
+
x
R
è
ø
From question, x = R (radius of disc)
E' =
s
2 Î0
æ
R
\ E ' = s ç1 2
2 Î0 çè
R + R2
=
ö
÷
÷
ø
é 5 r3 1 r 4 ù
3æ5 r ö
= 4pr0 ê . - . ú = pr0r ç - ÷
è3 Rø
êë 4 3 R 4 úû
Electric field at r,
1
q
E=
. 2
4p Î0 r
1 pr0 r 3 æ 5 r ö r0 r æ 5 r ö
.
ç - ÷
=
ç - ÷=
4pÎ0 r 2 è 3 R ø 4 Î0 è 3 R ø
41. (c) Electric lines of force due to a positive charge is
spherically symmetric.
All the charges are positive and equal in magnitude. So
repulsion takes place. Due to which no lines of force are
present inside the equilateral triangle and the resulting
lines of force obtained as shown:
s æ 2R - R ö
ç
÷
2 Î0 çè
2R ÷ø
4
E
14
\ % reduction in the value of electric field
4 ö
æ
çè E - E÷ø ´ 100
1000
14
=
=
% ; 70.7%
E
14
39. (b) F = qE = mg (q = 6e = 6× 1.6 × 10–19)
+q
=
mass
m
=
volume 4 3
pr
3
+
+
+
+
+ dE
+
r
æ5 xö
q = ò dq = 4 pr0 ò ç - ÷ x 2 dx
è4 Rø
0
j
+
m
4
pd
3
æ qE ö
Putting the value of d and m ç =
and solving we get r
è g ÷ø
= 7.8 × 10–7 m
40. (a) Let us consider a spherical shell of radius x and
thickness dx.
Charge on this shell
æ5 x ö
2
dq = r.4px 2 dx = r0 ç 4 - R ÷ .4px dx
è
ø
\ Total charge in the spherical region from centre to r (r < R ) is
or r3 =
42. (c) Let us consider a differential element dl subtending at angle
dQ at the centre Q as shown in the figure. Linear charge density
q
l=
Qr
dq
cos q
q
dl
+
+
Density (d) =
+q
+q
O
i
dE
dE
sin q
æ q ö
Charge on the element, dq = ç ÷ dl
è pr ø
q
(rd q)
=
pr
æqö
= ç ÷dq
è pø
(Q dl = rdq)
P-250
Physics
=
4p2 r 2 Î0
q
q
ò 4 p2 r 2 Î
0
sin qd q
The charge enclosed in a sphere of radius r1 can be
calculated by
0
p
0
[ - cos q]
4Q
Q = ò dq =
q
(+1 + 1) =
4p2 r 2 Î0
2p2 r 2 Î0
The direction of E is towards negative y-axis.
r
q
ˆj
\ E=2 2
2p r Î0
43. (a) Let us consider a spherical shell of radius x and
thickness dx.
Due to shpherically symmetric charge distribution, the
chrge on the spherical surface of radius x is
æ5 xö
2
dq = dVr×4px2dx = r0 ç - ÷ × 4px dx
4
R
è
ø
\ Total charge in the spherical region from centre to r (r < R) is
r
ò
0
0
é Q 4ù
r
1 êë R 4 1 úû
1 Q 2
Þ E=
=
r1
4p Î0
4p Î0 R 4
r12
46. (a) The electric field inside a thin spherical shell of radius
R has charge Q spread uniformly over its surface is zero.
Q ++
+
R
E=0
+
dx
3
\ The electric field at point P inside the sphere at a distance
r1 from the centre of the sphere is
1 Q
E=
4pE r12
E=k
Q
2
r
+ + +
Q
Outside the shell the electric field is E = k 2 . These
r
characteristics are represented by graph (a).
+Q2
+Q
++
x
r1
4Q é x 4 ù
Q 4
x dx =
ê ú = 4 r1
4 4
R
R êë úû
+++
++
+
æ5 xö
q = ò dq = 4 pr0 ò ç - ÷ x 2 dx
è4 Rø
0
R4
r1
+
+ +
+ + +
=
q
p
é Q.x ù
4Q
[4px2dx] = 4 x3dx
4ú
R
ë pR û
dq = ê
+ +
\ E = ò dE sin q =
Let us consider a spherical shell of thickness dx and radius
x. The area of this spherical shell = 4px2.
The volume of this spherical shell = 4px2dx. The charge
enclosed within shell
+
Electric field at the center O due to dq is
1
q
1
dq
× 2 dq
dE =
× 2 =
4
p
Î
4 p Î0 r
0 pr
Resolving dE into two rectangular component, we find
the component dE cos q will be counter balanced by
another element on left portion. Hence resultant field at O
is the resultant of the component dE sin q only.
1
47. (c)
é 5 r3 1 r4 ù
3æ5 r ö
= 4 pr0 ê × - × ú = pr0r ç - ÷
è3 Rø
ëê 4 3 R 4 ûú
\ Electric field intensity at a point on this spherical surface
1
q
E=
× 2
4 p Î0 r
44.
1
pr0 r 3 æ 5 r ö r0 r æ 5 r ö
×
=
ç - ÷=
4p Î0 r 2 è 3 R ø 4 Î0 çè 3 R ÷ø
(a)
45.
(b)
R
x
dx
r1
r2
A
B
When the two conducting spheres are connected by a
conducting wire, charge will flow from one to other till both
acquire same potential.
\ After connection, V1 = V2
Q
Q
Q Q
Þk 1 =k 2 Þ 1 = 2
r1
r2
r1
r2
The ratio of electric fields
Q
k 21
E1
r
E
Q r2
= 1 Þ 1 = 21 ´ 2
E2 k Q2
E2 r1 Q2
r22
Þ
E1 r1 ´ r22
E
r
2
=
Þ 1 = 2 =
E2 r12 ´ r2
E2 r1 1
P-251
Electric Charges and Fields
48.
(b) At P
- K 2q
( x - L)2
1
Þ
+
K 8q
x2
52. (b) Let v be the speed of dipole.
Using energy conservation
=0
Ki + U i = K f + U f
4
=
( x - L) 2 x 2
1
2
=
or
x-L x
Þ x = 2x – 2L or x = 2L
+8q
x=0
Þ 0-
P
x=L
Þ mv 2 =
x
49.
(c)
T cos q
v=
T
F = Eq =
T sin q
s
q
e0K
2 E1 + E2 = E3
a
2
+
kQ
( 2 a)
2
E3
E1
E2
2 E1
51.
=
kq
æ a ö
çè
÷
2ø
Þv=
2kp1 p2
mr 3
p
1
a 2p Î0 ma
–Q
A –Q
\ f1 = ò E. A cos 90° = 0
r uur
For surface BCGF fn = ò E . dA
\ f11 = é 4 ´ iˆ – ( y 2 + 1) ˆj ù .4iˆ = 16 x
ë
û
Nm 2
C
f1 – f11= – 48
54. (a)
55. (b) Surface charge density depends only due to Q. Also
f11 = 48
®
Ñò E . d A =
–Q B
A
E
Here, q = angle between electric field and area vector of a
surface
For surface ABCD Angle, q = 90°
®
2
–Q
Q
Q 2 Q
Þ
+ = 2q Þ q = (2 2 + 1)
4
1
2
(c) Given, Electric field, E = 3 × 104
Mass of the drop, m = 9.9 × 10–15 kg
At equilibrium, coulomb force on drop balances weight of drop.
qE = mg
Þ q=
r
3
f = ò E. A cos q
T sin q = qE
.... (i)
T cos q = mg
.... (ii)
Dividing (i) by (ii),
qE
q æ s ö sq
tan q =
=
mg mg çè e0K ÷ø e0 K . mg
\ s µ tan q
50. (b) For the system to be equilibrium, net field at A should
be zero
kQ ´ 2
2kp1 p2
53. (–48)
r
r
Flux of electric field E through any area A is defined as
mg
\
1 2 1 2
mv + mv + 0
2
2
p2 cos (180°) =
When p1 = p2 = p and r = a
P
q
r3
æ
ç
çQ Potential energy of interaction between dipole
ç
è
-2 p1 p2 cos q ö
=
÷
4p Î0 r 3 ÷ø
–2q
L
2k × p1
9.9 ´ 10-15 ´ 10
mg
q
=
= 3.3 ´ 10 -18 C
Þ
4
E
3 ´ 10
q1l
e0
or E × 4pr2 =
56. (a)
®
®
Q
1 Q
ÞE=
,r³R
4pe 0 r 2
e0
qin
Ñò E . d A = e 0
2Q
–Q
–Q
a
a
P-252
Physics
or E × 4pr2 =
1
S (4pr 2 )dr
ò
e0
or E × 4pr2 =
1
(kr )(4 pr 2 )dr
e0 ò
\ At point P, = +
r
At Point P1, F1 = +
0
or
E × 4pr2 =
\ E=
k 2
r
4e 0
...(i)
R
r4
2
Also 2Q = ò (kr ) (4pr )dr = 4pk
4
0
R
0
4
pkR
2
From above equations,
E=
....(ii)
Qr 2
....(iii)
2pe0 R 4
According to given condition
= EQ
Q4
....(iv)
4pe0 (20) 2
From equations (iii) and (iv), we have
a = 8–1/4 R.
57. (b) t = – PE sin q
or Ia = – PE (q)
PE
a=
( -q )
I
On comparing with
a = – w2q
w=
58.
59.
PE
=
I
qdE
ædö
2m ç ÷
è2ø
2
=
KPQ
( y / 3)3
= 27 F.
60. (b) When cube is of side a and point charge Q is at the
center of the cube then the total electric flux due to this
charge will pass evenly through the six faces of the cube.
So, the electric flux through one face will be equal to 1/6
of the total electric flux due to this charge.
4pk æ r 4 ö
ç ÷
e0 çè 4 ÷ø
Q=
KP
Q
y3
2qE
md
(b) Potential energy of a dipole is given by
rr
U = – P.E
= – PE cos q
[Whereq = angle between dipole and perpendicular to the
field]
= – (10–29) (103) cos 45°
= – 0.707 × 10–26 J = – 7 × 10–27J
(d) Electric field of equitorial plane of dipole
r
KP
=– 3
r
Flux through 6 faces =
Q
Îo
Q
6 Îo
61. (a) T = PE sin q Torque experienced by the dipole in an
r r r
electric field, T = P ´ E
r
p = p cosq iˆ + p sin q ĵ
r
r
E1 = Ei
r r r
T1 = p ´ E1 = (p cos q iˆ + p sin q ĵ ) × E( iˆ )
\
Flux through 1 face, =
t k̂ = pE sinq (– k̂ )
...(i)
r
ˆ
E2 = 3 E1 j
r
T2 = p cos qiˆ + p sin qˆj ) ´ 3 E1 ˆj
...(ii)
tkˆ = 3 pE1 cos qkˆ
From eqns. (i) and (ii)
pE sinq = 3 pE cosq
tanq = 3 \ q = 60°
62. (c) The net flux linked with closed surfaces S1, S2, S3 & S4
are
1
For surface S1, f1 = (2q)
e0
1
1
2q
For surface S2, f 2 = (q + q + q - q) =
e0
e0
1
1
For surface S3, f3 = (q + q) = (2q)
e0
e0
1
1
For surface S4, f 4 = (8q - 2q - 4q) = (2q)
e0
e0
Hence, f1 = f2 = f3 = f4 i.e. net electric flux is same for all
surfaces.
Keep in mind, the electric field due to a charge outside (S3
and S4), the Gaussian surface contributes zero net flux
through the surface, because as many lines due to that
charge enter the surface as leave it.
63. (c) Applying Gauss’s law
r uur Q
Ñò S E × ds = Î0
\ E × 4pr2 =
Q + 2pAr 2 - 2pAa 2
Î0
P-253
Electric Charges and Fields
r=
66. (b) We know that,
dr
dV
r dr
Q a
Q = r4pr2
r
Q=
ò
a
E=
Gaussiam
surface
b
A
4pr 2dr = 2pA[r2 – a2]
r
f=
Ñò E.dS = E Ñò dS
In case of hemisphere
fcurved = fcircular
Therefore, fcurved = E pa 2 .
ù
1 é Q - 2pAa 2
+ 2pA ú
ê
2
4p Î0 ëê
r
ûú
67. (c)
(a)
F1
E1
F2
–q
®
E = E 0 iˆ + 2E 0 ˆj
Given, E 0 = 100N / c
®
So, E = 100iˆ + 200ˆj
Radius of circular surface = 0.02 m
2
Area = pr =
22
´ 0.02 ´ 0.02
7
= 1.25 ´10-3 ˆi m 2 [Loop is parallel to Y-Z plane]
Now, flux (f) = EA cosq
-3
= 100iˆ + 200ˆj .1.25 ´ 10 ˆi cos q° [q = 0°]
(
65.
)
= 125 × 10–3 Nm2/c
= 0.125 Nm2/c
(c) Force of interaction
E2
As the dipole is placed in non-uniform field, so the force
acting on the dipole will not cancel each other. This will
result in a force as well as torque.
68. (a) The electric flux f1 entering an enclosed surface is
taken as negative and the electric flux f2 leaving the surface
is taken as positive, by convention. Therefore the net flux
leaving the enclosed surface, f = f2 – f1
According to Gauss theorem
q
f=
Þ q = Î0f = Î0(f2 – f1)
Î0
69. (None) Electric flux due to charge placed outside is zero.
But for the charge inside the cube, flux due to each face is
1é q ù
ê ú which is not in option.
6 ëÎ0 û
C
1 6p1p2 +q
F=
.
4p Î0 r 4
D
+q
p1
–q
E pa 2
1
=
2
2
+q
For E to be independent of ‘r’
Q – 2pAa2 = 0
Q
\ A=
2pa 2
64.
cos 45°
p2
r
–q
.
q
.
q
B
A
16
P-254
Physics
Electrostatic Potential
and Capacitance
4.
Electrostatic Potential and
TOPIC 1
Equipotential Surfaces
1.
Ten charges are placed on the circumference of a circle of
radius R with constant angular separation between
successive charges. Alternate charges 1, 3, 5, 7, 9 have
charge (+q) each, while 2, 4, 6, 8, 10 have charge (–q)
each. The potential V and the electric field E at the centre
of the circle are respectively :
(Take V = 0 at infinity)
[Sep. 05, 2020 (II)]
(a) V =
r
R
(a)
5.
10 q
4pe 0 R 2
(c) V = 0; E = 0
(d) V =
2.
10 q
10 q
;E=
4 pe0 R
4 pe 0 R 2
Two isolated conducting spheres S1 and S2 of radius
2
R
3
1
R have 12 mC and –3 mC charges, respectively, and
3
are at a large distance from each other. They are now
connected by a conducting wire. A long time after this is
done the charges on S1 and S2 are respectively :
and
3.
1
(R + r )
Q
4pe 0 2( R 2 + r 2 )
(b)
1 (2 R + r )
Q
4 pe0 ( R 2 + r 2 )
1
(R + r )
1 ( R + 2 r )Q
Q
(d)
2
2
4pe0 ( R 2 + r 2 )
4pe 0 2( R + r )
ur
A point dipole = p – po $x kept at the origin. The potential
and electric field due to this dipole on the y-axis at a
distance d are, respectively : (Take V = 0 at infinity)
[12 April 2019 I]
ur
ur
ur
p
-p
p
,
(a)
(b) 0,
4pe 0 d 3
4pe 0 d 2 4pe 0 d 3
ur
ur
ur
p
p
-p
,
(c) 0,
(d)
4pe 0 d 3
4pe 0 d 2 4pe 0 d 3
A uniformly charged ring of radius 3a and total charge q
is placed in xy-plane centred at origin. A point charge q is
moving towards the ring along the z-axis and has speed v
at z = 4a. The minimum value of v such that it crosses the
origin is :
[10 April 2019 I]
(c)
10q
;E=0
4pe 0 R
(b) V = 0; E =
A charge Q is distributed over two concentric conducting
thin spherical shells radii r and R (R > r). If the surface
charge densities on the two shells are equal, the electric
potential at the common centre is : [Sep. 02, 2020 (II)]
[Sep. 03, 2020 (I)]
(a) 4.5 mC on both
(b) +4.5 mC and –4.5 mC
(c) 3 mC and 6 mC
(d) 6 mC and 3 mC
Concentric metallic hollow spheres of radii R and 4R hold
charges Q1 and Q2 respectively. Given that surface charge
densities of the concentric spheres are equal, the potential
difference V(R) – V(4R) is :
[Sep. 03, 2020 (II)]
(a)
3Q1
16pe 0 R
(b)
3Q2
4pe0 R
(c)
Q2
4pe0 R
(d)
3Q1
4pe0 R
6.
1/2
(a)
2 æ 4 q2 ö
ç
÷
m è 15 4pe0a ø
1/2
7.
1/2
(b)
2 æ 1 q2 ö
ç
÷
m è 5 4pe0a ø
1/2
2 æ 2 q2 ö
2 æ 1 q2 ö
(c)
(d)
ç
÷
ç
÷
m è 15 4pe0a ø
m è 15 4pe0a ø
A solid conducting sphere, having a charge Q, is
surrounded by an uncharged conducting hollow spherical
shell. Let the potential difference between the surface of
the solid sphere and that of the outer surface of the hollow
shell be V. If the shell is now given a charge of – 4 Q, the
new potential difference between the same two surfaces
is :
[8 April 2019 I]
(a) – 2V
(b) 2 V
(c) 4 V
(d)
V
P-255
Electrostatic Potential and Capacitance
8.
9.
r
The electric field in a region is given by E = ( Ax + B ) iˆ ,
where E is in NC–1 and x is in metres. The values of
constants are A = 20 SI unit and B = 10 SI unit. If the
potential at x = 1 is V1 and that at x = –5 is V2, then
V1 – V2 is :
[8 Jan. 2019 II]
(a) 320 V
(b) – 48V (c) 180 V
(d) – 520 V
The given graph shows variation (with distance r from centre )
of :
[11 Jan. 2019 I]
rO
r
rO
(a) Electric field of a uniformly charged sphere
(b) Potential of a uniformly charged spherical shell
(c) Potential of a uniformly charged sphere
(d) Electric field of a uniformly charged spherical shell
10. A charge Q is distributed over three concentric spherical
shells of radii a, b, c (a < b < c) such that their surface
charge densities are equal to one another.
The total potential at a point at distance r from their
common centre, where r < a, would be:
[10 Jan. 2019 I]
Q ab + bc + ca
(a) 12pÎ
abc
0
(b)
Q (a 2 + b 2 + c 2 )
4pÎ0 (a 3 + b 3 + c3 )
Q (a + b + c)
Q
(c) 4pÎ (a + b + c)
(d) 4pÎ (a 2 + b 2 + c2 )
0
0
11. Two electric dipoles, A, B with respective dipole
r
r
moments d A = – 4 qa iˆ and d B = – 2 qa iˆ are placed on
the x-axis with a separation R, as shown in the figure
The distance from A at which both of them produce the
same potential is:
[10 Jan. 2019 I]
R
2R
(a) 2 + 1
(b)
2 +1
(c)
R
(d)
2R
2 -1
2 -1
12. Consider two charged metallic spheres S1 and S2 of radii
R1 and R2, respectively. The electric fields E1 (on S1) and
E2 (on S2) on their surfaces are such that E1/E2 = R1/R2.
Then the ratio V1(on S1)/V2(on S2) of the electrostatic
potentials on each sphere is:
[8 Jan. 2019 II]
(a) R1/R 2
(b) (R1/R2)2
3
æR ö
(c) (R2/R1)
(d) ç 1 ÷
è R2 ø
13. Three concentric metal shells A, B and C of respective
radii a, b and c (a < b < c) have surface charge densities
+s, –s and +s respectively. The potential of shell B is:
[2018]
s é a2 - b2 ù
(a) Î ê a +c ú
0 êë
ûú
s é a 2 - b2 ù
(b) Î ê b +cú
0 êë
ûú
s é b2 - c2 ù
(c) Î ê b +a ú
0 ëê
ûú
s é b 2 - c2 ù
(d) Î ê c +a ú
0 ëê
ûú
14. There is a uniform electrostatic field in a region. The
potential at various points on a small sphere centred at P,
in the region, is found to vary between in the limits 589.0 V
to 589.8 V. What is the potential at a point on the sphere
whose radius vector makes an angle of 60° with the direction
of the field ?
[Online April 8, 2017]
(a) 589.5 V (b) 589.2 V (c) 589.4 V (d) 589.6 V
15. Within a spherical charge distribution of charge density
r(r), N equipotential surfaces of potential V0, V0 + DV, V0
+ 2DV, .........V0 + NDV (DV > 0), are drawn and have
increasing radii r0, r1, r2,......... rN, respectively. If the
difference in the radii of the surfaces is constant for all
values of V0 and DV then :
[Online April 10, 2016]
1
(a) r(r) = constant
(b) r(r) µ 2
r
1
(c) r(r) µ
(d) r(r) µ r
r
16. The potential (in volts) of a charge distribution is given by
V(z) = 30 – 5z2 for |z| £ 1m
V(z) = 35 – 10 |z| for |z| ³ 1 m.
V(z) does not depend on x and y. If this potential is
generated by a constant charge per unit volume r0 (in
units of e0) which is spread over a certain region, then
choose the correct statement.
[Online April 9, 2016]
(a) r0 = 20 e0 in the entire region
(b) r0 = 10 e0 for |z| £ 1 m and p0 = 0 elsewhere
(c) r0 = 20 e0 for |z| £ 1 m and p0 = 0 elsewhere
(d) r0 = 40 e0 in the entire region
17. A uniformly charged solid sphere of radius R has potential
V0 (measured with respect to ¥) on its surface. For this
sphere the equipotential surfaces with potentials
3V0 5V0 3V0
V0
,
,
and
have radius R1, R2, R3 and R4
2
4
4
4
respectively. Then
[2015]
(a) R1 = 0 and R2 < (R4 – R3)
(b) 2R = R4
(c) R1 = 0 and R2 > (R4 – R3)
(d) R1 ¹ 0 and (R2 – R1) > (R4 – R3)
r
-1 exists in a region of
$
18. An electric field E = (25i$ + 30j)NC
space. If the potential at the origin is taken to be zero
then the potential at x = 2 m, y = 2 m is :
[Online April 11, 2015]
(a) –110 J (b) –140 J (c) –120 J (d) –130 J
r
19. Assume that an electric field E = 30x 2 ˆi exists in space.
Then the potential difference VA - VO , where VO is the
potential at the origin and VA the potential at x = 2 m is:
(a) 120 J/C
(b) –120 J/C
[2014]
(c) –80 J/C
(d) 80 J/C
P-256
Physics
20. Consider a finite insulated, uncharged conductor placed
near a finite positively charged conductor. The uncharged
body must have a potential : [Online April 23, 2013]
(a) less than the charged conductor and more than at
infinity.
(b) more than the charged conductor and less than at
infinity.
(c) more than the charged conductor and more than at
infinity.
(d) less than the charged conductor and less than at
infinity.
21. Two small equal point charges of magnitude q are
suspended from a common point on the ceiling by
insulating mass less strings of equal lengths. They come
to equilibrium with each string making angle q from the
vertical. If the mass of each charge is m, then the
electrostatic potential at the centre of line joining them will
æ 1
ö
be ç
= k÷.
è 4p Î0
ø
[Online April 22, 2013]
(a) 2 k mg tan q
(b)
k mg tan q
(c) 4 k mg / tan q
(d)
k mg / tan q
(
)
1 ( R + r) Q
(c) 4pe
0 R2 + r 2
(
)
( R + r) Q
1
(b)
4pe 0 2 R 2 + r 2
(
( R - r) Q
1
(d)
4pe 0 2 R 2 + r 2
(
q
A
q
B
D
-q
C
-q
ur
E
ur changes, V remains unchanged
unchanged, V changes
E remains
ur
both E and V change
ur
(d) E and V remain unchanged
28. The potential at a point x (measured in m m) due to some
charges situated on the x-axis is given by V(x) = 20/(x2 – 4)
volt. The electric field E at x = 4 m m is given by [2007]
(a) (10/9) volt/ m m and in the +ve x direction
(b) (5/3) volt/ m m and in the –ve x direction
(c) (5/3) volt/ m m and in the +ve x direction
(d) (10/9) volt/ m m and in the –ve x direction
29. Two thin wire rings each having a radius R are placed at a
distance d apart with their axes coinciding. The charges
on the two rings are +q and -q. The potential difference
between the centres of the two rings is
[2005]
(a)
(b)
(c)
22. A point charge of magnitude + 1 mC is fixed at (0, 0, 0). An
isolated uncharged spherical conductor, is fixed with its
center at (4, 0, 0). The potential and the induced electric
field at the centre of the sphere is :[Online April 22, 2013]
(a) 1.8 × 105 V and – 5.625 × 106 V/m
(b) 0 V and 0 V/m
(c) 2.25 × 105 V and – 5.625 × 106 V/m
(d) 2.25 × 105 V and 0 V/m
23. A charge of total amount Q is distributed over two
concentric hollow spheres of radii r and R (R > r) such that
the surface charge densities on the two spheres are equal.
The electric potential at the common centre is
[Online May 19, 2012]
1 ( R - r) Q
(a)
4pe 0 R2 + r 2
26. An electric charge 10–3 m C is placed at the origin (0, 0) of
X – Y co-ordinate system. Two points A and B are situated
at ( 2, 2) and (2, 0) respectively. The potential
difference between the points A and B will be
[2007]
(a) 4.5 volts
(b) 9 volts
(c) Zero
(d) 2 volt
27. Charges are placed on the vertices of a square as shown.
r
Let E be the electric field and V the potential at the
centre. If the charges on A and B are interchanged with
those on D and C respectively, then
[2007]
24. The electric potential V(x) in a region around the origin is
given by V(x) = 4x2 volts. The electric charge enclosed in
a cube of 1 m side with its centre at the origin is (in coulomb)
[Online May 7, 2012]
(a) 8e0
(b) – 4e0 (c) 0
(d) – 8e0
25. The electrostatic potential inside a charged spherical ball is
given by f = ar2 + b where r is the distance from the centre
and a, b are constants. Then the charge density inside the
ball is:
[2011]
(a) –6ae0r
(b) –24pae0
(c) –6ae0
(d) –24pae0r
ù
ú (b)
R 2 + d 2 úû
q
2 p Î0
(c)
q é1
1
ê 2
4 p Î0 ê R
R + d2
ë
)
)
é1
ê êë R
(a)
1
ù
ú (d)
úû
qR
4p Î0 d 2
zero
30. A thin spherical conducting shell of radius R has a charge
q. Another charge Q is placed at the centre of the shell.
R
2
[2003]
The electrostatic potential at a point P , a distance
from the centre of the shell is
2Q
(a) 4pe R
o
2Q
2q
(b) 4pe R - 4pe R
o
o
2Q
q
(c) 4pe R + 4pe R
o
o
(q + Q)2
(d) 4pe R
o
P-257
Electrostatic Potential and Capacitance
Electric Potential Energy and
TOPIC 2 Work Done in Carrying a
Charge
31. A solid sphere of radius R carries a charge Q + q distributed
uniformaly over its volume. A very small point like piece of
it of mass m gets detached from the bottom of the sphere
and falls down vertically under gravity. This piece carries
charge q. If it acquires a speed v when it has fallen through
a vertical height y (see figure), then : (assume the remaining
portion to be spherical).
[Sep. 05, 2020 (I)]
Q
R
33. Hydrogen ion and singly ionized helium atom are
accelerated, from rest, through the same potential
difference. The ratio of final speeds of hydrogen and
helium ions is close to :
[Sep. 03, 2020 (II)]
(a) 1 : 2
(b) 10 : 7
(c) 2 : 1
(d) 5 : 7
34. In free space, a particle A of charge 1 mC is held fixed at a
point P. Another particle B of the same charge and mass 4
mg is kept at a distance of 1 mm from P. If B is released,
then its velocity at a distance of 9 mm from P is :
é
ù
1
= 9 ´ 109 Nm 2C -2 ú
êTake
[10 April 2019 II]
4pe 0
ë
û
(a) 1.0m/s
(b) 3.0×104 m/s
3
(c) 2.0×10 m/s
(d) 1.5×102 m/s
35. A system of three charges are placed as shown in the
figure:
q
y
v
é
ù
qQ
(a) v 2 = y ê
+
g
ú
2
êë 4pe 0 R ym
úû
é
ù
qQ
+ gú
(b) v 2 = y ê
ë 4pe 0 R( R + y )m
û
é
ù
Qq R
+ gú
(c) v 2 = 2 y ê
3
êë 4pe0 ( R + y ) m
úû
é
ù
qQ
2
+ gú
(d) v = 2 y ê
4
(
)
pe
R
R
+
y
m
ë
0
û
32. A two point charges 4q and –q are fixed on the x-axis at
If D >> d, the potential energy of the system is best given
by
[9 April 2019 I]
(a)
1 é - q 2 - qQd ù
1 é - q 2 2qQd ù
+
ê
ú
ê
2 ú (b)
4p Î0 ë d 2 D û
4p Î0 ë d
D2 û
(c)
1 é q 2 qQd ù
1 é q 2 qQd ù
+ 2 ú (d)
- 2 ú
ê+
ê4p Î0 ë d
4p Î0 ë d
D û
D û
36. A positive point charge is released from rest at a distance
r0 from a positive line charge with uniform density. The
speed (v) of the point charge, as a function of
instantaneous distance r from line charge, is proportional
to :
[8 April 2019 II]
d
d
and x = , respectively. If a third point charge
ge
2
2
‘q’ is taken from the origin to x = d along the semicircle as
shown in the figure, the energy of the charge will :
[Sep. 04, 2020 (I)]
x=-
4q
(a) increase by
3q 2
4pe0 d
(b) increase by
2q 2
3pe0 d
(c) decrease by
q2
4pe0 d
(d) decrease by
4q 2
3pe0 d
–q
+r /r
(a) v µ e 0
(b) v µ
ærö
ln ç ÷
è r0 ø
ærö
ærö
(c) v µ ln ç ÷
(d) v µ ç ÷
r
è 0ø
è r0 ø
37. There is a uniform spherically symmetric surface charge
density at a distance Ro from the origin. The charge
distribution is initially at rest and starts expanding because
of mutual repulsion. The figure that represents best the
speed V (R(t)) of the distribution as a function of its
instantaneous radius R(t) is:
[12 Jan. 2019 I]
P-258
Physics
V(R(t))
V(R(t))
(a)
(b)
R (t)
Ro
Ro
R (t)
V(R(t))
V(R(t))
Vo
(c)
(d)
Ro
R (t)
Ro
R (t)
38. Three charges Q, + q and + q are placed at the vertices of
a right-angle isosceles triangle as shown below. The net
electrostatic energy of the configuration is zero, if the value
of Q is :
[11 Jan. 2019 I]
Q
+q
+q
- 2q
(b)
2 +1
(a) + q
-q
(d) –2q
(c)
1+ 2
39. Four equal point charges Q each are placed in the xy
plane at (0, 2), (4, 2), (4, – 2) and (0, – 2). The work
required to put a fifth charge Q at the origin of the
coordinate system will be:
[10 Jan. 2019 II]
(a)
Q2
4 pe 0
(c)
Q2
2 2 pe0
1 ö
æ
ç1 +
÷
3ø
è
(b)
Q2
4 pe 0
(d)
Q2
4pe 0
1 ö
æ
ç1 +
÷
5ø
è
40. Statement 1 : No work is required to be done to move a
test charge between any two points on an equipotential
surface.
Statement 2 : Electric lines of force at the equipotential
surfaces are mutually perpendicular to each other.
[Online April 25, 2013]
(a) Statement 1 is true, Statement 2 is true, Statement 2 is
the correct explanation of Statement 1.
(b) Statement 1 is true, Statement 2 is true, Statement 2 is
not the correct explanation of Statement 1.
(c) Statement 1 is true, Statement 2 is false.
(d) Statement 1 is false, Statement 2 is true.
41. An insulating solid sphere of radius R has a uniformly
positive charge density r. As a result of this uniform charge
distribution there is a finite value of electric potential at
the centre of the sphere, at the surface of the sphere and
also at a point outside the sphere. The electric potential
at infinite is zero.
[2012]
Statement -1 When a charge q is taken from the centre
to the surface of the sphere its potential energy changes
qr
.
by
3e0
Statement -2 The electric field at a distance r (r <R) from
rr
the centre of the sphere is
.
3e0
(a) Statement 1 is true, Statement 2 is true; Statement 2 is
not the correct explanation of statement 1.
(b) Statement 1 is true Statement 2 is false.
(c) Statement 1 is false Statement 2 is true.
(d) Statement 1 is true, Statement 2 is true, Statement 2 is
the correct explanation of Statement 1
42. Two positive charges of magnitude ‘q’ are placed, at the
ends of a side (side 1) of a square of side ‘2a’. Two negative
charges of the same magnitude are kept at the other corners.
Starting from rest, if a charge Q moves from the middle of
side 1 to the centre of square, its kinetic energy at the
centre of square is
[2011 RS]
(a) zero
(c)
1 2qQ æ
2 ö
1÷
4pe 0 a çè
5ø
(b)
1 2qQ æ
1 ö
1+
÷
4pe 0 a çè
5ø
(d)
1 2qQ æ
1 ö
1÷
4pe 0 a çè
5ø
43. Two points P and Q are maintained at the potentials of 10
V and – 4 V, respectively. The work done in moving 100
electrons from P to Q is:
[2009]
(a) 9.60 × 10–17J
(b) –2.24 × 10–16 J
(c) 2.24 × 10–16 J
(d) –9.60× 10–17 J
44. Two insulating plates are both uniformly charged in such
a way that the potential difference between them is V2 –
V1 = 20 V. (i.e., plate 2 is at a higher potential). The plates
are separated by d = 0.1 m and can be treated as infinitely
large. An electron is released from rest on the inner surface
of plate 1. What is its speed when it hits plate 2? (e = 1.6 ×
10–19 C, me = 9.11 × 10–31 kg)
[2006]
Y
0.1 m
X
1
2.65 × 106 m/s
(a)
(c) 1.87 × 106 m/s
2
(b) 7.02 × 1012 m/s
(d) 32 × 10–19 m/s
P-259
Electrostatic Potential and Capacitance
45. A charged particle ‘q’ is shot towards another charged
particle ‘Q’ which is fixed, with a speed ‘v’. It approaches
‘Q’ upto a closest distance r and then returns. If q were
given a speed of ‘2v’ the closest distances of approach
would be
[2004]
(a) r/2
(b) 2 r
(c) r
(d) r/4
46. On moving a charge of 20 coulomb by 2 cm, 2 J of work
is done, then the potential difference between the points
is
[2002]
(a) 0.1 V
(b) 8 V
(c) 2 V
(d) 0.5 V
Capacitors, Grouping of
TOPIC 3 Capacitors and Energy Stored
in a Capacitor
47. Two capacitors of capacitances C and 2C are charged to
potential differences V and 2V, respectively. These are then
connected in parallel in such a manner that the positive
terminal of one is connected to the negative terminal of the
other. The final energy of this configuration is :
[Sep. 05, 2020 (I)]
25 2
3 2
(a)
CV
(b)
CV
6
2
9 2
(c) zero
(d)
CV
2
48. In the circuit shown, charge on the 5 mF capacitor is :
[Sep. 05, 2020 (II)]
2 mF
4 mF
(a) 450 mC
(b) 590 mC
(c) 160 mC
(d) 650 mC
51. A 5 mF capacitor is charged fully by a 220 V supply. It is
then disconnected from the supply and is connected in
series to another uncharged 2.5 mF capacitor. If the energy
change during the charge redistribution is
X
J then value
100
of X to the nearest integer is ________.
[NA Sep. 02, 2020 (I)]
52. A 10 mF capacitor is fully charged to a potential difference
of 50 V. After removing the source voltage it is connected
to an uncharged capacitor in parallel. Now the potential
difference across them becomes 20 V. The capacitance of
the second capacitor is :
[Sep. 02, 2020 (II)]
(a) 15 mF
(b) 30 mF
(c) 20 mF
(d) 10 mF
53. Effective capacitance of parallel combination of two
capacitors C1 and C2 is 10 mF. When these capacitors are
individually connected to a voltage source of 1 V, the
energy stored in the capacitor C2 is 4 times that of C1. If
these capacitors are connected in series, their effective
capacitance will be:
[8 Jan. 2020 I]
(a) 4.2 mF
(b) 3.2 mF (c) 1.6 mF
(d) 8.4 mF
54. A capacitor is made of two square plates each of side ‘a’
making a very small angle a between them, as shown in
figure. The capacitance will be close to: [8 Jan. 2020 II]
V1
a
5 mF
d
O
6V
(a) 18.00 mC
(b) 10.90 mC
(c) 16.36 mC
(d) 5.45 mC
49. A capacitor C is fully charged with voltage V0 . After
disconnecting the voltage source, it is connected in parallel
C
with another uncharged capacitor of capacitance . The
2
energy loss in the process after the charge is distributed
between the two capacitors is :
[Sep. 04, 2020 (II)]
1
1
(b) CV02
CV02
3
2
1
1
CV02
(c)
(d)
CV02
4
6
50. In the circuit shown in the figure, the total charge is 750 mC
and the voltage across capacitor C2 is 20 V. Then the
charge on capacitor C2 is :
[Sep. 03, 2020 (I)]
(a)
C2
C1 = 15 mF
C3 = 8 m F
+
V
a
6V
–
Î0 a 2
(a)
d
æ aa ö
çè1 - ÷ø
2d
V2
Î0 a 2
(b)
d
æ aa ö
çè1 - ÷ø
4d
Î0 a 2 æ aa ö
Î0 a 2 æ 3aa ö
(d)
çè1 + ÷ø
ç1 ÷
d
d
d è
2d ø
55. A parallel plate capacitor has plates of area A separated
by distance ‘d’ between them. It is filled with a dielectric
which has a dielectric constant that varies as k(x) = K(1 +
ax) where ‘x’ is the distance measured from one of the
plates. If (ad) << l, the total capacitance of the system is
best given by the expression:
[7 Jan. 2020 I]
AK Î0 æ ad ö
(a)
ç1 +
÷
d è
2 ø
2
A Î0 K æ æ ad ö ö
ç1 + ç
÷
(b)
÷
d ç è 2 ø ÷
è
ø
æ
2 2ö
A Î0 K ç a d ÷
1+
(c)
2 ÷
d ç
è
ø
AK Î0
(1 + ad )
(d)
d
(c)
P-260
Physics
56. A 60 pF capacitor is fully charged by a 20 V supply. It is
then disconnected from the supply and is connected
to another uncharged 60 pF capacitor in parallel. The
electrostatic energy that is lost in this process by the
time the charge is redistributed between them is (in nJ)
[NA 7 Jan. 2020 II]
57. The parallel combination of two air filled parallel plate
capacitors of capacitance C and nC is connected to a
battery of voltage, V. When the capacitors are fully
charged, the battery is removed and after that a dielectric
material of dielectric constant K is placed between the
two plates of the first capacitor. The new potential
difference of the combined system is: [9 April 2020 II]
nV
(a)
K+n
(b) V
60. Figure shows charge (q) versus voltage (V) graph for
series and parallel combination of two given capacitors.
The capacitances are :
[10 April 2019 I]
61.
(n + 1) V
V
(d)
(K + n)
K+n
58. Two identical parallel plate capacitors, of capacitance C
each, have plates of area A, separated by a distance d.
The space between the plates of the two capacitors, is
filled with three dielectrics, of equal thickness and dielectric
constants K1, K2 and K3. The first capacitors is filled as
shown in Fig. I, and the second one is filled as shown in
Fig. II.
If these two modified capacitors are charged by the same
potential V, the ratio of the energy stored in the two, would
be (E1 refers to capacitors (I) and E2 to capacitors (II) :
[12 April 2019 I]
(c)
62.
63.
64.
K1 K 2 K 3
E1
(a) E = ( K + K + K )( K K + K K + K K
2
1
2
3
2 3
3 1
1 2
(a) 40 mF and 10 mF
(b) 60 mF and 40 mF
(c) 50 mF and 30 mF
(d) 20 mF and 30 mF
A capacitor with capacitance 5mF is charged to 5 mC. If
the plates are pulled apart to reduce the capacitance to 2
¼F, how much work is done?
[9 April 2019 I]
(a) 6.25 × 10–6 J
(b) 3.75 × 10–6 J
(c) 2.16 × 10–6 J
(d) 2.55 × 10–6 J
Voltage rating of a parallel plate capacitor is 500 V. Its
dielectric can withstand a maximum electric field of 106 V/
m. The plate area is 10–4 m2. What is the dielectric constant
if the capacitance is 15 pF ?
[8 April 2019 I]
–12
2
2
(given “0 = 8.86 × 10 C /Nm )
(a) 3.8
(b) 8.5
(c) 4.5
(d)
6.2
A parallel plate capacitor has 1mF capacitance. One of its
two plates is given + 2mC charge and the other plate,
+4mC charge. The potential difference developed across
the capacitor is :
[8 April 2019 II]
(a) 3 V
(b) 1 V
(c) 5 V
(d) 2 V
In the figure shown, after the switch ‘S’ is turned from
position ‘A’ to position ‘B’, the energy dissipated in the
circuit in terms of capacitance ‘C’ and total charge ‘Q’ is:
[12 Jan. 2019 I]
A
E1 ( K1 + K 2 + K3 )( K 2 K3 + K3 K1 + K1 K 2
(b) E =
K1 K 2 K 3
2
9 K1 K 2 K3
E1
(c) E = ( K + K + K )( K K + K K + K K )
2
1
2
3
2 3
3 1
1 2
E1 ( K1 + K 2 + K3 )(K 2 K 3 + K 3 K1 + K1 K 2
(d) E =
9 K1 K 2 K3
2
59. In the given circuit, the charge on 4 mF capacitor will be :
[12 April 2019 II]
S
e
(b) 9.6 mC (c) 13.4 mC
(d) 24 mC
C
3C
1 Q2
3 Q2
5 Q2
3 Q2
(b)
(c)
(d)
8 C
8 C
8 C
4 C
65. A parallel plate capacitor with plates of area 1 m2 each, are
at a separation of 0.1 m. If the electric field between the
plates is 100 N/C, the magnitude of charge on each plate is :
(a)
(Take Î0 = 8.85
(a) 5.4 mC
B
× 10–12
C2
)
N – M2
[12 Jan. 2019 II]
(a) 7.85 × 10–10 C
(b) 6.85 × 10–10 C
(c) 8.85 × 10–10 C
(d) 9.85 × 10–10 C
P-261
Electrostatic Potential and Capacitance
66. In the circuit shown, find C if the effective capacitance of
the whole circuit is to be 0.5 µF. All values in the circuit are
in µF.
[12 Jan. 2019 II]
A
C
2
2
2
1
2
2
2
B
7
µF
11
6
7
µF (c) 4 µF
(d)
µF
5
10
67. In the figure shown below, the charge on the left plate of
the 10 mF capacitor is –30mC. The charge on the right plate
of the 6mF capacitor is :
[11 Jan. 2019 I]
(a)
(b)
6 mF
10 m F
4 mF
2 mF
(a) –12 m C
(b) +12 m C
(c) –18 m C
(d) +18 m C
68. Seven capacitors, each of capacitance 2 µF, are to be
connected in a configuration to obtain an effective
69. A parallel plate capacitor having capacitance 12 pF is
charged by a battery to a potential difference of 10 V
between its plates. The charging battery is now
disconnected and a porcelain slab of dielectric constant
6.5 is slipped between the plates. The work done by the
capacitor on the slab is:
[10 Jan. 2019 II]
(a) 692 pJ
(b) 508 pJ
(c) 560 pJ
(d) 600 pJ
70. A parallel plate capacitor is of area 6 cm2 and a
separation 3 mm. The gap is filled with three dielectric
materials of equal thickness (see figure) with dielectric
constants K1 = 10, K2 = 12 and K3 = 1(4) The dielectric
constant of a material which when fully inserted in
above capacitor, gives same capacitance would be:
[10 Jan. 2019 I]
(a) 4
(b) 14
(c) 12
(d) 36
71. A parallel plate capacitor is made of two square plates of
side ‘a’, separated by a distance d (d<<a). The lower
triangular portion is filled with a dielectric of dielectric
constant K, as shown in the figure. Capacitance of this
capacitor is:
[9 Jan. 2019 I]
d
æ 6ö
capacitance of ç ÷ µF. Which of the combinations,
è 13 ø
shown in figures below, will achieve the desired value?
[11 Jan. 2019 II]
K
a
(a)
(a)
K Î0 a 2
2d (K + 1)
K Î0 a 2
1 K Î0 a 2
In K
(d)
2
d
d
72. A parallel plate capacitor with square plates is filled with
four dielectrics of dielectric constants K1, K2, K3, K4
arranged as shown in the figure. The effective dielectric
constant K will be:
[9 Jan. 2019 II]
(a) K =
(b) K =
(d)
K Î0 a 2
In K
d (K – 1)
(c)
(b)
(c)
(b)
K1
K2
L/2
K3
K4
L/2
d/2
d/2
( K1 + K3 ) ( K 2 + K 4 )
K1 + K 2 + K 3 + K 4
( K1 + K 2 ) ( K3 + K 4 )
2(K1 + K 2 + K 3 + K 4 )
P-262
Physics
( K1 + K 2 ) ( K 3 + K 4 )
(c) K =
K1 + K 2 + K 3 + K 4
( K1 + K 4 ) ( K 2 + K3 )
(d) K =
2(K1 + K 2 + K 3 + K 4 )
73. A parallel plate capacitor of capacitance 90 pF is connected
to a battery of emf 20V. If a dielectric material of dielectric
5
constant k = is inserted between the plates, the
3
magnitude of the induced charge will be:
[2018]
(a) 1.2 n C (b) 0.3 n C (c) 2.4 n C (d) 0.9 n C
74. In the following circuit, the switch S is closed at t = 0. The
charge on the capacitor C1 as a function of time will be
æ
CC ö
given by ç Ceq = 1 2 ÷ .
C1 + C2 ø
è
[Online April 16, 2018]
C1
(a) CeqE[1 – exp(–t/RCeq)]
C2
(b) C1E[1 – exp(–tR/C1)]
S
(c) C2E[1 – exp(–t/RC2)]
R
E
(d) CeqE exp(–t/RCeq)
75. The equivalent capacitance between A and B in the circuit
given below is:
6 µF
2 µF
A
5 µF
5 µF
4 µF
2 µF
B
[Online April 15, 2018]
(a) 4.9 µF
(b) 3.6 µF (c) 5.4 µF
(d) 2.4 µF
76. A parallel plate capacitor with area 200cm2 and separation
between the plates 1.5cm, is connected across a battery of
emf V. If the force of attraction between the plates is 25 × 10–
6N, the value of V is approximately: [Online April 15, 2018]
78. A capacitance of 2m F is required in an electrical circuit
across a potential difference of 1.0 kV. A large number of
1m F capacitors are available which can withstand a
potential difference of not more than 300 V. The minimum
number of capacitors required to achieve this is [2017]
(a) 24
(b) 32
(c) 2
(d) 16
79. A combination of parallel plate capacitors is maintained at
a certain potential difference.
C1
A
D
When a 3 mm thick slab is introduced between all the
plates, in order to maintain the same potential difference,
the distance between the plates is increased by 2.4 mm.
Find the dielectric constant of the slab.
[Online April 9, 2017]
(a) 3
(b) 4
(c) 5
(d) 6
80. The energy stored in the electric field produced by a metal
sphere is 4.5 J. If the sphere contains 4 mC charge, its
radius will be : [Take : 1 = 9 ´ 109 N - m 2 / C 2 ]
4 pe0
[Online April 8, 2017]
(a) 20mm (b) 32mm (c) 28mm (d) 16mm
81. A combination of capacitors is set up as shown in the
figure. The magnitude of the electric field, due to a point
charge Q (having a charge equal to the sum of the charges
on the 4 mF and 9 mF capacitors), at a point distance 30 m
from it, would equal :
[2016]
3m F
4m F
9m F
2m F
+ –
8V
(a) 150V
(b) 100V (c) 250V
(d) 300V
77. A capacitor C1 is charged up to a voltage V = 60V by
connecting it to battery B through switch (1), Now C1 is
disconnected from battery and connected to a circuit
consisting of two uncharged capacitors C2 = 3.0mF and C3 =
6.0mF through a switch (2) as shown in the figure. The sum
of final charges on C2 and C3 is: [Online April 15, 2018]
(a) 36mC
(b) 20mC
B
E
2 ö
æ
-12 C
ç e 0 = 8.85 ´ 10
÷
N.m 2 ø
è
(1)
B
60 V
C3
C2
C
(2)
(c) 54mC
C
1
A
8
C2
C1
(a) 420 N/C
(b) 480 N/C
(c) 240 N/C
(d) 360 N/C
82. Figure shows a network of capacitors where the numbers
indicates capacitances in micro Farad. The value of
capacitance C if the equivalent capacitance between point
A and B is to be 1 mF is :
[Online April 10, 2016]
C3
2
(d) 40mC
6
2
4
12
B
P-263
Electrostatic Potential and Capacitance
32
31
33
34
mF (b)
mF (c)
mF (d)
mF
23
23
23
23
83. Three capacitors each of 4 mF are to be connected in such
a way that the effective capacitance is 6mF. This can be
done by connecting them :
[Online April 9, 2016]
(a) all in series
(b) all in parallel
(c) two in parallel and one in series
(d) two in series and one in parallel
84. In the given circuit, charge Q2 on the 2µF capacitor
changes as C is varied from 1µF to 3µF. Q2 as a function of
'C' is given properly by: (figures are drawn schematically
and are not to scale)
[2015]
(a)
1µF
C
2µF
E
Charge
dielectric whose permittivity varies linearly from Î1 at one
plate to Î2 at the other. The capacitance of capacitor is:
[Online April 19, 2014]
(a) Î0 (Î1 + Î2 ) A / d
(b) Î0 ( Î2 + Î1 ) A / 2d
(c) Î0 A / éëd ln ( Î2 / Î1 ) ùû
(d) Î0 ( Î2 - Î1 ) A / ëé d ln ( Î2 / Î1 ) ûù
88. The space between the plates of a parallel plate capacitor
is filled with a ‘dielectric’ whose ‘dielectric constant’ varies
with distance as per the relation:
K(x) = Ko + lx (l = a constant)
The capacitance C, of the capacitor, would be related to its
vacuum capacitance Co for the relation :
[Online April 12, 2014]
(a) C =
ld
Co
ln (1 + K o ld )
(b) C =
l
C
d.ln (1 + K o ld ) o
(c) C =
ld
C
ln (1 + ld / K o ) o
(d) C =
l
Co
d.ln (1 + Ko / ld )
Charge
Q2
Q2
(a)
(b)
1µF
3µF
C
1µF
3µF
89. A parallel plate capacitor is made of two plates of length l,
width w and separated by distance d. A dielectric slab
(dielectric constant K) that fits exactly between the plates
is held near the edge of the plates. It is pulled into the
C
Charge
Charge
Q2
Q2
(c)
3µF
C
1µF
3µF
C
85. In figure a system of four capacitors connected across
a 10 V battery is shown. Charge that will flow from
switch S when it is closed is : [Online April 11, 2015]
2mF a
¶U
where U is the energy of
¶x
the capacitor when dielectric is inside the capacitor up to
distance x (See figure). If the charge on the capacitor is Q
then the force on the dielectric when it is near the edge is:
[Online April 11, 2014]
capacitor by a force F = -
(d)
1µF
87. The gap between the plates of a parallel plate capacitor of
area A and distance between plates d, is filled with a
3mF
x
l
S
d
3mF b 2mF
2
(a)
10 V
(a) 5 µC from b to a
(b) 20 µC from a to b
(c) zero
(d) 5 µC from a to b
86. A parallel plate capacitor is made of two circular plates
separated by a distance 5 mm and with a dielectric of
dialectric constant 2.2 between them. When the electric
field in the dielectric is 3 ´ 104 V m the charge density of
the positive plate will be close to:
[2014]
(a) 6 ´ 10-7 C m 2
(c) 3 ´ 104 C m 2
(b) 3 ´10-7 C m2
(d) 6 ´104 C m 2
Q d
2wl 2 e o
Q 2d
K
(b)
( K - 1)
(d)
Q2 w
2dl 2 e0
Q2w
( K - 1)
K
2wl 2 e o
2dl 2 e o
90. Three capacitors, each of 3 mF, are provided. These cannot
be combined to provide the resultant capacitance of:
[Online April 9, 2014]
(a) 1 mF
(b) 2 mF
(c) 4.5 mF
(d) 6 mF
91. A parallel plate capacitor having a separation between the
plates d, plate area A and material with dielectric constant
K has capacitance C0. Now one-third of the material is
replaced by another material with dielectric constant 2K,
so that effectively there are two capacitors one with area
(c)
P-264
Physics
1
2
A, dielectric constant 2K and another with area
A
3
3
and dielectric constant K. If the capacitance of this new
C
capacitor is C then
is
[Online April 25, 2013]
C0
4
2
1
(c)
(d)
3
3
3
92. To establish an instantaneous current of 2 A through a 1
mF capacitor ; the potential difference across the capacitor
plates should be changed at the rate of :
[Online April 22, 2013]
(a) 2 × 104 V/s
(b) 4 × 106 V/s
(c) 2 × 106 V/s
(d) 4 × 104 V/s
ur
93. A uniform electric field E exists between the plates of a
charged condenser. A charged particle enters the space
ur
between the plates and perpendicular to E . The path of
the particle between the plates is a :
[Online April 9, 2013]
(a) straight line
(b) hyperbola
(c) parabola
(d) circle
94. The figure shows an experimental plot discharging of a
capacitor in an RC circuit. The time constant t of this circuit
lies between :
[2012]
(b)
Potential difference
V in volts
(a) 1
25
20
15
10
5
0
50 100 150
200
250
Time in seconds
300
(a) 150 sec and 200 sec (b) 0 sec and 50 sec
(c) 50 sec and 100 sec
(d) 100 sec and 150 sec
95. The capacitor of an oscillatory circuit is enclosed in a
container. When the container is evacuated, the resonance
frequency of the circuit is 10 kHz. When the container is
filled with a gas, the resonance frequency changes by 50
Hz. The dielectric constant of the gas is
[Online May 26, 2012]
(a) 1.001
(b) 2.001 (c) 1.01
(d) 3.01
96. Statement 1: It is not possible to make a sphere of
capacity 1 farad using a conducting material.
Statement 2: It is possible for earth as its radius is
6.4 × 106 m.
[Online May 26, 2012]
(a) Statement 1 is true, Statement 2 is true, Statement 2 is
the correct explanation of Statement 1.
(b) Statement 1 is false, Statement 2 is true.
(c) Statement 1 is true, Statement 2 is true, Statement 2 is
not the correct explanation of Statement 1.
(d) Statement 1 is true, Statement 2 is false.
97. A series combination of n1 capacitors, each of capacity
C1 is charged by source of potential difference 4 V. When
another parallel combination of n2 capacitors each of
capacity C2 is charged by a source of potential difference
V, it has the same total energy stored in it as the first
combination has. The value of C2 in terms of C1 is then
[Online May 12, 2012]
n2
2 C1
(a) 16 n C1
(b) n n
1
1 2
n2
16 C1
(c) 2 n C1
(d) n n
1
1 2
98. Two circuits (a) and (b) have charged capacitors of
capacitance C, 2C and 3C with open switches. Charges on
each of the capacitor are as shown in the figures. On closing
the switches
[Online May 7, 2012]
S
S
Q
C
2Q
3C
L
2Q
2C
Q
2C
L
R
Circuit (a)
R
Circuit (b)
(a) No charge flows in (a) but charge flows from R to L in (b)
(b) Charges flow from L to R in both (a) and (b)
(c) Charges flow from R to L in (a) and from L to R in (b)
(d) No charge flows in (a) but charge flows from L to R in (b)
99. Let C be the capacitance of a capacitor discharging through
a resistor R. Suppose t1 is the time taken for the energy
stored in the capacitor to reduce to half its initial value and
t2 is the time taken for the charge to reduce to one-fourth
its initial value. Then the ratio t1/ t2 will be
[2010]
(a) 1
(b)
1
2
(c)
1
4
(d) 2
100. A parallel plate capacitor with air between the plates has
capacitance of 9 pF. The separation between its plates is
‘d’. The space between the plates is now filled with two
dielectrics. One of the dielectrics has dielectric constant
d
k1 = 3 and thickness
while the other one has dielectric
3
2d
. Capacitance of the
constant k2 = 6 and thickness
3
capacitor is now
[2008]
(a) 1.8 pF
(b) 45 pF (c) 40.5 pF (d) 20.25 pF
101. A parallel plate condenser with a dielectric of dielectric
constant K between the plates has a capacity C and is
charged to a potential V volt. The dielectric slab is slowly
removed from between the plates and then reinserted. The
net work done by the system in this process is [2007]
P-265
Electrostatic Potential and Capacitance
1
( K - 1) CV 2
2
(a) zero
(b)
2
(c) CV ( K - 1)
K
(d) ( K - 1) CV 2
102. A parallel plate capacitor is made by stacking n equally
spaced plates connected alternatively. If the capacitance
between any two adjacent plates is ‘C’ then the resultant
capacitance is
[2005]
(a) (n + 1) C
(b) (n – 1) C
(c) nC
(d) C
103. A fully charged capacitor has a capacitance ‘C’. It
is discharged through a small coil of resistance wire
embedded in a thermally insulated block of specific heat
capacity ‘s’ and mass ‘m’. If the temperature of the block is
raised by ‘DT’, the potential difference ‘V’ across the
capacitance is
[2005]
(a)
(c)
mC DT
s
2ms DT
C
(b)
(d)
2mC DT
s
ms DT
C
104. A sheet of aluminium foil of negligible thickness is
introduced between the plates of a capacitor. The
capacitance of the capacitor
[2003]
(a) decreases
(b) remains unchanged
(c) becomes infinite
(d) increases
-18
105. The work done in placing a charge of 8 ´ 10 coulomb
on a condenser of capacity 100 micro-farad is [2003]
(a) 16 ´ 10 -32 joule
(b) 3.1´10 -26 joule
(c) 4 ´10-10 joule
(d) 32 ´ 10-32 joule
106. If there are n capacitors in parallel connected to V volt
source, then the energy stored is equal to
[2002]
(a) CV
(b)
1
nCV2 (c) CV2
2
(d)
1
CV2
2n
107. Capacitance (in F) of a spherical conductor with radius 1 m
is
[2002]
(a) 1.1´ 10 -10
(b) 10 -6
(c) 9 ´ 10 -9
(d) 10 -3
P-266
1.
Physics
4.
KQnet
R
(c) Potential at the centre, VC =
(d) Let s be the surface charge density of the shells.
s
QQnet = 0
s
\ VC = 0
Let E be electric field produced by each charge at the centre,
then resultant electric field will be EC = 0, since equal electric
field vectors are acting at equal angle so their resultant is
equal to zero.
2E
72°
2E
72°
72°
Charge on the inner shell, Q1 = s 4pr 2
Charge on the outer shell, Q2 = s 4pR 2
\ Total charge, Q = s 4p (r 2 + R 2 )
2E
72°
Þs=
72°
2E
(d) Total charge Q1 + Q2 = Q '1 + Q '2
=
KQ '1 KQ '2
=
= 12 – 3 = 9 µC
2
R
R
3
3
=
Q '1 = 2Q '2 Þ 2Q '2 + Q '2 = 9mC
(a) We have given two metallic hollow spheres of radii R
and 4R having charges Q1 and Q2 respectively.
Potential on the surface of inner sphere (at A)
kQ1 kQ2
+
R
4R
Potential on the surface of outer sphere (at B)
5.
æ
1 ö
ç Here, k =
÷
4pe0 ø
è
kQ1 kQ2
+
4R 4R
1 ( r + R )Q
4pe 0 (r 2 + R 2 )
(b) The electric potential at the bisector is zero and
electric field is antiparallel to the dipole moment.
æ ®ö
ç-P ÷
ç d3 ÷
ç
÷
è
ø
(c) Potential at any point of the charged ring
1
\ V = 0 and E =
4pe0
6.
Vp =
Q2
Q1
4R
4p(r 2 + R 2 )
®
VA =
R
KQ4p(r + R)
=
\ Q '1 = 6 mC and Q '2 = 3 mC
VB =
A
æ
1 ö
ç where K =
÷
4pe0 ø
è
K s 4pr 2 K s4pR 2
+
r
R
= K s4p ( r + R )
Two isolated conducting sphres S1 and S2 are now
connected by a conducting wire.
3.
4p ( r + R 2 )
KQ1 KQ2
+
r
R
VC =
= 12mC - 3mC = 9mC
\ V1 = V2 =
Q
2
Potential at the common centre,
2E
2.
C
r
R
B
Potential difference,
Q
3 kQ
3
DV = VA - VB = × 1 =
× 1
4 R
16p Î0 R
R = 3a
Z = 4a
Kq
R 2 + Z2
P-267
Electrostatic Potential and Capacitance
l = R 2 + Z2 = 5a
The minimum velocity (v0) should just sufficient to reach
the point charge at the center, therefore
7.
kQa kQ b kQ c
+
+
a
b
c
Since surface charge densities are equal to one another
i.e., sa = sb = sc
1
ö2
2 æ
2q 2
ç
÷
m è 15 ´ 4pe 0 a ø
\ Qa : Qb : Qc :: a 2 : b 2 : c 2
(d) When charge Q is on inner solid conducting sphere
+Q
–Q
+Q
E
Electric field between spherical surface
r KQ
r r
E = 2 Soò E.dr = V given
r
Now when a charge – 4Q is given to hollow shell
é
ù
a2
\ Qa = ê 2 2 2 ú Q
ëê a + b + c úû
é
ù
b2
Qb = ê 2 2 2 ú Q
êë a + b + c úû
é
ù
c2
Qc = ê 2 2 2 ú Q
êë a + b + c úû
Q é ( a + b + c) ù
\ V=
ê
ú
4p Î0 ë a 2 + b2 + c2 û
11. (d) Let at a distance ‘x’ from point B, both the dipoles
produce same potential
R
+Q
–Q
2qa
4qa
–3Q
\
Electric field between surface remain unchanged.
r KQ
E= 2
r
as, field inside the hollow spherical shell = 0
\ Potential difference between them remain unchanged
r r
i.e. ò E.dr = V
8.
uur
(c) Given, E = ( Ax + B ) iˆ
4qa
2qa
= 2
(R + x) x
Þ
( )
2x = R + x Þ x =
=
R
1
ò ( 20x + 10) dx
-5
= – 180 V
or V1 – V2 = 180 V
9.
(b) Electric potential is constant inside a charged spherical
shell.
2R
+R =
2 –1
2 –1
12. (b) Electric field at a point outside the sphere is given by
E=
Using V = ò Edx , we have
R
2 –1
Therefore distance from A at which both of them produce
the same potential
1Q
4p Î0 r
or E = 20x + 10
V2 – V1 =
P
Potential at point P, V =
4Kq 2
4 1 q2
=
15ma 15 4pe 0 ma
Þ v0 =
r
10. (d) c
1
mv 20 = q [VC - VP ]
2
é Kq Kq ù
= qê
ë 3a 5a úû
v02 =
a
b
\E =
2
Q
But r = 4
pR 3
3
rR 3
3 Î0 r 2
At surface r = R
rR 3
3 Î0
Let r1 and r2 are the charge densities of two sphere.
rR
r R
E1 = 1 and E2 = 2 2
3e 0
3e 0
\E =
P-268
Physics
E1 r1 R1
R
=
= 1
E2 r2 R2 R2
This gives r1 = r2 = r
Potential at a point outside the sphere
- dv
= 10 | z |
dr
- dv
S2 =
= 10 (constant : E)
dr
\ The source is an infinity large non conducting thick
plate of thickness 2 m.
16. (b) S1 =
Q
æ
ö
3 ç
÷
1 Q = rR çQ r = Q ÷
V=
4 3÷
3e0 r ç
4pe 0 r
pR ÷
ç
3
è
ø
At surface, r = R
V=
\
\10Z ×10A =
rR12
rR 2
rR 2
and V2 = 2
so, V1 =
3e 0
3e 0
3e 0
V1 æ R1 ö
=
V2 çè R2 ÷ø
r 0 = 10e0 for | z | £ 1m .
17. (a) We know, V0 =
2
KQ
r
where r is distance of point from the centre of shell
KQ
Potential inside the shell, Vinside =
R
where ‘R” is radius of the shell
s
C
–s
B
s
a A
13. (b) Potential outside the shell, Voutside =
b c
VB =
Kq A Kq B KqC
+
+
rb
rb
rc
VB =
1
4p Î0
15. (c) As we know electric field, E =
- dv
dr
E = constant \ dv and dr same
Kf
E= 2 =c
r
0
1
Þrµ
r
Now,
Kq
5 Kq
=
(3R 2 – r 2 )
4 R
2R 3
Þ R2 =
R
2
3 Kq Kq
= 3
4 R
R
Also, R1 = 0 and R2 < (R4 – R3)
ù
s é a 2 - b2
+ cú
ê
Î0 êë b
úû
14. (c) Potential gradient is given by,
DV = E.d
0.8 = Ed (max)
DV = Ed cos q = 0.8 × cos 60 = 0.4
Hence, maximum potential at a point on the sphere
= 589.4 V
r
Kq
(3R 2 – r 2 ) [For r < R]
2R 3
At the centre of sphare r = 0. Here
3
V = V0
2
Now, Vi =
R4 = 4R
VB =
\ f = ò r4 pr 2 dr
Kq
= V surface
R
1 Kq Kq
=
4 R
R4
é s 4pa 2 s4 pb 2 s 4pc2 ù
+
ê
ú
b
c ûú
ëê b
Þ f µ r2
r×A µ Z
e0
v0
V
2, 2
0
0
ò dV = –
v0 + 2D v
r
ò
(25dx + 30dy )
on solving we get,
V = – 110 volt.
19. (c) Potential difference between any two points in an
electric field is given by,
r uur
dV = –E × dx
VA
v0 + v
r
dv
dx
Potential at the point x = 2m, y = 2m is given by :
18. (a) As we know, E = –
ò
VO
2
dV = - ò 30 x 2 dx
0
V A - VO = -[10 x 3 ]20 = -80 J/ C
20. (a) The potential of uncharged body is less than that of
the charged conductor and more than at infinity.
P-269
Electrostatic Potential and Capacitance
21. (c)
By Gauss's theorem
O
E=
q
Tcosq
q
C Tsinq q
x
Fe
q
Charge density, r =
In equilibrium, Fe = T sin q
mg = T cos q
Y
Fe
q2
=
mg 4p Î0 x 2 ´ mg
\ x=
26. (c)
q2
4p Î0 tan q mg
kq
kq
V=
+
= 4 kmg / tan q
x/2 x/2
22. (c) q = 1µC = 1 × 10–6C
r = 4 cm = 4 ×10–2 m
kq 9 ´ 109 ´10 -6
=
Potential V =
= 2.25 × 105 V.
r
4 ´ 10-2
=
9 ´109 ´1´10-6
16 ´10-4
\
r pr 2
So, q1 =
=
4pR 2
Qr 2
R2 + r 2
and q2 =
Now, potential, V =
=
=
QR 2
R2 + r 2
1 é q1 q 2 ù
+
ê
ú
4 pe 0 ë r
R û
1 é Qr
QR ù
+
4pe 0 êë R 2 + r 2 R 2 + r 2 úû
Q( R + r )
2
2
1
4pe0
A(Ö2,Ö2)
X
(0,0) ®
rB B (2,0)
rB =
(
)
2, 2 from the origin,
4 = 2 units.
(2) 2 + (0) 2 = 2 units.
Now, potential at A, due to charge q = 10 –3 mC
Q
1
×
4 p Î0 ( rA )
Potential at B, due to charge Q = 10–3 QC VB =
Q
1
×
4 p Î0 ( rB )
\ Potential difference between the points A and B is given
by
VA – VB =
1 10 –3
1 10 –3
×
×
4p Î0 rA
4p Î0 rB
10–3 æ 1 1 ö 10 –3 æ 1 1 ö
ç - ÷=
4p Î0 è rA rB ø 4 p Î0 çè 2 2 ÷ø
Q
´ 0 = 0.
4p Î0
27. (a) As shown in the figure, the resultant electric fields
before and after interchanging the charges will have the
same magnitude, but opposite directions.
As potential is a scalar quantity, So the potential will be
same in both cases.
q
q
A
B
=
R +r
24. (c) Charges reside only on the outer surface of a
conductor with cavity.
25. (c) Electric field
df
E== – 2ar
dr
= – 6e0a
The distance of point B(2, 0) from the origin,
=
q2
4pr 2dr
rA = ( 2)2 + ( 2 ) 2 =
r2
= –5.625 × 106 V/m
dq
The distance of point A
VA =
kq
23. (c) Let q1 and q2 be charge on two spheres of radius
'r' and 'R' respectively
As, q1 + q2 = Q
and s1 = s2 [Surface charge density are equal]
q1
®
rA
O
Electric potential at the centre of the line
Induced electric field E = –
....(ii)
From (i) and (ii),
Q = –8 pe0ar3
Þdq = – 24pe0ar2 dr
mg
tan q =
1 q
4 pe 0 r 2
®
E
....(i)
D
-q
C
-q
P-270
Physics
-q
-q
A
B
®
E
Electric potential due to charge q inside the shell is
1 q
4pe o R
\ The net electric potential at point P is
V2 =
V = V1 + V2 =
C
D
q
q
20
28. (a) Given, potential V(x) = 2
volt
x -4
d æ 20 ö
dV
Electric field E = =- ç
dx è x 2 - 4 ÷ø
dx
q
1 2Q
1
+
4pe o R 4pe o R
31. (d) By using energy conservation,
DKE + (DPE )Electro + ( DPE )gravitational = 0
Q
40 x
ÞE= +
q
2
( x - 4) 2
At x = 4 mm ,
40 ´ 4
y
160
10
= + volt / mm.
144
9
r
Positive sign indicates that E is in +ve x-direction.
29. (a)
q
q
E=+
(42 - 4) 2
=+
Þ
æ1
1
1 ö
mV 2 = mgy + kQq ç 2
è R R + y ø÷
R
R
2
1
d
Potential at the center of ring of charge +q = potential due
to iteself + potential due to other ring of charge –q.
éq
ù
q
ê ú
êë R
R 2 + d 2 úû
Potential at the centre of ring of charge –q = potential due
to itself + potential due to other ring of charge +q.
Þ V2 =
1
4pe 0
1
4pe 0
DV = V1 – V2
é- q
q
+
ê
êë R
R2 + d 2
1 éq q
ê + =
4 pe 0 ëê R R
q
R2 + d 2
-
Q
R/2
R
V1 =
1
Q
1 2Q
=
4pe o R / 2 4pe o R
P
q
Potential of – q is same as initial and final point of the path.
Y
q
O
d/2
éq
ù
q
ê ú
êë R
R 2 + d 2 úû
30. (c) Electric potential due to charge Q at point P is
1
=
2pe 0
32. (d) Change in potential energy, Du = q(V f - Vi )
4q
ù
ú
R 2 + d 2 ûú
2kQq
y
m R( R + y)
é
ù
qQ
or, V 2 = 2 y ê
+ gú
4
pe
R
(
R
+
y
)
m
ë
0
û
ù
ú
úû
q
v
æ Qq
Qq ö
1
mV 2 + ç k
-k
+ (- mgy ) = 0
R ø÷
2
è R+ y
Þ V 2 = 2 gy +
Þ V1 =
R
d/2
–q
X
d
4q 2
æ k 4q k 4q ö
Du = q ç
=÷
è 3d / 2 d / 2 ø
3pe 0 d
–ve sign shows the energy of the charge is decreasing.
33. (c) According to work energy theorem, gain in kinetic
energy is equal to work done in displacement of charge.
1 2
mv = q DV
2
Here, DV = potential difference between two positions of
charge q.
For same q and DV.
\
vµ
1
m
P-271
Electrostatic Potential and Capacitance
Mass of hydrogen ion mH = 1
Mass of helium ion mHe = 4
\
vH
4
=
= 2 :1.
vHe
1
34. (c) Using conservation of energy
1
U i = U F + mv 2
2
kq1q2 kq1q2 1 2
=
+ mv
r1
r2
2
é1 1 ù
1
Þ mv 2 = kq1q2 ê - ú
2
ë r1 r2 û
v2 =
2kq1q2
m
38. (b) Net electrostatic energy for the system
é q 2 Qq Qq ù
U = Kê +
+
=0
a
a a 2ú
êë
úû
é
1 ù
Þ q = -Q ê1 +
ú
2
ë
û
ÞQ=
(O,2 ) Q
(O, – 2 ) Q
2 ´ 9 ´ 109 ´ 10-12 é 1 ù
1 - ú = 4 ´ 10+6
-6
-3 ê
ë 9û
4 ´ 10 ´ 10
3
v = 2 × 10 m/s
0 + Vq = mv2 + v’q
or
mv2 = (V – V’)q
r
=
r
- q ò Edr = q ò
r0
r0
l
lq æ ln 3 ö
dr =
ç
÷
2p Î0 r
2p Î0 è r0 ø
r
Þ vµ lnr
0
37. (c) Total energy of charge distribution is constant at any
instant t.
Uf + Kf = Ui + Ki
i.e.,
\
1
KQ2
KQ2
mV 2 +
= 0+
2
2R
2R 0
1
KQ 2 KQ2
mV 2 =
2
2R 0
2R
\ V=
\ V=
2 KQ 2 æ 1
1ö
- ÷
ç
m 2 è R0 R ø
2
KQ æ 1
1ö
1
1
- ÷=C
ç
m è R0 R ø
R0 R
Also the slope of V – R curve will go on decreasing.
a
+q
+q
Q(4, + 2)
Q(4, – 2)
Potential at origin
=
or
a 2
a
39. (b)
é1 1 ù
ê - ú
ë r1 r2 û
é
ù
ê
1
( - q )Q ú
q(- q)
qQ
ê
ú
U=
+
+
35. (d)
dö æ
dö ú
4 p Î0 ê d
æ
çè D + ÷ø çè D - ÷ø ú
êë
2
2 û
2
2
1 é q
qQd ù
=
- 2 ú , Ignoring d
ê4p Î0 êë d
D úû
4
36. (b) Using, [K + U]i = [K + U]f
-q 2
2+1
Q
KQ KQ KQ KQ
+
v= 2 + 2 +
20
20
1 ö
æ
and potential at ¥ = 0 =KQ ç 1+
÷
5ø
è
\ Work required to put a fifth charge Q at origin W =
Q2 æ
1 ö
1+
VQ = 4
ç
÷
pe0 è
5ø
40. (c) The work done in moving a charge along an
equipotential surface is always zero.
The direction of electric field is perpendicular to the
equipotential surface or lines.
41. (c) The potential energy at the centre of the sphere
3 KQ q
2 R
The potential energy at the surface of the sphere
K qQ
Us =
R
Now change in the energy
Uc =
DU = U c -U s
K Qq é 3 ù KQq
-1 =
R êë 2 úû
2R
4 3
Where Q = r.V = r. pR
3
2K pR 3 rq
DU =
3
R
3
2
1 pR rq
DU = ´
3 4 p Î0 R
=
R 2 rq
6 Î0
Using Gauss’s law
DU =
P-272
Physics
kqQ
r
1
m(2v )2 =
Þ r'=
2
r'
4
46. (a) By using
W = q(VB – VA)
4
b ´ pR 3
r uur qen
ò E × dA = E0 = E30
Þ
b ´ 4 pR 3
3E0
4 3 1
Þ E(4pR2) = b ´ 3 pR ´ E
0
br
Þ E=
(r < R)
3E0
42. (d) Initial potential of the charge,
ò EdA(cos q) =
Þ
VA =
\ VB – VA =
47. (b) When capacitors C and 2C capacitance are charged
to V and 2V respectively.
+
(Here potential due to each q =
- kq
a 5
2C–
+
2V
Q1 = CV Q2 = 2C ´ 2V = 4CV
1 2q æ
1 ö
1–
ç
÷
4 pE a è
5ø
to each – q =
C–
V
2 kq 2 kq
a
a 5
Þ VA =
2J
= 0.1J/C = 0.1V
20C
kq
and potential due
a
When connected in parallel
Q1 = CV
+ –
)
2a
A
q
q
–
+
Q2 = 4CV
By conservation of charge
4CV - CV = (C + 2C )Vcommon
2a
3CV
=V
3C
Therefore final energy of this configuration,
Vcommon =
B
–q
–q
Final potential of the charge
VB = 0
(Q Point B is equidistant from all the four charges)
\ Using work energy theorem,
(WAB)electric = Q(VA – VB)
=
2qQ é
1 ù
1ú
4pE0 a êë
5û
= (–100 × 1.6 × 10–19)(– 4 – 10)
= +2.24 × 10–16J
44. (a) Gain in kinetic energy = work done by potential
difference
=
45. (d)
2 ´ 1.6 ´ 10-19 ´ 20
9.1 ´ 10 -31
1 2 kQq
mv =
2
r
+q1
–
V0
2 mF
+
–
–
+q24 m F
Q
5 mF
0V
6V
Let q1 and q2 be the charge on the capacitors of 2mF and
4mF. Then charge on capacitor of 5mF
43. (c) Work done, WPQ = q(VQ – VP)
1 2
mv Þ v =
2
48. (a)
6V
æ 1 ö 2Qq é
1 ù
= ç 4 pe ÷ a ê1 5 úû
ë
0ø
è
eV =
1
æ1
ö 3
U f = ç CV 2 + ´ 2CV 2 ÷ = CV 2
è2
ø 2
2
2eV
m
= 2.65 × 106 m/s
Q = q1 + q2
Þ 5V0 = 2(6 - V0 ) + 4(6 - V0 )
Þ 5V0 = 12 - 2V0 + 24 - 4V0
Þ 11V0 = 36 Þ V0 =
36
V
11
180
mC
11
49. (d) When two capacitors with capacitance C1 and C2 at
potential V1 and V2 connected to each other by wire, charge
begins to flow from higher to lower potential till they
acquire common potential. Here, some loss of energy takes
place which is given by.
Þ Q = 5V0 =
P-273
Electrostatic Potential and Capacitance
C1C2
(V1 - V2 )2
2(C1 + C2 )
In the equation, put V2 = 0, V1 = V0
Heat loss, H =
C1 = C, C2 =
C
2
C
C´
2 (V - 0) 2 = C V 2
Loss of heat =
0
Cö 0
6
æ
2çC + ÷
è
2ø
1
2
H = CV0
6
50. (b) According to question,
Q = 750mC = q2 + q3
C1 = 15 mF
750m C
C2 q2
C3 = 8 mF
Q
q3
V1
V2 = 20V
If C2 be the capacitance of uncharged capacitor, then
common potential is
V=
C1V1 + C2V2
C1 + C2
Þ 20 =
53. (c) In parallel combination, Ceq = C1 + C2 = 10 mF
When connected across 1 V battery, then
c1
æ1
2ö
CV
c2
U1 çè 2 1 ÷ø 1
C
1
=
= Þ 1 =
U2 æ 1
4 C2 4
2ö
çè C2V ÷ø
2
1v
\ C2 = 8 mF and C1 = 2 mF
Now C1 and C2 are connected in series combination,
CC
2 ´ 8 16
\ Cequivalent = 1 2 =
=
= 1.6mF
C1 + C2 2 + 8 10
54. (a)
Capacitors C2 and C3 are in parallel hence,
Voltage across C2 = voltage across C3 = 20 V
Change on capacitor C3,
q3 = C3 ´ V3 = 8 ´ 20 = 160mC
\ q2 = 750mC - 160mC = 590mC
51. (4)
Given, C1 = 5 mF and V1 = 220 Volt
When capacitor C1 fully charged it is disconnected from
the supply and connected to uncharged capacitor C2.
C2 = 2.5 mF, V2 = 0
Energy change during the charge redistribution,
DU = U i - U f =
1 C1C2
(V1 - V2 ) 2
2 C1 + C2
1 5 ´ 2.5
´
(220 - 0)2 mJ
2 (5 + 2.5)
5
=
´ 22 ´ 22 ´ 100 ´ 10 -6 J
2´3
5 ´ 11 ´ 22
55 ´ 22
=
´ 10 -4 J =
´ 10 -4 J
3
3
1210
1210
=
´ 10-4 J =
´ 10-3 J ; 4 ´ 10 -2 J
3
3
x
= 4 ´ 10 -2
According to questions,
100
\x = 4
52. (a) Given,
Capacitance of capacitor, C1 = 10 mF
Potential difference before removing the source voltage,
V1 = 50 V
=
10 ´ 50 + 0
Þ C = 15 mF
20 + C
dx
a
x
xtan a
d
a
x= 0
Consider an infinitesimal strip of capacitor of thickness
dx at a distance x as shown.
Capacitance of parallel plate capacitor of area A is given
e0 A
by C =
t
[Here t = seperation between plates]
So, capacitance of thickness dx will be
e 0 adx
\ dC =
d + x tan a
Total capacitance of system can be obtained by
integrating with limits x = 0 to x = a
\ Ceq = ò dC = ae 0
x=a
ò
x=0
dx
x tan a + d
[By Binomial expansion]
a
Þ Ceq
ae æ x tan a ö
ae
= 0 ò ç1 –
dx = 0
÷
d è
d ø
d
0
Þ Ceq =
a 2 e 0 æ a tan a ö e0 a 2
= ç1 –
÷=
è
d
2d ø
d
55. (a) Given, K (x) = K(1 + ax)
Capacitance of element, Cel =
K e0 A
dx
a
æ
x 2 tan a ö
x
–
ç
÷
2d ø
è
0
æ aa ö
çè1 – ÷ø
2d
P-274
Physics
Þ Cel =
e0 K (1 + ax) A
dx
9k1k2 k3
E1
= E = ( k + k + k )(k k + k k + k k )
2
1
2
3 1 2
2 3
3 1
d
æ
ö
1
dx
æ1ö
\ òdç ÷ =
= òç
÷
C
C
e
KA
(1
+
a
x
)
è ø
el
ø
0è 0
Þ
=
10V
1 é
a2d 2 ù
êad ú
2 úû
e0 KAa ëê
1 é ad ù
1e 0 KA êë
2 úû
x dx
e 0 KA
e KA æ
ad ö
\C =
ÞC= 0
ç 1+
÷
d è
2 ø
æ ad ö
d ç1 ÷
è
2 ø
56. (6) In the first condition, electrostatic energy is
Ui =
1
1
CV02 = ´ 60 ´ 10 –12 ´ 400 = 12 ´ 10 –9 J
2
2
In the second condition U F =
Uf =
1
æV ö
2C . ç 0 ÷
è 2ø
2
2
1
C 'V ' 2
2
V0 ö
æ
çèQ C ' = 2C , V ' = 2 ÷ø
1
´ 60 ´ 10 -12 ´ (20) 2 = 6 × 10–9 J
4
Energy lost = Ui – Uf = 12×10–9J –6 × 10–9J = 6 nJ
=
57. (d) V ¢=
CV + (nC )V
kC + nC
(n + 1)V
k+ n
1
d /3
d /3
d /3
58. (c) C = k e A + k e A + k e A
1
1 0
2 0
3 0
3k1k2 k3 e 0 A
or C1 = d (k k + k k + k k )
1 2
2 3
3 1
C2
k1e 0 ( A / 3) k 2 e 0 ( A / 3) k3 e 0 ( A / 3)
+
+
d
d
d
(k1 + k2 + k3 )e 0 A
=
3d
=
1
C1V 2
U1
2
=
U2 1
C V2
2 2
6µF
4µF
1
1
[ln(1 + ax )]d0
=
C e 0 KAa
1
1
Þ
=
ln(1 + ad )[a d << 1]
C e 0 KAa
=
59. (d) V1 + V2 = 10
and 4V1 = 6V2
On solving above equations, we get
V1 = 6 V
Charge on 4 mf,
q = CV1 = 4 × 6 = 24 mC.
60. (a) Equivalent capacitance in series combination (C’) is
given by
C1C2
1
1
1
=
+
Þ C' =
C ' C1 C2
C1 + C2
For parallel combination equivalent capacitance
C” = C1 + C2
For parallel combination
q = 10(C1 + C2)
q1 = 500 µC
500 = 10(C1 + C2)
C1 + C2 = 50µF
....(i)
For Series Combination–
q 2 = 10
C1C 2
( C1 + C2 )
C1C 2
From equation
50
C1C2 = 400
From equation (i) and (ii)
C1 = 10µF
C2 = 40µF
80 = 10
....(ii)
....(iii)
qæ 1
1 ö
61. (b) w = w f - vi = 2 çç C - C ÷÷
i ø
è f
(5 ´ 10)2 æ 1 1 ö
6
çè - ÷ø ´ 10
2
2 5
= 3.75 × 10–6J
62. (b) Capacitance of a capacitor with a dielectric of dielectric
constant k is given by
k Î0 A
C=
d
k Î0 AE
V
\ C=
Q E=
V
d
-12
k
´
´
´ 10-4 ´ 106
8.86
10
15 ´ 10-12 =
500
k = 8.5
=
P-275
Electrostatic Potential and Capacitance
63. (b) V =
Q
C
68. (b) As required equivalent capacitance should be
Q2
Q1
Ceq =
æ Q - Q2 ö
=ç 1
è 2C ÷ø
Q1 – Q2 (Q1 – Q2
æ 4 - 2ö
–
=ç
=1V
÷
2
2
è 2 ´1ø
64. (b) Energy stored in the system initially
1
Ui = CE 2
2
1 Q 2 (CE)2 1 CE 2
Uf =
=
=
2 Ceq 2 ´ 4C 2 4
[As Q = CE, and Ceq = 4C]
(
1
3 3
3 Q2
CE 2 ´ = CE 2 =
2
4 8
8 C
s
Q
65. (c) E = e = Ae
0
0
\Q = e0. E. A = 8.85 × 10–12 × 100 × 1
= 8.85 × 10–10C
C
1
66. (a) A
6
mF
13
Therefore three capacitors must be in parallel and 4 must
be in series with it.
1 é 1 ù é1 1 1 1ù
=
+ + + +
Ceq êë 3C úû êë C C C C úû
3C 6
= mF [ as C = 2 m F]
13 13
So, desired combination will be as below:
Ceq =
DU =
69. (b) W = – Du
= ( -1)
A
For series combination
e2 c k - 1
2 k
= 508 J
k1 Î0 A1 k 2 Î0 A 2 k 3 Î0 A 3 k Î0 A
+
+
=
d
d
d
d
or
B
71. (b)
7C
1
Þ 73 =
2
+C
3
d
y k
xa
Þ 14 C = 7 + 3 C
7
mF
11
- +
10mF
10 Î0 A/3 12 Î0 A/3 14 Î0 A/3 K Î0 A
+
+
=
d
d
d
d
Î0 A æ 10 12 14 ö K Î0 A
ç + + ÷=
d è 3 3 3ø
d
\ K = 12
1
1
1
= C +C
Ceq
1
2
67. (d)
2c
=
B
7
3
C
30mC
2kc
70. (c) Let dielectric constant of material used be K.
4
3
ÞC=
( ce ) 2 - ( ce ) 2
dx
a
y d
d
= Þy= x
x a
a
1
y
(d - y)
d
=
+
dy = (dx) Þ
e
e0 adx
dc
K
adx
a
0
1
y æy
ö
=
çè + d - y÷ø
dc e 0 abx k
From figure,
6mF
- +
- +
4mF
- +
2mF
As given in the figure, 6µF and 4µF are in parallel. Now
using charge conservation
6
´ 30 = 18µC
6+4
Since charge is asked on right plate therefore is +18µC
e 0 adx
ò dc = ò y
k
Charge on 6µF capacitor =
+d-y
d
or,
c = e 0 a.
a
dò
dy
æ1 ö
0 d + y ç - 1÷
èk ø
P-276
Physics
74. (a) During charging charge on the capacitor increases
with time. Charge on the capacitor C1 as a function of time,
Q = Q0(1 – e–t/RC)
Ceq
d
e0a 2 é æ
æ 1 ööù
=
êl n ç d + y çè - 1÷ø ÷ ú
ø û0
k
æ1 ö ë è
çè - 1÷ø d
k
- t RCeq ù
Q = Ceq E éë1 - e
û
æ
æ1 öö
ç d + d ç k - 1÷ ÷
k Î0 a 2
è
ø÷
=
lnç
d
÷
(1 - k ) d ç
ç
÷
è
ø
=
(Q Q0 = Ceq E)
2
k Î0 a 2
æ 1 ö k Î0 a lnk
lnç ÷ =
(1 - k ) d è k ø ( k - 1) d
k1
k1 k2 L/2
72. (Bonus)
k3 k4 L/2
C12
Þ
Þ
75. (d) The simplified circuit of the circuit given in question
as follows:
k2
C1
k3
C2
k4
C3
C4
Ceq
C34
é L
ù
L
´ L k 2 êÎ0 ´ L ú
2
û
2
. ë
d/2
d/2
é L
ù
êÎ0 2 ´ L ú
(k1 + k 2 ) ê
ú
ë d/2 û
k1 Î0
C12 =
C1C2
=
C1 + C2
k1k 2 Î0 L2
k1 + k 2 d
in the same way we get,
C12 =
C34 =
R
E
Both capacitor will have charge as they are connected in
series
k3k 4 Î0 L2
k3 + k 4 d
é kk
k k ùÎ L2
.. (i)
\ C eq = C12 + C34 = ê 1 2 + 3 4 ú 0
ë k1 + k 2 k3 + k 4 û d
k Î0 L2
... (ii)
d
on comparing equation (i) to equation (ii), we get
k1k 2 (k3 + k 4 ) + k3 k 4 (k1 + k 2 )
(k1 + k 2 )(k 3 + k 4 )
This does not match with any of the options so this must
be a bonus.
73. (a) Charge on Capacitor, Qi = CV
After inserting dielectric of dielectric constant
= K Qf = (kC) V
Induced charges on dielectric
Qind = Qf – Qi = KCV – CV
æ5 ö
( K - 1)CV = ç - 1 ÷ × 90 pF × 2V = 1.2nc
è3 ø
C
2m F
D E 4m F
5m F
2m F
B
5m F
The equivalent capacitance between C & D capacitors of
2mF, 5mF and 5 mF are in parallel.
\ CCD= 2 + 5 + 5 = 12 mF (Q In parallel grouping
Ceq = C1 + C2 +.... + Cn)
Similarly equivalent capacitance between E & B CEB
= 4 + 2 = 6mF
Now equivalent capacitance between A & B
1
1 1 1 5
= + + =
Ceq 6 12 6 12
12
= 2.4 mF
(Q In series grouping,
5
1
1
1
1
=
+
+ .......... +
)
C eq C1 C 2
Cn
76. (c) Given area of Parallel plate capacitor, A = 200 cm2
Separation between the plates, d = 1.5 cm
Force of attraction between the plates, F = 25 × 10–6N
F = QE
Þ C eq =
F=
Now if keq = K, Ceq =
k eq =
6m F
A
s
Q
Q2
=
(E due to parallel plate =
)
2 Î0 A2 Î0
2 A Î0
But Q = CV =
\
=
F=
(Î0 AV )2
+ –
d 2 ´ 2 A Î0
2
(Î0 A) ´ V
2
2
d ´ 2 ´ ( A Î0 )
or, 25 ´ 10-6 =
Þ V=
d = 1.5 cm
Î0 A(V )
d
=
V
(Î0 A) ´ V
2
d2 ´ 2
(8.85 ´ 10-12 ) ´ (200 ´ 10 -4 ) ´ V 2
2.25 ´ 10-4 ´ 2
25 ´ 10-6 ´ 2.25 ´ 10-4 ´ 2
8.85 ´ 10-12 ´ 200 ´ 10-4
» 250 V
P-277
Electrostatic Potential and Capacitance
77. (a) The sum of final charges on C2 and C3 is 36 µC.
78. (b) To get a capacitance of 2 m F arrangement of
capacitors of capacitance 1 mF as shown in figure 8
capacitors of 1mF in parallel with four such branches in
series i.e., 32 such capacitors are required.
é æ 12 ö
ù
Charge on C1 is q1 = ê ç
÷ø ´ 8 ú ´ 4 = 24mC
è
4
12
+
ë
û
The voltage across CP is VP =
4
×8 = 2V
4 + 12
\ Voltage across 9mF is also 2V
\ Charge on 9mF capacitor = 9 × 2 = 18mC
\ Total charge on 4 mF and 9mF = 42mC
m
m
1
1 1 1 1
= + + +
Ceq 8 8 8 8
m
m
\E =
\ Ceq = 2 mF
79. (c) Before introducing a slab capacitance of plates
e A
C1 = 0
3
If a slab of dielectric constant K is introduced between
plates then
e A
Ke 0A
then C1' = 0
C=
2.4
d
'
C1 and C1 are in series hence,
C1'
e A e A
k 0 . 0
e0A
3 2.4
=
e0 A e 0 A
3
+
k
3
2.4
Slab
3 k = 2.4 k + 3
0.6 k = 3
Hence, the dielectric constant of slap is given by,
30
k=
=5
6
Q2
80. (d) Energy of sphere =
2C
16 ´ 10 -12
4.5 =
2C
16 ´ 10-12
C=
= 4pe 0R
9
(capacity of spherical conductor)
R=
16 ´ 10-12
1
´
9
4pe 0
= 9 ´ 109 ´
1
9
Q 4pe = 9 ´ 10
0
16
´ 10-12 = 16 mm
9
3µF
81. (a) 4µF
9µF
C1 = 4µF
r2
= 9 × 109 ×
42 ´ 10-6
= 420 NC–1
30 ´ 30
82. (a) Capacitors 2mF and 2mF are parallel, their equivalent
= 4 mF
6mF and 12 mF are in series, their equivalent = 4 mF
3
mF
8
And 4mF (12 & 6 mF) and 4mF in parallel = 4 + 4 = 8mF
Now 4mF (2 and 2 mF) and 8mF in series =
8mF in series with 1mF =
Now Ceq =
With C –
1
8
+ 1 Þ mF
8
9
8 8 32
+ =
9 3
9
Ceq of circuit =
32
9
1
1 9
32
= +
=1Þ C =
C eq C 32
23
83. (d) To get effective capacitance of 6 mF two capacitors of
4 mF each connected in sereies and one of 4 mF capacitor
in parallel with them.
4mF
4mF
4mF
Two capacitances in series
1
1
1
1 1 1
\ =
+
= + =
C C1 C2 4 4 2
1 capacitor in parallel
\ C eq = C3 + C = 4 + 2 = 6 mF
84. (d)
12µF = CP
KQ
Q1
1mF
Q2
2mF
Q
C
Þ
2µF
8V
8V
From figure, Q2 =
2
2
Q = Q
2 +1
3
P-278
Physics
æ C ´3 ö
Q= Eç
÷
è C + 3ø
2 æ 3CE ö 2CE
\ Q2 = ç
÷=
3è C + 3ø C +3
Therefore graph d correctly dipicts.
Charge
1mF
85. (a) when switch is closed
–10 C
5V
3mF
C
15 C
c=
ld
æ
ld ö
ln ç 1 +
÷
è K0 ø
. C0
sö
æ
çè here, C0 = ÷ø
d
89. (c)
90. (d) Possible combination of capacitors
(i) Three capacitors in series combination
5V
+10 C
–15 C
When switch is open
–12 C
s æ
ld ö
= l ln ç1 + K ÷
0ø
è
Now it is given that capacitance of vacuum = C0.
Q
Thus, C =
V
s.s
=
(Let surface area of plates = s)
v
s.s
=
s æ
ld ö
ln ç1 +
÷
l è K0 ø
d
1
= sl.
(Q in vacuum e0 =1)
d æ
ld ö
ln ç1 +
÷
è K0 ø
+12 C
3µF
3µF
3µF
1
1 1 1
= + +
Ceq 3 3 3
1
= 1mF
Ceq
(ii) Three capacitors in parallel combination
\
6V
4V
4V
6V
3µF
+12 C
–12 C
Charge of 5mc flows from b to a
86. (a) Electric field in presence of dielectric between the two
plates of a parallel plate capaciator is given by,
s
E=
Ke 0
Then, charge density
s = Ke0E
= 2.2 × 8.85 × 10–12 × 3 × 104
» 6 × 10–7 C/m2
87. (d)
88. (c) The value of dielectric constant is given as,
3µF
3µF
Ceq = 3 + 3 + 3 = 9 µF
(iii) Two capacitors in parallel and one is in series
3µF
3µF
3µF
Ceq = 2µF
(iv) Two capacitors in series and one is in parallel
K = K 0 + lx
d
d
0
0
And, V = ò Edr Þ V = ò
d
s
dx
K
1
s
= s ò ( K + l x ) dx = éëln ( K 0 + ld ) - ln K 0 ùû
0
l
0
Ceq = 4.5 µF
91. (b) C0 =
k Î0 A
d
P-279
Electrostatic Potential and Capacitance
C=
k Î0 2 2k Î0 A 4 k Î0 A
+
=
3d
3d
3 d
4 k Î0 A
C
4
\
= 3 d
=
k Î0 A
C0
3
d
Q It
92. (c) As, C = =
V V
V I
2
= =
Þ
t C 1 ´ 10 -6
= 2 × 106 V/s
93. (c) When charged particle enters perpendicularly in an
electric field, it describes a parabolic path
2
1 æ QE öæ x ö
y= ç
÷ç ÷
2 è m øè 4 ø
This is the equation of parabola.
y
E
P(x, y)
u
x
94. (d) The discharging of a capacitor is given as
q = q 0 exp [ - t / RC ]
RC = time constant = t
q = q 0 e-t/ t
If e is the capacitance of the capacitor
q = CV and q = CV0
Thus, CV = CV0 e t / t
-t / t
…(i)
V = V0 e
From the graph (given in the problem
when t = 0.5, V = 25 i.e.,
V0 = 25 volt.
and when t = 200, V = 5 volt
Thus equation (i) becomes
5 = 25e -200/
t
Þ 1 / 5 = e -200/ t
Taking loge on both sides
1
loge = -200 / t Þ – 200 = log e 5
5
t
200
t=
log e 5
200
200
=
or t =
æ 10 ö log e 10 - log e 2
log e ç ÷
è 2ø
200
200
t=
=
= 124.300
2.302 - 0.693 1.609
Which lies between 100 s and 150 s
95. (c) The dielectric constant of the gas is 1.01
96. (d) Capacitance of sphere is given by :
C = 4 p Î0 r
If, C = 1F then radius of sphere needed:
C
1
=
r=
4p Î0 4p´ 8.85 ´10 -12
1012
= 9 × 109 m
4p´ 8.85
9 × 109 m is very large, it is not possible to obtain such
a large sphere. Infact earth has radius 6.4 × 106 m only
and capacitance of earth is 711mF.
97. (d) Equivalent capacitance of n2 number of capacitors
each of capacitance C2 in parallel = n2C2
Equivalent capacitance of n 1 number of capacitors each
of capacitances C1 in series.
C
Capacitance of each is C1 = 1
n1
According to question, total energy stored in both the
combinations are same
or, r =
i.e.,
1 æ C1 ö
1
( 4V ) 2 = ( n2 C2 ) V 2
ç
÷
2 è n1 ø
2
16C1
n1 n2
98. (c) Charge (or current) always flows from higher potential
to lower potential.
\ C2 =
Potential=
Charge
Capacitance
q2
99. (c) Initial energy of capacitor, E1 = 1
2C
Final energy of capacitor,
2
æ q1 ö
q12
1
E2 = E1 =
=ç 2÷
2
4C ç
÷
è 2C ø
\ t1 = time for the charge to reduce to
value
1
of its initial
2
1
and t2 = time for the charge to reduce to of its initial
4
value
t
æ q2 ö
We have, q2 = q1e-t / CR Þ ln ç q ÷ = CR
è 1ø
1
t
æ
ö
1
...(1)
\ ln ç
÷=
è 2 ø CR
æ 1 ö -t2
and ln ç ÷ =
...(2)
è 4 ø CR
1
æ 1 ö
ln æç ö÷
ln ç
÷
1
1
t1
è
2ø
2
ø =
By (1) and (2) ,
=
= è
1
2
4
æ
ö
1
t2
2ln ç ÷
ln æç ö÷
è2ø
è4ø
P-280
Physics
103. (c) Applying conservation of energy,
Electric potential energy of capacitor = heat absorbed
100. (c)
The capacitance with air between the plates
e0 A
= 9pF
d
On introducing two dielectric between the plates, the given
capacitance is equal to two capacitances connected in
series where
k Î A 3Î A
C1 = 1 0 = 0
d /3
d /3
3 ´ 3Î0 A
9 Î0 A
=
=
d
d
and
3k2Î0 A
kÎ A
C2 = 2 0 =
2d / 3
2d
3 ´6 Î0 A
9 Î0 A
=
=
2d
d
The equivalent capacitance Ceq is
C=
1
1
1
=
+
Ceq C1 C2
=
d
d
2d
+
=
9 Î0 A 9 Î0 A 9 Î0 A
9 Î0 A 9
= ´ 9 pF = 40.5 pF
2 d
2
101. (a) The potential energy of a charged capacitor is given
\ Ceq =
(
)
-18 2
8 ´ 10
= 32 × 10–32 J
\ U = 1´
2 100 ´ 10 -6
106. (b) In parallel, equivalent capacitance of n capacitor of
capacitance C
C¢ = nC
Energy stored in this capacitor
1 1 2
E= C V
2
1
1
2
2
Þ E = (nC )V = nCV
2
2
C
C
n times
C
2
Q
.
2C
When a dielectric slab is introduced between the plates
by U =
2m. s. Dt
1
CV 2 = m. s Dt ; V =
C
2
104. (b) The capacitance without aluminium foil is
e A
C= 0
d
Here, d is distance between the plates of a capacitor
A = Area of plates of capacitor
When an aluminium foil of thickness t is introduced
between the plates.
e A
Capacitance, C¢ = 0
d –t
If thickness of foil is negligible 50 d – t ~ d. Hence, C = C¢.
105. (d) The work done is stored in the form of potential energy
which is given by
1 Q2
U=
2 C
the energy is given by
Q2
, where K is the dielectric
2KC
constant.
Again, when the dielectric slab is removed slowly its
energy increases to initial potential energy. Thus, work
done is zero.
102. (b) As n plates are joined alternately positive plate of all
(n – 1) capacitor are connected to one point and negative
plate of all (n – 1) capacitors are connected to other point.
It means (n – 1) capacitors joined in parallel.
\ Resultant capacitance = (n – 1)C
V
V
Alternatively
Each capacitor has a potential difference of V between the
plates.
So, energy stored in each capacitor
1
CV 2 .
2
\ Energy stored in n capacitor
=
é1
2ù
= ê CV ú ´ n
ë2
û
107. (a) Capacitance of spherical conductor = 4pE0R
Here, R is radius of conductor
1
´ 1 = 1.1 ´ 10-10 F
\ C = 4p Î0 R =
9 ´109
17
P-281
Current Electricity
Current Electricity
Electric Current, Drift of
TOPIC 1 Electrons, Ohm's Law,
Resistance and Resistivity
1.
2.
A circuit to verify Ohm’s law uses ammeter and voltmeter
in series or parallel connected correctly to the resistor.
In the circuit :
[Sep. 06, 2020 (II)]
(a) ammeter is always used in parallel and voltmeter is
series
(b) Both ammeter and voltmeter must be connected in
parallel
(c) ammeter is always connected in series and voltmeter
in parallel
(d) Both, ammeter and voltmeter must be connected in
series
Consider four conducting materials copper, tungsten,
mercury and aluminium with resistivity rC, rT, rM and rA
respectively. Then :
[Sep. 02, 2020 (I)]
(a) rC > r A > rT
(b) r M > r A > rC
(c)
3.
r A > rT > rC
4.
5.
(d) r A > rM > rC
(a) The emf of the battery is 1.5 V and its internal
resistance is 1.5 W
(b) The value of the resistance R is 1.5 W
(c) The potential difference across the battery is 1.5 V when
it sends a current of 1000 mA
(d) The emf of the battery is 1.5 V and the value of R is 1.5 W
A current of 5 A passes through a copper conductor
(resistivity) = 1.7×10 – 8Wm) of radius of cross-section
5 mm. Find the mobility of the charges if their drift
velocity is 1.1×10 – 3 m/s.
[10 Apr. 2019 I]
(a) 1.8m2/Vs
(b) 1.5 m2/Vs
(c) 1.3 m2/Vs
(d)
1.0m2/Vs
In an experiment, the resistance of a material is plotted as
a function of temperature (in some range). As shown in
the figure, it is a straight line.
[10 Apr. 2019 I]
One may canclude that:
To verify Ohm’s law, a student connects the voltmeter
across the battery as, shown in the figure. The measured
voltage is plotted as a function of the current, and the
following graph is obtained :
[12 Apr. 2019 I]
(a)
6.
If Vo is almost zero, identify the correct statement:
R0
2
(b) R(T) = R 0 e - T0 /T
T2
- T 2 /T02
2
2
2
(d) R(T) = R 0 eT /T0
Space between two concentric conducting spheres of radii
a and b (b > a) is filled with a medium of resistivity r. The
resistance between the two spheres will be :
[10 Apr. 2019 II]
(c)
R(T) = R 0 e
(a)
r æ1 1ö
ç - ÷
4p è a b ø
(b)
r æ1 1ö
ç - ÷
2p è a b ø
r æ1 1ö
r æ1 1ö
(d)
ç + ÷
ç + ÷
4p è a b ø
2p è a b ø
In a conductor, if the number of conduction electrons
per unit volume is 8.5 × 1028 m –3 and mean free time is
25 fs (femto second), it’s approximate resistivity is:
(m e = 9.1 × 10–31 kg)
[9 Apr. 2019 II]
(a) 10–6 W m
(b) 10–7 Wm
(c) 10–8 Wm
(d) 10–5 Wm
(c)
7.
R(T) =
P-282
Physics
A 200 W resistor has a certain color code. If one replaces
the red color by green in the code, the new resistance will
be :
[8 April 2019 I]
(a) 100 W (b) 400 W (c) 300 W
(d) 500 W
9. The charge on a capacitor plate in a circuit, as a function of
time, is shown in the figure:
[12 Jan. 2019 II]
6
5
4
q(µc) 3
2
0
6
8
2
4
t(s)
What is the value of current at t = 4 s ?
(a) Zero
(b) 3 µA
(c) 2 µA
(d) 1.5 µA
10. A resistance is shown in the figure. Its value and tolerance
are given respectively by:
[9 Jan. 2019 I]
8.
16. A uniform wire of length l and radius r has a resistance of
r
100 W. It is recast into a wire of radius . The resistance
2
of new wire will be :
[Online April 9, 2017]
(a) 1600 W (b) 400 W (c) 200 W
(d) 100 W
17. When 5V potential difference is applied across a wire of
length 0.1 m, the drift speed of electrons is 2.5 × 10–4 ms–1.
If the electron density in the wire is 8 × 1028 m–3, the
resistivity of the material is close to :
[2015]
(a) 1.6 × 10–6 Wm
(b) 1.6 × 10–5 Wm
(c) 1.6 × 10–8 Wm (d)
1.6 × 10–7 Wm
18. Suppose the drift velocity nd in a material varied with the
applied electric field E as nd µ E . Then V – I graph for
a wire made of such a material is best given by :
[Online April 10, 2015]
V
V
(a)
(b)
I
(a) 270 W, 10%
(b) 27 kW, 10%
(c) 27 kW, 20%
(d) 270 W, 5%
11. Drift speed of electrons, when 1.5 A of current flows in a
copper wire of cross section 5 mm2, is v. If the electron
density in copper is 9 × 1028/m3 the value of v in mm/s
close to (Take charge of electron to be = 1.6 × 10–19C)
[9 Jan. 2019 I]
(a) 0.02
(b) 3
(c) 2
(d)
0.2
12. A copper wire is stretched to make it 0.5% longer. The
percentage change in its electrical resistance if its volume
remains unchanged is:
[9 Jan. 2019 I]
(a) 2.0%
(b) 2.5%
(c) 1.0%
(d)
0.5%
13. A carbon resistance has following colour code. What is
the value of the resistance?
[9 Jan. 2019 II]
GOY Golden
(a) 530 kW ± 5% (b)
5.3 kW ± 5%
(c) 6.4 MW ± 5%(d)
64 MW ± 10%
14. A heating element has a resistance of 100W at room
temperature. When it is connected to a supply of 220 V,
a steady current of 2 A passes in it and temperature is
500°C more than room temperature. What is the
temperature coefficient of resistance of the heating
element?
[Online April 16, 2018]
(a) 1 × 10–4°C–1
(b) 5 × 10–4°C–1
(c) 2 × 10–4°C–1
(d) 0.5 × 10–4°C–1
15. A copper rod of cross-sectional area A carries a uniform
current I through it. At temperature T, if the volume charge
density of the rod is r, how long will the charges take to
travel a distance d ?
[Online April 15, 2018]
(a)
2rdA
IT
(b)
2rdA
I
(c)
rdA
I
(d)
rdA
IT
I
V
V
(c)
(d)
I
I
19. Correct set up to verify Ohm’s law is :
[Online April 23, 2013]
A
V
(a)
(b)
A
V
A
(c)
V
V
(d)
A
P-283
Current Electricity
20. The resistance of a wire is R. It is bent at the middle by 180°
and both the ends are twisted together to make a shorter wire.
The resistance of the new wire is [Online May 26, 2012]
(a) 2 R
(b) R/2
(c) R/4
(d) R/8
21. If a wire is stretched to make it 0.1% longer, its resistance
will :
[2011]
(a) increase by 0.2%
(b) decrease by 0.2%
(c) decrease by 0.05%
(d) increase by 0.05%
DIRECTIONS : Question No. 22 and 23 are based on the
following paragraph.
Consider a block of conducting material of resistivity ‘r’ shown in
the figure. Current ‘I’ enters at ‘A’ and leaves from ‘D’. We apply
superposition principle to find voltage ‘DV’ developed between ‘B’
and ‘C’. The calculation is done in the following steps:
(i) Take current ‘I’ entering from ‘A’ and assume it to spread
over a hemispherical surface in the block.
(ii) Calculate field E(r) at distance ‘r’ from A by using Ohm’s
law E = r j, where j is the current per unit area at ‘r’.
(iii) From the ‘r’ dependence of E(r), obtain the potential V(r) at r.
(iv) Repeat (i), (ii) and (iii) for current ‘I’ leaving ‘D’ and
superpose results for ‘A’ and ‘D’.
I
a
A
I
DV
a
b
B
C
27. The length of a given cylindrical wire is increased by 100%.
Due to the consequent decrease in diameter the change in
the resistance of the wire will be
[2003]
(a) 200%
(b) 100%
(c) 50%
(d) 300%
TOPIC 2 Combination of Resistances
28. In the given circuit diagram, a wire is joining points B and
D. The current in this wire is:
[9 Jan. 2020 I]
(a) 0.4A
(b) 2A
(c) 4A
(d) zero
29. The series combination of two batteries, both of the same
emf 10 V, but different internal resistance of 20 W and 5
W, is connected to the parallel combination of two
resistors 30 W and R W. The voltage difference across
the battery of internal resistance 20 W is zero, the value
of R (in W) is _________.
[NA. 8 Jan. 2020 II]
30. The current I1 (in A) flowing through 1 W resistor in the
following circuit is:
[7 Jan. 2020 I]
D
22. DV measured between B and C is
[2008]
rI
rI
rI
rI
–
–
(a)
(b)
pa p(a + b)
a (a + b)
rI
rI
rI
–
(c)
(d)
2pa 2p(a + b)
2p(a - b)
23. For current entering at A, the electric field at a distance ‘r’
from A is
[2008]
rI
rI
rI
rI
(a)
(b)
(c)
(d)
2
2
2
8pr
r
2pr
4pr 2
24. The resistance of a wire is 5 ohm at 50°C and 6 ohm at
100°C. The resistance of the wire at 0°C will be
[2007]
(a) 3 ohm (b) 2 ohm (c) 1 ohm
(d) 4 ohm
25. A material 'B' has twice the specific resistance of 'A'. A
circular wire made of 'B' has twice the diameter of a wire
made of 'A'. then for the two wires to have the same
resistance, the ratio lB/lA of their respective lengths must
be
[2006]
1
1
(a) 1
(b)
(c)
(d) 2
2
4
26. An electric current is passed through a circuit containing
two wires of the same material, connected in parallel. If the
4
2
lengths and radii are in the ratio of and , then the ratio
3
3
of the current passing through the wires will be [2004]
(a) 8/9
(b) 1/3
(c) 3
(d) 2
(a) 0.4
(b) 0.5
(c) 0.2
(d) 0.25
31. A wire of resistance R is bent to form a square ABCD as
shown in the figure. The effective resistance between E
and C is: (E is mid-point of arm CD) [9 April 2019 I]
A
B
D
(a) R
(b)
C
E
7
R
64
(c)
3
R
4
(d)
1
R
16
32. A metal wire of resistance 3 W is elongated to make a uniform
wire of double its previous length. This new wire is now
bent and the ends joined to make a circle. If two points on
the circle make an angle 60° at the centre, the equivalent
resistance between these two points will be: [9 Apr. 2019 II]
(a)
12
W
5
(b)
5
W
2
(c)
5
W
3
(d)
7
W
2
P-284
Physics
33. In the figure shown, what is the current (in Ampere) drawn
from the battery? You are given :
[8 Apr. 2019 II]
R1 = 15 W, R2 = 10 W, R3 = 20 W, R4 = 5 W, R5 = 25 W,
R6 = 30 W, E = 15 V
A
B
5R
5R
(c)
(d) 3 R
2
3
38. In the given circuit diagram when the current reaches steady
state in the circuit, the charge on the capacitor of
capacitance C will be :
[2017]
r2
(a) CE +
(r r2 )
r1
(b) CE
(r1 + r)
(c) CE
r
(d) CE 1
(r2 + r)
(a) 2 R
(a) 13/24
(b) 7/18
(c) 9/32
(d) 20/3
34. A uniform metallic wire has a resistance of 18 W and is
bent into an equilateral triangle. Then, the resistance
between any two vertices of the triangle is:
[10 Jan. 2019 I]
(a) 4 W
(b) 8 W
(c) 12 W
(d) 2 W
35. The actual value of resistance R, shown in the figure is 30
W. This is measured in an experiment as shown using the
V
standard formula R = , where V and I are the reading of
I
the voltmeter and ammeter, respectively. If the measured
value of R is 5% less, then the internal resistance of the
voltmeter is:
[10 Jan. 2019 II]
V
A
39.
In the above circuit the current in each resistance is
[2017]
(a) 0.5A
(b) 0 A
(c) 1 A
(d) 0.25 A
1W
A1
40.
1W
V
R3
R4
R1
A3
4W
1W
(a) 600 W (b) 570 W (c) 35 W
(d) 350 W
36. In the given circuit the internal resistance of the 18 V cell
is negligible. If R1 = 400W, R3 = 100 W and R4 = 500 W
and the reading of an ideal voltmeter across R4 is 5 V,
then the value of R2 will be:
[9 Jan. 2019 II]
1W
A2
B
9V
0.5W
R
(b)
4W
4W
1W
1W
A 9 V battery with internal resistance of 0.5 W is connected
across an infinite network as shown in the figure. All
ammeters A1, A2, A3 and voltmeter V are ideal.
Choose correct statement.
[Online April 8, 2017]
(a) Reading of A1 is 2 A (b) Reading of A1 is 18 A
(c) Reading of V is 9 V
(d) Reading of V is 7 V
41. Six equal resistances are connected between points P, Q
and R as shown in figure. Then net resistance will be
maximum between :
[Online April 25, 2013]
P
R2
(a) 300 W
(c) 550 W
18 V
r
r
(b) 450 W
r
r
(d) 230 W
37. In the given circuit all resistances are of value R ohm each.
The equivalent resistance between A and B is :
[Online April 15, 2018]
¥
r
Q
(a) P and R
(c) Q and R
r
R
(b) P and Q
(d) Any two points
P-285
Current Electricity
42. A letter 'A' is constructed of a uniform wire with resistance
1.0 W per cm, The sides of the letter are 20 cm and the cross
piece in the middle is 10 cm long. The apex angle is 60 . The
resistance between the ends of the legs is close to:
[Online April 9, 2013]
(a) 50.0 W (b) 10 W
(c) 36.7 W (d) 26.7 W
43. Two conductors have the same resistance at 0°C but their
temperature coefficients of resistance are a1 and a2. The
respective temperature coefficients of their series and
parallel combinations are nearly
[2010]
a1 + a 2
a1 + a 2
, a1 + a 2
(a)
(b) a1 + a 2 ,
2
2
a1a 2
a
+
a
a
+
2 , 1 a2
(c) a1 + a 2 ,
(d) 1
a1 + a 2
2
2
44. The current I drawn from the 5 volt source will be [2006]
10W
5W
10W
I
20W
10W
+–
48.
5 volt
5W
6W
(a) 0.71 A from positive to negative terminal
(b) 0.42 A from positive to negative terminal
(c) 0.21 A from positive to negative terminal
(d) 0.36 A from negative to positive terminal
49. In the circuit, given in the figure currents in different branches
and value of one resistor are shown. Then potential at point
B with respect to the point A is :
[Sep. 05, 2020 (II)]
2V
D
E
B
1A
2W
F
C
2A
2A
1V
(a) + 2 V
(b) – 2 V
(c) – 1 V
(d) + 1 V
50. The value of current i1 flowing from A to C in the circuit
diagram is :
[Sep. 04, 2020 (II)]
8V
3W
i
(a) 4 A
(b) 2 A
(c) 1 A
(d) 6 A
46. The resistance of the series combination of two resistances
is S. when they are joined in parallel the total resistance is
P. If S = nP then the minimum possible value of n is
[2004]
(a) 2
(b) 3
(c) 4
(d) 1
47. A 3 volt battery with negligible internal resistance is
connected in a circuit as shown in the figure. The current
I, in the circuit will be
[2003]
3W
3W
B
2W
2W
5W
4W
4W i1
A
C
2W
(a) 2 A
51.
i
2W
(b) 4 A
B
D
(c) 1 A
(d) 5 A
60W
40W
A
C
90W
110W
D
3W
(b) 1.5 A
10V
2W
4W
In the figure shown, the current in the 10 V battery is
close to :
[Sep. 06, 2020 (II)]
1.5W
(a) 1 A
10W
20V
2W
3V
Kirchhoff 's Laws, Cells,
Thermo e.m.f. & Electrolysis
A
(a) 0.33 A (b) 0.5 A
(c) 0.67 A
(d) 0.17 A
45. The total current supplied to the circuit by the battery is
[2004]
6V
TOPIC 3
(c) 2 A
(d) 1/3 A
40 V
Four resistances 40 W, 60 W, 90 W and 110 W make the
arms of a quadrilateral ABCD. Across AC is a battery of
emf 40 V and internal resistance negligible. The potential
difference across BD in V is __________.
[NA. Sep. 04, 2020 (II)]
P-286
Physics
52. An ideal cell of emf 10 V is connected in circuit shown in
figure. Each resistance is 2 W. The potential difference (in
V) across the capacitor when it is fully charged is
__________.
[Sep. 02, 2020 (II)]
R1
R3
R2
C
R5
57. In the circuit shown, the potential difference between A
and B is :
[11 Jan. 2019 II]
1V
1W
M
5W
A
1W
2V
1W
3V
D
C
B
N
R4
10 V
53. In the given circuit, an ideal voltmeter connected across
the 10 W resistance reads 2V. The internal resistance r, of
each cell is :
[10 Apr. 2019 I]
(a) 1 W
(b) 0.5 W
(c) 1.5 W
(d) 0 W
54. For the circuit shown, with R1 = 1.0 W, R2 = 2.0 W, E1 = 2 V
and E2 = E3 = 4 V, the potential difference between the
points ‘a’ and ‘b’ is approximately (in V) : [8 April 2019 I]
(a) 2.7
(b) 2.3
(c) 3.7
(d)
3.3
55. A cell of internal resistance r drives current through an
external resistance R. The power delivered by the cell to
the external resistance will be maximum when :
[8 Apr. 2019 II]
(a) R = 0.001 r
(b) R = 1000 r
(c) R = 2r
(d) R = r
56. In the given circuit diagram, the currents, I1 = – 0.3 A, I4 = 0.8
A and I5 = 0.4 A, are flowing as shown. The currents I 2, I3
and I6, respectively, are :
[12 Jan. 2019 II]
I
Q
P
6
I3
(a) 1 V
(b) 2 V
(c) 3 V
(d) 6 V
58. In the given circuit the cells have zero internal resistance.
The currents (in Amperes) passing through resistance
R1 and R2 respectively, are:
[10 Jan. 2019 I]
(a) 1, 2
(b) 2, 2
(c) 0.5, 0
(d) 0, 1
59. When the switch S, in the circuit shown, is closed then
the valued of current i will be:
[9 Jan. 2019 I]
(a) 3A
(b) 5A
(c) 4A
(d)
2A
60. Two batteries with e.m.f. 12 V and 13 V are connected in
parallel across a load resistor of 10W. The internal
resistances of the two batteries are 1W and 2W respectively.
The voltage across the load lies between:
[2018]
(a) 11.6 V and 11.7 V
(b) 11.5 V and 11.6 V
(c) 11.4 V and 11.5 V
(d) 11.7 V and 11.8 V
61. In the circuit shown, the current in the 1W resistor is:
[2015]
6V
P 2W
1W
I5
10W
I2
I1
S
I4
(a) 1.1 A, – 0.4 A, 0.4 A
R
(b) 1.1 A, 0.4 A, 0.4 A
(c) 0.4 A, 1.1 A, 0.4 A
(d) –0.4 A, 0.4 A, 1.1 A
3W
(a) 0.13 A, from Q to P
(c) 1.3A from P to Q
9V
W 3W
(b) 0.13 A, from P to Q
(d) 0A
P-287
Current Electricity
62. In the electric network shown, when no current flows
through the 4W resistor in the arm EB, the potential
difference between the points A and D will be :
[Online April 11, 2015]
2W
F
D
E
r = 0.5 W
E1
E2
2V
4W
2W
R
R
4V
A
B
9V
3V
C
(a) 6 V
(b) 3 V
(c) 5 V
(d) 4 V
63. The circuit shown here has two batteries of 8.0 V and 16.0
V and three resistors 3 W, 9 W and 9 W and a capacitor of
5.0 mF.
[Online April 12, 2014]
I
3W
5 mF
P2
5V
2W
9W
9W
8.0 V
(a) 5.5 W
(b) 3.5 W
(c) 4.5 W
(d) 2.5 W
67. A 5V battery with internal resistance 2W and a 2V battery
with internal resistance 1W are connected to a 10W resistor
as shown in the figure.
[2008]
68.
5W
69.
20 W 10 W
2V
1W
16.0 V
I2
I1
How much is the current I in the circuit in steady state?
(a) 1.6 A
(b) 0.67 A
(c) 2.5 A
(d) 0.25 A
64. In the circuit shown, current (in A) through 50 V and 30 V
batteries are, respectively.
[Online April 11, 2014]
50 V
10W
30 V
5W
(a) 2.5 and 3
(b) 3.5 and 2
(c) 4.5 and 1
(d) 3 and 2.5
65. A d.c. main supply of e.m.f. 220 V is connected across a
storage battery of e.m.f. 200 V through a resistance of 1W.
The battery terminals are connected to an external resistance
‘R’. The minimum value of ‘R’, so that a current passes
through the battery to charge it is: [Online April 9, 2014]
(a) 7 W
(b) 9 W
(c) 11 W
(d) Zero
66. A dc source of emf E1 = 100 V and internal resistance r = 0.5
W, a storage battery of emf E2 = 90 V and an external
resistance R are connected as shown in figure. For what
value of R no current will pass through the battery ?
[Online April 22, 2013]
70.
71.
The current in the 10W resistor is
(a) 0.27 A P2 to P1
(b) 0.03 A P1 to P2
(c) 0.03 A P2 to P1
(d) 0.27 A P1 to P2
A battery is used to charge a parallel plate capacitor till the
potential difference between the plates becomes equal to
the electromotive force of the battery. The ratio of the
energy stored in the capacitor and the work done by the
battery will be
[2007]
(a) 1/2
(b) 1
(c) 2
(d) 1/4
The Kirchhoff's first law (Si = 0) and second law (SiR = SE),
where the symbols have their usual meanings, are
respectively based on
[2006]
(a) conservation of charge, conservation of momentum
(b) conservation of energy, conservation of charge
(c) conservation of momentum, conservation of charge
(d) conservation of charge, conservatrion of energy
A thermocouple is made from two metals, Antimony and
Bismuth. If one junction of the couple is kept hot and the
other is kept cold, then, an electric current will
[2006]
(a) flow from Antimony to Bismuth at the hot junction
(b) flow from Bismuth to Antimony at the cold junction
(c) now flow through the thermocouple
(d) flow from Antimony to Bismuth at the cold junction
Two sources of equal emf are connected to an external
resistance R. The internal resistance of the two sources are
R1and R2 (R1 > R1). If the potential difference across the
source having internal resistance R2 is zero, then
[2005]
P-288
(a) R = R2 - R1
(b) R = R2 ´ ( R1 + R2 ) /( R2 - R1 )
(c) R = R1R2 /( R2 - R1 )
(d) R = R1R2 /( R1 - R2 )
72. Two voltameters, one of copper and another of silver, are
joined in parallel. When a total charge q flows through the
voltameters, equal amount of metals are deposited. If the
electrochemical equivalents of copper and silver are Z1
and Z2 respectively the charge which flows through the
silver voltameter is
[2005]
Z
q
q
Z2
(a)
(d) q 1
(c) q
`(b)
Z
Z
Z2
Z1
1+ 1
1+ 2
Z2
Z1
73. An energy source will supply a constant current into the
load if its internal resistance is
[2005]
(a) very large as compared to the load resistance
(b) equal to the resistance of the load
(c) non-zero but less than the resistance of the load
(d) zero
74. The thermo emf of a thermocouple varies with the
temperature q of the hot junction as E = aq + bq2 in volts
where the ratio a/b is 700°C. If the cold junction is kept at
0°C, then the neutral temperature is
[2004]
(a) 1400°C
(b) 350°C
(c) 700°C
(d) No neutral temperature is possible for this termocouple.
75. The electrochemical equivalent of a metal is 3.35 × 10–7 kg
per Coulomb. The mass of the metal liberated at the cathode
when a 3A current is passed for 2 seconds will be
[2004]
(a) 6.6×1057kg
(b) 9.9×10–7 kg
(c) 19.8×10–7 kg
(d) 1.1×10–7 kg
76. The thermo e.m.f. of a thermo-couple is 25 mV/°C at room
temperature. A galvanometer of 40 ohm resistance, capable
of detecting current as low as 10–5 A, is connected with
the thermo couple. The smallest temperature difference that
can be detected by this system is [2003]
(a) 16°C
(b) 12°C
(c) 8°C
(d) 20°C
77. The negative Zn pole of a Daniell cell, sending a constant
current through a circuit, decreases in mass by 0.13g in 30
minutes. If the electeochemical equivalent of Zn and Cu
are 32.5 and 31.5 respectively, the increase in the mass of
the positive Cu pole in this time is
[2003]
(a) 0.180 g (b) 0.141g (c) 0.126 g (d) 0.242 g
78. The mass of product liberated on anode in an
electrochemical cell depends on
[2002]
(a) (It)1/2
(b) It
(c) I/t
(d) I2 t
(where t is the time period for which the current is passed).
Physics
TOPIC 4 Heating Effect of Current
79. An electrical power line, having a total resistance of 2 W,
delivers 1 kW at 220 V. The efficiency of the transmission
line is approximately :
[Sep. 05, 2020 (I)]
(a) 72%
(b) 91%
(c) 85%
(d) 96%
80. Model a torch battery of length l to be made up of a thin
cylindrical bar of radius ‘a’ and a concentric thin cylindrical
shell of radius ‘b’ filled in between with an electrolyte of resistivity
r (see figure). If the battery is connected to a resistance of
value R, the maximum Joule heating in R will take place for :
[Sep. 03, 2020 (I)]
r
l
a
b
r
æbö
ln ç ÷
2pl è a ø
2r æ b ö
r
b
(c) R = l n æç ö÷
(d) R =
ln ç ÷
pl è a ø
pl è a ø
81. In a building there are 15 bulbs of 45 W, 15 bulbs of 100 W,
15 small fans of 10 W and 2 heaters of 1 kW. The voltage
of electric main is 220 V. The minimum fuse capacity (rated
value) of the building will be:
[7 Jan. 2020 II]
(a) 10 A
(b) 25 A
(c) 15 A
(d) 20 A
82. The resistive network shown below is connected to a D.C.
source of 16 V. The power consumed by the network is 4
Watt. The value of R is :
[12 Apr. 2019 I]
(a)
r æbö
R=
ç ÷
2pl è a ø
(b) R =
(a) 6W
(b) 8W
(c) 1W
(d) 16W
o
83. One kg of water, at 20 C, is heated in an electric kettle
whose heating element has a mean (temperature averaged)
resistance of 20 W. The rms voltage in the mains is 200 V.
Ignoring heat loss from the kettle, time taken for water to
evaporate fully, is close to :
[Specific heat of water = 4200 J/(kgoC), Latent heat of water
= 2260 kJ/kg]
[12 Apr. 2019 II]
(a) 16 minutes
(b) 22 minutes
(c) 3 minutes
(d) 3 minutes
Current Electricity
84. Two electric bulbs, rated at (25 W, 220 V) and (100 W, 220 V),
are connected in series across a 220 V voltage source. If
the 25 W and 100 W bulbs draw powers P1 and P2
respectively, then:
[12 Jan. 2019 I]
(a) P1=16 W, P2=4 W
(b) P1=16 W, P2=9 W
(c) P1=9 W, P2=16 W
(d) P1=4 W, P2=16 W
85. Two equal resistances when connected in series to a battery,
consume electric power of 60 W. If these resistance are now
connected in parallel combination to the same battery, the
electric power consumed will be :
[11 Jan. 2019 I]
(a) 60 W
(b) 240 W (c) 120 W
(d) 30 W
86. A 2 W carbon resistor is color coded with green, black,
red and brown respectively. The maximum current which
can be passed through this resistor is: [10 Jan. 2019 I]
(a) 20 mA (b) 100 mA (c) 0.4 mA (d) 63 mA
87. A current of 2 mA was passed through an unknown resistor
which dissipated a power of 4.4 W. Dissipated power when
an ideal power supply of 11 V is connected across it is:
[10 Jan. 2019 II]
(a) 11 × 10–5 W
(b) 11 × 10–3 W
–4
(c) 11 × 10 W
(d) 11 × 105 W
88. A constant voltage is applied between two ends of a metallic
wire. If the length is halved and the radius of the wire is
doubled, the rate of heat developed in the wire will be:
[Online April 15, 2018]
(a) Increased 8 times
(b) Doubled
(c) Halved
(d) Unchanged
89. The figure shows three circuits I, II and III which are
connected to a 3V battery. If the powers dissipated by the
configurations I, II and III are P1, P2 and P3 respectively,
then :
[Online April 9, 2017]
1W
1W
1W
1W
1W
1W
1W
1W
(I)
3V
3V 1W
1W
1W
1W
1W
1W
1W
1W
(II)
3V
(a)
5
400ln J
6
(c) 300 J
2
(b) 200ln J
3
1.5
J
(d) 400 ln
1.3
91.
R
r
In the circuit shown, the resistance r is a variable resistance.
If for r = fR, the heat generation in r is maximum then the
value of f is :
[Online April 9, 2016]
1
1
3
(b) 1
(c)
(d)
2
4
4
92. In a large building, there are 15 bulbs of 40 W, 5 bulbs of
100 W, 5 fans of 80 W and 1 heater of 1 kW. The voltage
of electric mains is 220 V. The minimum capacity of the
main fuse of the building will be:
[2014]
(a) 8 A
(b) 10 A
(c) 12 A
(d) 14 A
93. Four bulbs B1, B2, B3 and B4 of 100 W each are connected
to 220 V main as shown in the figure.
[Online April 19, 2014]
(a)
220 V
B1
B2
B3
B4
Ammeter
The reading in an ideal ammeter will be:
(a) 0.45 A (b) 0.90 A (c) 1.35 A
(d) 1.80 A
94. The supply voltage to room is 120V. The resistance of the
lead wires is 6W. A 60 W bulb is already switched on.
What is the decrease of voltage across the bulb, when a
240 W heater is switched on in parallel to the bulb?[2013]
(a) zero
(b) 2.9 Volt
(c) 13.3 Volt
(d) 10.04Volt
95. Which of the four resistances P, Q, R and S generate the
greatest amount of heat when a current flows from A to
B?
[Online April 23, 2013]
P = 2W
(III)
(a) P1 > P2 > P3
(b) P1 > P3 > P2
(c) P2 > P1 > P3
(d) P3 > P2 > P1
90. The resistance of an electrical toaster has a temperature
dependence given by R(T) = R0 [1 + a(T – T0)] in its range
of operation. At T0 = 300K, R = 100 W and at T = 500 K, R =
120 W. The toaster is connected to a voltage source at 200
V and its temperature is raised at a constant rate from 300
to 500 K in 30 s. The total work done in raising the
temperature is :
[Online April 10, 2016]
P-289
R
Q = 4W
A
B
R = 1W
S = 2W
(a) Q
(b) S
(c) P
(d) R
96. Two electric bulbs rated 25W – 220 V and 100W – 220V are
connected in series to a 440 V supply. Which of the bulbs
will fuse?
[2012]
(a) Both
(b) 100 W (c) 25 W
(d) Neither
97. A 6.0 volt battery is connected to two light bulbs as shown in
figure. Light bulb 1 has resistance 3 ohm while light bulb 2
has resistance 6 ohm. Battery has negligible internal
resistance. Which bulb will glow brighter?
[Online May 19, 2012]
P-290
Physics
Bulb 1
Bulb 2
+
–
6.0 V
(a) Bulb 1 will glow more first and then its brightness will
become less than bulb 2
(b) Bulb 1
(c) Bulb 2
(d) Both glow equally
98. Three resistors of 4 W, 6 W and 12 W are connected in
parallel and the combination is connected in series with a
1.5 V battery of 1 W internal resistance. The rate of Joule
heating in the 4 W resistor is [Online May 12, 2012]
(a) 0.55 W (b) 0.33 W (c) 0.25 W (d) 0.86 W
99. This question has Statement 1 and Statement 2. Of the
four choices given after the Statements, choose the one
that best describes the two Statements.
Statement 1: The possibility of an electric bulb fusing is
higher at the time of switching ON.
Statement 2: Resistance of an electric bulb when it is not lit
up is much smaller than when it is lit up.
[Online May 7, 2012]
(a) Statement 1 is true, Statement 2 is false
(b) Statement 1 is false, Statement 2 is true, Statement
2 is not a correct explanation of Statement 1.
(c) Statement 1 is true, Statement 2 is true, Statement 2
is a correct explanation of Statement 1.
(d) Statement 1 is false, Statement 2 is true.
100. The resistance of a bulb filmanet is 100W at a temperature
of 100°C. If its temperature coefficient of resistance be
0.005 per °C, its resistance will become 200 W at a
temperature of
[2006]
(a) 300°C
(b) 400°C (c) 500°C
(d) 200°C
101. An electric bulb is rated 220 volt - 100 watt. The power
consumed by it when operated on 110 volt will be [2006]
(a) 75 watt (b) 40 watt (c) 25 watt (d) 50 watt
102. A heater coil is cut into two equal parts and only one part
is now used in the heater. The heat generated will now
be
[2005]
(a) four times
(b) doubled
(c) halved
(d) one fourth
103. The resistance of hot tungsten filament is about 10 times
the cold resistance. What will be the resistance of 100 W
and 200 V lamp when not in use ?
[2005]
(a) 20 W
(b) 40 W
(d) 400W
(c) 200 W
104. The thermistors are usually made of
[2004]
(a) metal oxides with high temperature coefficient of
resistivity
(b) metals with high temperature coefficient of
resistivity
(c) metals with low temperature coefficient of resistivity
(d) semiconducting materials having low temperature
coefficient of resistivity
105. Time taken by a 836 W heater to heat one litre of water
from 10°C to 40°C is
[2004]
(a) 150 s
(b) 100 s
(c) 50 s
(d) 200 s
106. A 220 volt, 1000 watt bulb is connected across a 110 volt
mains supply. The power consumed will be
[2003]
(a) 750 watt
(b) 500 watt
(c) 250 watt
(d) 1000 watt
107. A wire when connected to 220 V mains supply has power
dissipation P1. Now the wire is cut into two equal pieces
which are connected in parallel to the same supply. Power
dissipation in this case is P2. Then P2 : P1 is
[2002]
(a) 1
(b) 4
(c) 2
(d) 3
108. If in the circuit, power dissipation is 150 W, then R is
[2002]
R
2W
(a) 2W
(b) 6W
15 V
(c) 5W
(d) 4W
Wheatstone Bridge and
TOPIC 5 Different Measuring
Instruments
109. Two resistors 400W and 800W are connected in series across
a 6 V battery. The potential difference measured by a
voltmeter of 10 kW across 400W resistor is close to:
[Sep. 03, 2020 (II)]
(a) 2 V
(b) 1.8 V
(c) 2.05 V
(d) 1.95 V
110. Which of the following will NOT be observed when a
multimeter (operating in resistance measuring mode)
probes connected across a component, are just reversed?
[Sep. 03, 2020 (II)]
(a) Multimeter shows an equal deflection in both cases
i.e. before and after reversing the probes if the chosen
component is resistor.
(b) Multimeter shows NO deflection in both cases i.e.
before and after reversing the probes if the chosen
component is capacitor.
(c) Multimeter shows a deflection, accompanied by a
splash of light out of connected and NO deflection
on reversing the probes if the chosen component is
LED.
(d) Multimeter shows NO deflection in both cases i.e.
before and after reversing the probes if the chosen
component is metal wire.
111. A potentiometer wire PQ of 1 m length is connected to a
standard cell E 1 . Another cell E 2 of emf 1.02 V is
connected with a resistance 'r' and switch S (as shown in
figure). With switch S open, the null position is obtained
at a distance of 49 cm from Q. The potential gradient in
the potentiometer wire is :
[Sep. 02, 2020 (II)]
P-291
Current Electricity
E1
J
P
r
Q
G
E2
S
(a) 0.02 V/cm
(b)
0.01 V/cm
(c) 0.03 V/cm
(d)
0.04 V/cm
112. In a meter bridge experiment S is a standard resistance. R
is a resistance wire. It is found that balancing length is
l = 25 cm. If R is replaced by a wire of half length and half
diameter that of R of same material, then the balancing
distance l¢ (in cm) will now be ____. [NA. 9 Jan. 2020 II]
113. The length of a potentiometer wire is 1200 cm and it carries
a current of 60 mA. For a cell of emf 5 V and internal
resistance of 20 W, the null point on it is found to be at
1000 cm. The resistance of whole wire is:
[8 Jan. 2020 I]
(a) 80 W
(b) 120 W (c) 60 W
(d) 100 W
114. Four resistances of 15 W, 12 W, 4 W and 10 W respectively
in cyclic order to form Wheatstone’s network. The
resistance that is to be connected in parallel with the
resistance of 10 W to balance the network is _____ W.
[NA. 8 Jan. 2020 I]
115. The balancing length for a cell is 560 cm in a potentiometer
experiment. When an external resistance of 10 W is
connected in parallel to the cell, the balancing length
N
changes by 60 cm. If the internal resistance of the cell is
10
W, where N is an integer then value of N is ________.
[NA. 7 Jan. 2020 II]
116. In a meter bridge experiment, the circuit diagram and the
corresponding observation table are shown in figure.
[10 Apr. 2019 I]
Sl.No.
1.
2.
3.
RW
1000
100
10
l (cm)
60
13
1.5
4.
1
1.0
Which of the reading is consistent ?
(a) 3
(b) 2
(c) 4
(d) 1
117. In the circuit shown, a four-wire potentiometer is made of
a 400 cm long wire, which extends between A and B. The
resistance per unit length of the potentiometer wire is
r = 0.01 W/cm. If an ideal voltmeter is connected as shown
with jockey J at 50 cm from end A, the expected reading of
the voltmeter will be :
[8 Apr. 2019 II]
(a) 0.50 V (b) 0.75 V (c) 0.25 V
(d) 0.20 V
118. In a meter bridge, the wire of length 1 m has a non-uniform
dR
cross-section such that, the variation
of its resistance
dl
dR
1
R with length l is
. Two equal resistances are
µ
dl
l
connected as shown in the figure. The galvanometer has
zero deflection when the jockey is at point P. What is the
length AP?
[12 Jan. 2019 I]
R'
R'
G
P
l
1 l
(a) 0.2 m
(b) 0.3 m
(c) 0.25 m
(d) 0.35 m
119. An ideal battery of 4 V and resistance R are connected in
series in the primary circuit of a potentionmeter of length
1 m and resistance 5W. The value of R, to give a potential
difference of 5 mV across 10 cm of potentiometer wire is:
[12 Jan. 2019 I]
(a) 490W
(b) 480W
(c) 395W
(d) 495W
120. The resistance of the meter bridge AB in given figure is 4W.
With a cell of emf e= 0.5 V and rheostat resistance Rh = 2W
the null point is obtained at some point J. When the cell is
replaced by another one of emf e =e2 the same null point
J is found for Rh = 6W. The emf e2 is:
[11 Jan. 2019 I]
P-292
Physics
e
A
B
J
Rh
6V
(a) 0.4 V
(b) 0.3 V
(c) 0.6 V
(d) 0.5 V
121. In a Wheatstone bridge (see fig.), Resistances P and Q are
approximately equal. When R = 400 W, the bridge is balanced.
On interchanging P and Q, the value of R, for balance, is
405W. The value of X is close to :
[11 Jan. 2019 I]
B
Q
P
A
G
K2
11
11
13
5
L
L
L
L
(b)
(c)
(d)
12
24
24
12
124. The Wheatstone bridge shown in Fig. here, gets balanced
when the carbon resistor used as R1 has the colour code
(Orange, Red, Brown). The resistors R2 and R4 are 80 W
and 40 W, respectively.
Assuming that the colour code for the carbon resistors
gives their accurate values, the colour code for the carbon
resistor, used as R3, would be:
[10 Jan. 2019 II]
(a)
C
R1
G
X
R
R3
D
K1
(a) 401.5 ohm
(b) 404.5 ohm
(c) 403.5 ohm
(d) 402.5 ohm
122. In the experimental set up of metre bridge shown in the
figure, the null point is obtaine data distance of 40 cm from
A. If a 10 W resistor is connected in series with R1, the null
point shifts by 10 cm. The resistance that should be
connected in parallel with (R1 + 10) W such that the null
point shifts back to its initial position is :
[11 Jan. 2019 II]
R1
R2
+
B
–
(b) Brown, Blue, Black
(d) Grey, Black, Brown
125. In a potentiometer experiment, it is found that no current
passes through the galvanometer when the terminals of
the cell are connected across 52 cm of the potentiometer
wire. If the cell is shunted by a resistance of 5 W, a balance
is found when the cell is connected across 40 cm of the
wire. Find the internal resistance of the cell. [2018]
(b) 1.5 W
(c) 2 W
(d) 2.5 W
126. On interchanging the resistances, the balance point of a
meter bridge shifts to the left by 10 cm. The resistance of
their series combination is 1kW. How much was the resistance
on the left slot before interchanging the resistances?
[2018]
(a) 990 W
(a) 20 W
(b) 40 W
(c) 60 W
(d) 30 W
123. A potentiometer wire AB having length L and resistance
12 r is joined to a cell D of emf e and internal resistance
r. A cell C having emf e/2 and internal resistance 3r is
connected. The length AJ at which the galvanometer as
shown in fig. shows no deflection is: [10 Jan. 2019 I]
R4
(a) Brown, Blue, Brown
(c) Red, Green, Brown
(a) 1 W
G
A
R2
(b) 505 W
(c) 550 W
(d) 910 W
127. In a meter bridge, as shown in the figure, it is given that
resistance Y=12.5 W and that the balance is obtained at a
distance 39.5 cm from end A (by jockey J). After
interchanging the resistances X and Y, a new balance point
is found at a distance l2 from end A. What are the values of
X and l2 ?
[Online April 15, 2018]
P-293
Current Electricity
X
Y
131. A 10V battery with internal resistance 1W and a 15V battery
with internal resistance 0.6 W are connected in parallel to
a voltmeter (see figure). The reading in the voltmeter will
be close to :
[Online April 10, 2015]
B
G
l1
A
(100 – l1)
C
39.5 J wire
10V
METER SCALE
1W
15V
Key
Battery
(a) 19.15 W and 39.5 cm
(b) 8.16 W and 60.5 cm
(c) 19.15 W and 60.5 cm
(d) 8.16 W and 39.5 cm
128. Which of the following statements is false ?
[2017]
(a) A rheostat can be used as a potential divider
(b) Kirchhoff's second law represents energy conservation
(c) Wheatstone bridge is the most sensitive when all the
four resistances are of the same order of magnitude
(d) In a balanced wheatstone bridge if the cell and the
galvanometer are exchanged, the null point is disturbed.
129. In a meter bridge experiment resistances are connected as
shown in the figure. Initially resistance P = 4 W and the
neutral point N is at 60 cm from A. Now an unknown resistance R is connected in series to P and the new position of
the neutral point is at 80 cm from A. The value of unknown
resistance R is :
[Online April 9, 2017]
Q
P
G
A
B
N
E
Rh
( )
K
33
20
W
(b) 6 W
(c) 7 W
(d)
W
5
3
130. A potentiometer PQ is set up to compare two resistances as
shown in the figure. The ammeter A in the circuit reads 1.0 A
when two way key K3 is open. The balance point is at a length
l1 cm from P when two way key K3 is plugged in between 2
and 1, while the balance point is at a length l2 cm from P when
key K3 is plugged in between 3 and 1. The ratio of two
(a)
resistances R1 , is found to be :
R2
(a)
l1
l1 + l2
(b)
l2
l2 - l1
(c)
[Online April 8, 2017]
l1
l1 - l2
(d)
l1
l2 - l1
0.6W
V
(a) 12.5 V (b) 24.5 V (c) 13.1 V
(d) 11.9 V
132. In an experiment of potentiometer for measuring the internal
resistance of primary cell a balancing length l is obtained on
the potentiometer wire when the cell is open circuit. Now the
cell is short circuited by a resistance R. If R is to be equal to
the internal resistance of the cell the balancing length on the
potentiometer wire will be
[Online May 26, 2012]
(a) l
(b) 2l
(c) l/2
(d) l/4
133. It is preferable to measure the e.m.f. of a cell by
potentiometer than by a voltmeter because of the following
possible reasons.
[Online May 12, 2012]
(i) In case of potentiometer, no current flows through
the cell.
(ii) The length of the potentiometer allows greater precision.
(iii) Measurement by the potentiometer is quicker.
(iv) The sensitivity of the galvanometer, when using a
potentiometer is not relevant.
Which of these reasons are correct?
(a) (i), (iii), (iv)
(b) (i), (iii), (iv)
(c) (i), (ii)
(d) (i), (ii), (iii), (iv)
134. In a sensitive meter bridge apparatus the bridge wire should
possess
[Online May 12, 2012]
(a) high resistivity and low temperature coefficient.
(b) low resistivity and high temperature coefficient.
(c) low resistivity and low temperature coefficient.
(d) high resistivity and high temperature coefficient.
135. In a metre bridge experiment null point is obtained at 40 cm
from one end of the wire when resistance X is balanced
against another resistance Y. If X < Y, then the new position
of the null point from the same end, if one decides to
balance a resistance of 3X against Y, will be close to :
[Online April 9, 2013]
(a) 80 cm
(b) 75 cm
(c) 67 cm
(d) 50 cm
136. The current in the primary circuit of a potentiometer is 0.2
A. The specific resistance and cross-section of the
potentiometer wire are 4 × 10–7 ohm metre and 8 × 10–7 m2,
respectively. The potential gradient will be equal to
[2011 RS]
(a) 1 V /m (b) 0.5 V/m (c) 0.1 V/m (d) 0.2 V/m
137. Shown in the figure below is a meter-bridge set up with null
deflection in the galvanometer.
P-294
Physics
R
55W
G
20 cm
The value of the unknown resistor R is
[2008]
(a) 13.75 W (b) 220 W (c) 110 W
(d) 55 W
138. In a Wheatstone's bridge, three resistances P, Q and R
connected in the three arms and the fourth arm is formed
by two resistances S1 and S2 connected in parallel. The
condition for the bridge to be balanced will be [2006]
(a)
P
2R
=
Q S1 + S2
(b)
P R ( S1 + S2 )
=
Q
S1 S2
(c)
P R ( S1 + S2 )
=
2S1S2
Q
(d)
P
R
=
Q S1 + S2
139. In a potentiometer experiment the balancing with a cell is at
length 240 cm. On shunting the cell with a resistance of 2W,
the balancing length becomes 120 cm. The internal
resistance of the cell is
[2005]
(a) 0.5W
(b) 1W
(c) 2W
(d) 4W
140. In a meter bridge experiment null point is obtained at 20 cm.
from one end of the wire when resistance X is balanced
against another resistance Y. If X < Y, then where will be the
new position of the null point from the same end, if one
decides to balance a resistance of 4 X against Y [2004]
(a) 40 cm
(b) 80 cm
(c) 50 cm
(d) 70 cm
141. The length of a wire of a potentiometer is 100 cm, and the
e. m.f. of its standard cell is E volt. It is employed to measure
the e.m.f. of a battery whose internal resistance is 0.5W. If
the balance point is obtained at l = 30 cm from the positive
end, the e.m.f. of the battery is
[2003]
(a)
30 E
100.5
30 ( E - 0.5i )
(b)
30 E
(100 - 0.5)
30 E
100
100
where i is the current in the potentiometer wire.
142. An ammeter reads upto 1 ampere. Its internal resistance is
0.81ohm. To increase the range to 10 A the value of the
required shunt is
[2003]
(a) 0.03 W (b) 0.3 W
(c) 0.9 W
(d) 0.09 W
143. If an ammeter is to be used in place of a voltmeter, then we
must connect with the ammeter a
[2002]
(a) low resistance in parallel
(b) high resistance in parallel
(c) high resistance in series
(d) low resistance in series.
(c)
(d)
P-295
Current Electricity
1.
2.
(c) Ammeter : In series connection, the same current
flows through all the components. It aims at measuring
the current flowing through the circuit and hence, it is
connected in series.
Voltmeter : A voltmeter measures voltage change between
two points in a circuit. So we have to place the voltmeter in
parallel with the cicuit component.
(b) rM = 98 ´ 10-8
r A = 2.65 ´ 10-8
rT = 5.65 ´ 10-8
\rM > rT > r A > rC
(a) When i = 0, V = e = 1.5 volt
(d) Charge mobility
( m) =
Vd
E
[ Where Vd = drift velocity ]
and resistivity ( r ) =
Þ m=
=
5.
E EA
I(r)
=
Þ E=
j
I
A
Vd Vd A
=
E
Ir
(
1.1 ´10 –3 ´ p ´ 5 ´10 –3
)
2
5 ´ 1.7 ´10 –8
m = 1.0
2
m
Vs
é æ 1 ö ù
ê- mç 2 ÷+ cú
= eë è T ø û
-T02
R(T) = R 0e
(a)
dR =
T2
(r)(dx)
4px 2
R = ò dR
b
ò dR = rò
a
7.
(c)
r=
m
ne 2 t
9.1 ´ 10-31
8.5 ´1028 ´ (1.6 ´10 -19 ) 2 ´ 25 ´ 10 -15
= 10–8 W-m
8.
(d) Number 2 is associated with the red colour. This
colour is replaced by green.
Q Colour code figure for green is 5
\ New resistance = 500 W
9.
(a) Clearly, from graph
dq
= 0at t = 4s [Since q is constant]
Current, I =
dt
10. (b) Color code : Bl, Br, R, O, Y, G, B, V, Gr, W
0, 1, 2, 3, 4, 5, 6, 7, 8 9
R = AB × C ± D% where D = tolerance
Dgold = ±5%, Dsilver = ±10%; Dno colour = ±20%
Red voilet orange silver
R = 27 × 103 W ± 10% = 27 kW ± 10%
11. (a) Using, I = neAvd
\Drift speed v d =
1
neA
= 0.02 mms–1
´1.6 ´ 10-19 ´ 5 ´ 10-6
rl
12. (c) Resistance, R =
A
2
l l rl
R=r ´ =
[?Volume (V) = A?.]
A l
V
Since resistivity and volume remains constant therefore
% change in resistance
DR 2Dl
=
= 2 ´ (0.5) = 1%
R
l
13. (a) Colour code for carbon resistor
Bl, Br, R, O, Y, G, Blue, V, Gr, W
0 1 2 3 4 5 6 7 8 9
Resistance, R = AB × C ± D
Bands A and B are the first two significant figures of
resistance
B and C indicates the decimal multiplier or the number of
zeros that follow A and B
B and D is tolerance: Gold = ± 5%,
Silver = ± 10 % No colour = ± 20%
9 ´ 10
æ 1 ö
Þ l nR = - m ç
÷+c
è T2 ø
here, m & c are constants
6.
æ r ö æ1 1ö
R = ç ÷ .ç - ÷
è 4p ø è a b ø
1.5
(b) Equation of straight line from graph
y = – mx + c
R
r é -1 ù
4p êë x úû a
=
rC = 1.724 ´ 10-8
3.
4.
b
ÞR=
dx
4 px 2
æ 1 ö
- mç
2÷
= e è T ø ´ ec
28
R = 53 ´ 104 ± 5% = 530kW ± 5%
P-296
14.
15.
Physics
(c) Resistance after temperature increases by 500°C i.e.,
V 220
Rt = =
= 110W
I
2
R0 = 100 (given) temperature coefficient of resistance, a = ?
using Rt = R0 (1 + at)
110 = 100 (1 + a500)
a=
10
100 ´ 500
or,
a = 2 ´ 10 -4°C -1
(c) Charge density r =
charge
q
=
Þ q = rAd
volume Ad
q rAd
=
I
I
(a) Given, R1 = 100 W, r' = r/2, R2 = ?
rl
Resistivity of wire, R =
Q Area × length = volume
A
rV
Hence, R = 2
A
Since, r ® constant, V ® constant
1
Rµ 2
A
1
Q A = pr2
or R µ
r4
R2
= 16 Þ R2 = 16 ´ 100 = 1600 W, Resistance of new wire.
R1
l
(b) V = IR = (neAvd )r
A
V
\ r = V lne
d
Here V = potential difference
l = length of wire
n = no. of electrons per unit volume of conductor.
e = no. of electrons
Placing the value of above parameters we get resistivity
Also, q = IT Þ T =
16.
17.
r=
5
8 ´ 1028 ´ 1.6 ´ 10-19 ´ 2.5 ´ 10-4 ´ 0.1
= 1.6 × 10–5Wm
18.
19.
(c) i = neAVd and Vd µ E (Given)
or, i µ E
i2 µ E
i2 µ V
Hence graph (c) correctly dipicts the V-I graph for a wire
made of such type of material.
(a) In ohm's law, we check V = IR where I is the corrent
flowing through a resistor and V is the potential difference
across that resistor. Only option (a) fits the above criteria.
Remember that ammeter is connected in series with
resistance and voltmeter parallel with the resistance.
l
A
If wire is bent in the middle then
l
l¢ = , A¢ = 2 A
2
l¢
\ New resistance, R¢ = r
A¢
l
r
rl R
2
= .
=
=
2A
4A 4
(a) Resistance of wire
20. (c) Resistance of wire (R) = r
21.
R=
rl rl 2
=
(Q V = Al)
A V
l2
= constant × l2
V
\ Fractional change in resistance
Hence, R = r
DR
Dl
=2
R
l
DR
æ dl ö
= 200 ´ ç ÷
R
è l ø
Q dl/l = 0.1%
100 ´
{ }
0.1 ù
é
\ % change in R = ê 200 ´
= 0.2%
100 úû
ë
\ Resistance will increase by 0.2%.
22. (a) Let j be the current density.
I
Then j ´ 2pr 2 = I Þ j =
2 pr 2
rI
\ E = rj =
2 pr 2
Now, VB – VC
a
=-
ò
r uur
E × dr = -
a +b
a
a
ò
rI
2 pr 2
a+b
dr
rI é 1 ù
rI
rI
- ú
=
ê
2 p ë r û a + b 2 pa 2 p ( a + b )
On applying superposition as mentioned we get
rI
rI
'
DVBC = 2 ´ DVBC
=
pa p ( a + b )
rI
23. (c) As shown in Answer (a) E =
2pr 2
24. (d) Resistance of a metal conductor at temperature t°C
is given by
Rt = R0 (1 + at),
R0 is the resistance of the wire at 0ºC
and a is the temperature coefficient of resistance.
Resistance at 50°C, R50 = R0 (1 + 50a)
.. (i)
Resistance at 100°C, R100 = R0 (1 + 100a) ... (ii)
From (i), R50 – R0 = 50aR0
... (iii)
From (ii), R100 – R0 = 100aR0
... (iv)
=-
P-297
Current Electricity
25.
Dividing (iii) by (iv), we get
R50 - R0
1
=
R100 - R0 2
Here, R50 = 5W and R100 = 6W
5 - R0 1
\
=
6 - R0 2
or, 6 – R0 = 10 – 2 R0 or, R0 = 4W.
(d) Let dA and dB are the diameter of wire A and B respectively.
Let rB and rA be the resistivity of wire A and B. We have given
rB = 2rA
dB = 2dA
If both resistances are equal
RB = RA
Þ
\
28. (b) From circuit diagram,
1 1 1
4
= + Þ R1 =
R1 1 4
5
1
1 1
6
= + Þ R2 =
R2 2 3
5
r B l B r Al A
=
AB
AA
lB rA
=
´
l A rB
d B2
d A2
=
rA
´
2r A
4d A2
d A2
Reff = R1 + R2 =
=2
i=
R1 i1
26.
(b)
R2 i2
V
Given,
l1
4
r 2
= and 1 =
l2
3
r2 3
rl
rl
R1 = 1 ; R2 = 2
2
pr1
pr22
When wires are in parallel to the circuit potential difference
across each wire is same
i1R1 = i2R2
l 2 r12
rl
pr 2
i1 R2
´
=
= 22 ´ 1 =
l1 r2 2
i2 R1
pr2 rr1
3 4 1
= ´ =
4 9 3
(d) Since volume of wire remains unchanged on
increasing length, hence
A × l = = A¢ × l¢
Þ l¢ = 2l
A´ l A´ l A
\ A¢ =
=
=
l¢
2l
2
Percentage change in resistance
l¢
l
r -b
R f - Ri
A ´ 100
=
´100 = A¢
l
Ri
r
A
éæ l¢ A ö ù
= êç ´ ÷ - 1ú ´ 100
ëè A¢ l ø û
\
27.
é æ 2l A ö ù
= ê ç ´ ÷ - 1ú ´ 100 = (4 – 1) × 100
çA
÷
ëê è 2 l ø ûú
= 300%
4 6
+ = 2W
5 5
v
20
=
= 10 A
Reff
2
\
I BC =
4i 3i i
- = = 2A
5 5 5
29. (30.00)
The resistance of 30W is in parallel with R. Their effective
resistance
1
1 1
= +
R ' 30 R
30 R
R' =
...(i)
30 + R
20 ´ 20
Also, V = IR Þ 10 =
R '+ 25
Þ R '+ 25 = 40 Þ R’ = 15
30 R
30 + R
Þ 30 + R = 2R
Þ R = 30 W
R ' = 15 =
30. (c)
31. (b)
R eq
æ 7R ö æ R ö
ç
÷ç ÷
7R
8 øè 8 ø
=è
=
64
R
Using (i)
P-298
32.
Physics
(c) When length becomes double its resistance becomes
V4
= 0.01A
R4
V3 = i1 R3 = 1V
V3 + V4 = 6V = V2
V1 + V3 + V4 = 18V
Þ V1 =12 V
Current, i4=
(R µ l 2 )
R = 4 × 3 = 12 W
Req =
33.
2 ´ 10 5
= W
12
3
(c) R3, R4 and R5 are in series so their equivalent
R = 20 + 5 + 25 = 50 W
This is parallel with R2, and so net resistance of the circuit
V1
= 0.03A
R1
i=
i = i1 + i2 Þ i 2 = i – i, = 0.03 – 0.01A = 0.02 A
V2
6
=
= 300W
i 2 0.02
37. (a) Rseries = R1 + R2 + ..... + Rn
\R 2 =
1
=
R parallel
1
1
1
+
+ ....... +
R1 R 2
Rn
R
Req
160
æ 10 ´ 50 ö
=ç
W
÷ + 15 + 30 =
3
è 10 + 50 ø
So, i =
34.
R
A
C
R
R
B
R
C
R
R
C
15
9
e
=
=
A
Req (100 / 3) 32
(a)
R
C
R
R
R
R
A
R
Req = 2R
B
R
C
R
C
38. (a) In steady state, flow fo current through capacitor will
be zero.
Current through the circuit,
E
r + r2
Potential difference through capacitor
i=
Resistance, R µ l so resistance of each side of the
equilateral triangle = 6 W
Resistance Req between any two vertices
1
1 1
= + Þ R eq. = 4 W
R eq 12 6
35.
1
1
1
(b) Using, R = R + R
eq
1
2
RRu
0.95 R = R + Ru (measured value 5% less then internal
resistance of voltmeter)
or, 0.95 × 30 = 0.05 Ru
\ Ru = 19 × 30 = 570 W
36.
R3= 100W
(a)
i
R4= 500W
i1
R1= 400W
i2
R2
18V
Across R4 reading of voltmeter, V4 = 5V
Vc =
æ E ö
Q
= E - ir = E - ç
r
C
è r + r2 ÷ø
\
Q = CE
r2
r + r2
39. (b) The potential difference in each loop is zero.
\ No current will flow or current in each resistance is Zero.
40. (a) The given circuit can be redrawn as,
1W
A
xW
B
4W
xW
1W
as 4 W and x W are parallel x' =
1 1 (4 + x)
+ =
4 x
4x
4x
4+x
& 1 W and 1 W are also parallel x" = 2 W
Now equivalent resistance of circuit
x' =
4x
8 + 6x
+2 =
4+ x
4+x
4x + x2 = 8 + 6x
x2 – 2x – 8 = 0
x=
P-299
Current Electricity
2±6
2 ± 4 - 4(1)(-8) 2 ± 36
=
= 4W
=
2
2
2
V
Reading of Ammeter A1 =
(R + r)
x=
41.
9
A1 =
= 2 Ampere
4 + 0.5
(b) Resistance between P and Q
5
r´ r
æ r rö
6 = 5r
rPQ = r P ç + ÷ =
è 3 2ø
5
11
r+ r
6
Resistance between Q and R
A
xW
20 –xW
D
60°
= R0 [2 + ( a1 + a2 ) D t ]
é æ a + a2 ö ù
= 2 R0 ê1 + ç 1
÷ D tú
ë è 2 ø û
a + a2
\ a eq = 1
2
1 1 1
In Parallel, = +
R R1 R2
1
1
+
R0 [1 + a1D t ] R0 [1 + a2 D t ]
1
R0
(1 + aeq D t )
2
1
1
=
+
R0 (1 + a1D t ) R0 (1 + a2 D t )
Þ
r 3
´ r
r ær ö 3 2
3
= r
rPR = P ç + r÷ =
r
3
è
ø
3 2
+ r 11
3 2
Hence, it is clear that rPQ is maximum
(d)
In Series, R = R1 + R2
=
r 4
´ r
r
r
4
rQR = P (r + ) = 2 3 = r
r
4
2
3
+ r 11
2 3
Resistance between P and R
42.
Here, R0 is the resistance of conductor at 0°C
2(1 - aeq Dt ) = (1 - a1Dt )(1 - a 2Dt )
a1 + a 2
2
44. (b) The network of resistors is a balanced wheatstone
bridge. Hence, no current will flow through centre resistor.
The equivalent circuit is
\ a eq =
10W
20W
xW
30W
10W
E
10 W
B
20 –x W
15W
5W
C
10W
5V
5V
1
1
1
=
+
For ADE
R ' 2x 10
or
R'=
20x
10 + 2x
20x
R BC =
+ 20 - x + 20 - x … (i)
10 + 2x
20x
+ 40 = 2x
or
10 + 2x
Solving we get
x = 10 W
Putting the value of x = 10 W in equation (i)
We get
R BC
20 ´ 10
=
+ 20 - 10 + 20 - 10
10 + 2 ´ 10
80
= 26.7 W
3
(d) Let R1 and R2 be the resistances of two conductors, then
=
43.
10W
R1 = R0 [1 + a1D t ]
R2 = R0 [1 + a 2 D t ]
5V
Req
V
5
15 ´ 30
= 0.5 A
= 10 W Þ I = =
=
R 10
15 + 30
6V
2W
1.5W
45. (a)
6W
3W
6V
6V
3W
3/2W 3/2W
3W
hence Req = 3/2;
3W
\I =
6
= 4A
3/ 2
P-300
46.
Physics
(c) Let R1 and R2 be the two given resistances
Resistance of the series combination,
S = R1 + R2
Resistance of the parallel combination,
(14i1 + 10i2 = 10) ´ 17
R1 R2
R1 + R2
As per question S = nP
n ( R1R2 )
Þ R1 + R2 =
( R1 + R2 )
Þ (R1 + R2)2 = nR1R2
Minimum value of n is 4 for that
(R1 + R2)2 = 4R1R2
Þ (R1 – R2)2 = 0
(b) In the given circuit, resistance of 3W is in parallel
with series combination of two 3W resistance.
P=
47.
3 ´ 6 18
=
= 2W
3+ 6 9
Using ohm’s law V = IR
Rp =
ÞI =
3W
3W
49. (d)
1A = i1 i
A
(c)
5W
A
Þ VB - VA = 3 - 2 = 1 volt
50. (c) The equivalent circuit can be drawn as
2W
A
6W 3V
i2
i2
2W
i 1 4W
4W
B
i1
C
2W
E
10W
20V 2W
C
2W
i1 + i2
D
i2 = 2A
3
F
C
2A
1V
Let us assume the potential at A = VA = 0
Using Kirchoff's junction rule at C, we get
i1 + i3 = i2
2W
48.
B
2W
I
3W
i3
Þ VA + 1 + i3 (1) - 2 = VB
3W
3V
2V
D
E
1A
1A + i3 = 2 A Þ i3 = 2 A
Now using Kirchoff's loop law along ACDB
VA + 1 + i3 (2) - 2 = VB
V 3
= = 1.5 A
R 2
3V
Þ 238i1 - 170i2 = 170
...(iv)
On solving equations (iii) and (iv), we get
30
-138i1 = 30 Þ i1 = = -0.217
138
i1 is negative it means current flows from positive to
negative terminal.
4W
Using Kirchoff's loop law in loop ABCD
-5i2 - 10(i1 + i2 ) - 2i2 + 20 = 0
Þ - 10i1 - 17i2 + 20 = 0
...(i)
Using Kirchoff's loop law in loop BEFC
10V
F
8V
Voltage across AC = 8 V
Resistance RAC = 4 + 4 = 8 W
V
8
i1 =
=
= 1 Amp
RAC 4 + 4
B
51. (2)
40W
60W
i1
A
C
i2
90W
Þ -10 + 4i1 + 10(i1 + i2 ) = 0
Þ 14i1 + 10i2 + 10 = 0
...(ii)
Multiplying equation (i) by 10, we have
(10i1 + 17i2 = 20) ´ 10
Þ 100 i1 - 170 i 2 = 200
...(iii)
Multiplying equation (ii) by 17, we have
110W
D
40 V
Current through AB, i1 =
40
= 0.4
40 + 60
Current through AD, i2 =
40
1
=
90 + 110 5
Using KVL in BAD loop
P-301
Current Electricity
54. (d) Applying parallel combination of batteries
VB + i1 (40) - i2 (90) = VD
Þ VB - VD =
52.
E1 E2 E3
+
+
1+1 2 1 +1
1
1
1
+ +
1 +1 2 1+1
1
4
(90) - (40)
5
10
Þ VB - VD = 18 - 16 = 2 V
(08.00)
2W
C
i2
D
A
B
2W
2W
i1
2W
i2
F
E i
2W
i
10 V
As capacitor is fully charged no current will flow through it.
2W
1A
A
1A
2W
2W
E
2A
3A
B
2W
10 V
We have the current distribution as shown in the figure.
2 4 4
+ +
2 2 2 = 5´ 2
1 1 1
3
+ +
2 2 2
10
= =3.3 Volt
3
æ e ö
55. (d) i = ç
÷
èR+rø
Power delivered to R.
2
æ e ö
P=i R= ç
÷ R
è R+r ø
2
æ 4 ´ 2ö
+2
Equivalent resistance, Req = ç
è 4 + 2 ÷ø
10
10 ´ 3
=
= 3 Amp
4
10
+2
3
i1 = 2 A and i2 = 1 A
VAEB = 1 × 2 + 3 × 2 = 8 V
(b) For the given circuit
Net current, i =
53.
P to be maximum,
or
or
dP
=0
dR
2
d éæ e ö ù
êç
÷ Rú = 0
dR êè R + r ø ú
ë
û
R=r
(I6)0.4A
56. (b)
3
8 + 2r
Now voltage across AB
i=
3
i´6 =
´6 = 2
8 + 2r
Þ 9 = 8 + 2r
Þr=
1
W
2
I3
(I5)0.4A
0.4A
0.3A
I2
0.8A(I4)
From KCL, I3 = 0.8 – 0.4 = 0.4 A
I2 = 0.4 + 0.4 + 0.3
= 1.1 A
and
I6 = 0.4 A
57. (b) Given, E1 = 1V, E2 = 2V, E3 = 3V, r1 = 1W,
r2 = 1W and r 3 = 1W
I1
P-302
Physics
VAB
58.
E1 E 2 E 3 1 2 3
+ +
+ +
r
r2 r3 1 1 1
6
= VCD = 1
=
1 1 1
1 1 1 = = 2V
+ +
+ +
3
r1 r2 r3
1 1 1
(c) Current passing through resistance R1,
v 10
i1 = =
= 0.5A
R1 20
and, i2 = 0
59.
(b)
6 = 3 I1 + I1 – I2 ; 4I1 – I2 = 6
...(1)
– 9 + 2I2 – (I1 – I2) + 3I2 = 0
– I1 + 6I2 = 9
...(2)
On solving (1) and (2)
I1 = 0.13A
Direction Q to P, since I1 > I2.
62. (c) As no current flows through arm EB then
VD = 0V
VE = 0V
VB = –4V
VA = 5V
So, potential difference between the points A and D
VA – VD = 5V
63. (b)
line 3
line 2
line 1
3W
8.0 V
Let voltage at C = xV
From kirchhoff’s current law,
KCL : i1 + i2 = i
60.
(b)
3W
V X 10
= = =5A
R R 2
T
S
R
V
12V
v–12
1W
8.0 V
U
2W
P
v–13
Q
0
10W
Using Kirchhoff’s law at P we get
V - 12 V - 13 V - 0
+
+
=0
1
2
10
[Let potential at P, Q, U = 0 and at R = V
V V V 12 13 0
Þ
+ +
= + +
1 2 10 1 2 10
Þ
64.
65.
10 + 5 + 1
24 + 13
æ 16 ö 37
Þ Vç ÷ =
V=
10
2
è 10 ø 2
37 ´10 370
=
= 11.56 volt
16 ´ 2
32
(a) From KVL
– 6 + 3I1 + 1 (Ii – I2) = 0
6V
P 2W
ÞV=
61.
66.
9V
1W
3W
q
I1
I
I2
9W
16.0 V
In steady state capacitor is fully charged hence no current
will flow through line 2.
By simplyfing the circuit
20 - x 10 - x x - 0
+
=
Þ x = 10
2
4
2
\i=
5m F
4W
67.
9W
16.0 V
Hence resultant potential difference across resistances will
be 8.0 V.
V
Thus current I =
R
8.0
8
=
=
3 + 9 12
2
or, I = = 0.67 A
3
(a) Current through 50 V and 30 V batteries are
respectively 2.5 A and 3 A.
(c) Given, emf of cell E = 200 V
Internal resistance of cells = 1 W
D. C. main supply voltage V = 220 V
External resistance R = ?
æE-Vö
r =ç
÷R
è V ø
æ 20 ö
1=ç
\ R = 11 W.
÷´ R
è 220 ø
R + r 10
0.5 10
100
90
=
=
=
(c)
Þ 1+
Þ
R
9
R
9
R+r R
0.5 1
= \ R = 4.5 W
Þ
R 9
(c) Applying Kirchoff’s second law in AB P2P1A, we get
-2i + 5 - 10 i1 = 0
2i + 10i1 = 5
.....(i)
P-303
Current Electricity
B
i
P2
i–i1
m = Zq
C
i1
10W
5V
1W
Þ Zµ
2V
2W
69.
70.
71.
D
P1
Again applying Kirchoff's second law in P2 CDP1P2 we
get,
10 i1 + 2 – i + i1= 0
2i – 22i1 = 4
....(ii)
From (i) and (ii)
32i1 = 1
1
Þ i1 =
A from P2 to P1
32
1
2
(a) Energy in capacitor = CV
2
Work done by battery = QV = CV2
where C = Capacitance of capacitor
V = Potential difference,
e = emf of battery
1
CV 2
1
2
= (Q V = e)
Required ratio =
2
2
CV
(d) Note : Kirchhoff's first law is based on conservation
of charge and Kirchhoff's second law is based on
conservation of energy.
(d) At cold junction, current flows from Antimony to
Bismuth because current flows from metal occurring later
in the series to metal occurring earlier in the thermoelectric
series. In thermoelectric series, Bismuth comes earlier than
Antimony so at cold junction, current. Flow from Antimony
to Bismuth.
(a)
R1
R2
E
E
I
R
Let E be the emf of each source of current
2E
Current in the circuit I =
R + R1 + R2
Potential difference across cell having internal resistance R2
V = E – iR2 = 0
2E
× R2 = 0
E–
R + R1 + R2
Þ R + R1 + R 2 - 2R 2 = 0
72.
.... (i)
Also q = q1 + q2
A
68.
1 Þ Z1 = q2
Z 2 q1
q
q
q
= 1 +1
q2 q2
Þ
.... (ii)
(Dividing (ii) by q2)
q
q
1+ 1
q2
From equation (i) and (iii),
Þ q2 =
.... (iii)
q
Z
1+ 2
Z1
(d) Current is given by
q2 =
73.
E
,
R+r
If internal resistance (r) is zero,
E
I=
= constant.
R
Thus, energy source will supply a constant current if its
internal resistance is zero.
74. (d) Given E = aq + bq2
dE
Þ
= a + 2bq
dq
At neutral temperature
dE
q = qn :
=0
dq
I=
d2E
-a
= -350 Þ
= 2b
2b
d q2
hence no q is possible for E to be maximum no neutral
temperature is possible.
75. (c) From the Faraday’s first law of electrolysis,
m = Zit
Þ m = 3.3 × 10–7 × 3 × 2
= 19.8 × 10–7 kg
76. (a) Let the smallest temperature difference be q°C that
can be detected by the thermocouple, then
Thermo emf = (25 × 10–6) q
Let I is the smallest current which can be detected by the
galvanometer of resistance R.
Potential difference across galvanometer
IR = 10–5 × 40
\ 10–5 × 40 = 25 × 10–6 × q
Þ q = 16°C.
77. (c) According to Faraday’s first law of electrolysis
m= Z×I×t
When I and t is same, m µ Z
Þ qn =
mCu ZCu Þ m = Z Cu ´ m
=
Cu
Zn
Z Zn
mZn Z Zn
Þ R + R1 - R 2 = 0
\
Þ R = R 2 - R1
(a) From Faraday’s first law of electrolysis, mass
deposited
Þ mCu =
31.5
´ 0.13 = 0.126 g
32.5
P-304
78.
79.
Physics
(b) From the Faraday’s first law of electrolysis
m = ZIt Þ m µ It
(b) Given : Power, P = 1 kW = 1000 W
R = 2W, V = 220 V
Current, I =
84. (a) As R =
Current flown i =
P 1000
=
V
220
P1 = i 2 R1 =
2
æ 1000 ö
´2
Ploss = I 2 R = ç
è 220 ÷ø
1000
\ Efficiency = 1000 + P ´ 100 = 96%.
loss
80.
(b) Maximum power in external resistance is generated
when it is equal to internal resistance of battery i.e., PR
maximum when r = R
The maximum Joule heating in R will take place for, the
resistance of small element
r
R
r
b
ln
2 pl a
(d) Net Power, P
= 15 × 45 + 15 × 100 + 15 × 10 + 2 × 1000
= 15 × 155 + 2000 W
Power, P = VI
P
V
15 ´155 + 2000
\ I main =
= 19.66 A » 20 A
220
(b) Equivalent resistance,
4 R ´ 4R
6 R ´ 12 R
+ R+
+R
4R + 4R
6 R + 12 R
= 2R + R + 4R + R
= 8R.
V2
162
Using, P = R Þ 4 =
eq
8R
\
83.
(b)
R=
162
=8W
4´8
e2
(R/2)
or P' = 4P = 240 W(Q P = 60 W)
(a) Colour code for carbon resistor
Bl, Br, R, O, Y, G,
Blue, V,
0 1
2 3
4 5
6
7
Resistance, R = AB × C ± D
\ Resistance, R = 50 × 102 W
Now using formula, Power, P = i 2R
Gr, W
8 9
P
2
= 20mA
=
R
50 ´102
87. (a) Power, P = I2R
4.4 = 4 × 10–6 × R
Þ R = 1.1 × 106W
When supply of 11 v is connected
\i=
or, R =
Req =
2202
= 16 W
25
e2
e2
=
R eq 2R
In parallel condition, Req = R/2.
86.
b
b
82.
æ 2202 2202 ö
ç 25 + 100 ÷
è
ø
´
New power, P¢ =
l
Þ I=
2202
Power consumed, P =
b
81.
220
R1 + R 2
Similarly, P2 = i2R2 = 4 W
85. (b) When two resistances are connected in series,
Req = 2R
rdr
r dr
DR =
ÞR=
2prl
2pl òa r
a
V2
2202
2202
, so R1 =
and R 2 =
P
25
100
v 2 112 112
´
´10 –6
=
R
1.1 1.1
=11´10 –5 W
(a) Rate of heat i.e., Power developed in the wire =
Power, P’=
88.
P=
V2
R
Resistance of the wire of length, L R1 =
rL rL
=
A pr 2
V2
R1
Resistance of the wire when length is halved i.e., L/2
\
Power, P1 =
L
2 = rL = R1
R2 =
8
p(2r )2 p8r 2
r
V
8V
=
R1 R1
8
or, P2 = 8P1 i.e., power
or original wire.
\
Power, P2 =
increased 8 times of previous
P-305
Current Electricity
89.
(c) From the given circuit, net resistances
RI = 1 W, RII = 1/2 W, RIII = 3/2 W
It is clear that R3 > R1 > R2
Hence, P3 < P1 < P2
As Power (P) =
(None) Work done in 30s, W =
or,
W=
ò
0
48
´120 = 106.66 volt
54
Hence decrease in voltage
V1 – V2 = 117.073 – 106.66 = 10.04 Volt (approximately)
V2 =
ò
0
30
240
´ 120 = 117.73 volt
246
Voltage across bulb after heater is switched on,
V1 =
V2
1
ÞPµ
R
R
30
90.
Voltage across bulb before heater is switched on,
V2
dt
R
(200)2
(200)2
dt =
20t
100
100(1 + a
)
3
30
dt
ò 20a
t
0 1+
3
æ 1 + 20a
ö
´ 30 ÷
400 ´ 3 ç
3
=
lnç
÷ = 60,000 ln
20a
1
è
ø
Q 120 = 100 éë1 + a ( 200 ) ùû
æ 6ö
çè ÷ø
5
92.
1
1000
(c) Heat energy will be maximum when resistance will be
minimum.
(c) Total power consumed by electrical appliances in
the building, Ptotal = 2500W
Watt = Volt × ampere
Þ 2500 = V × I Þ 2500 = 220 I
Þ
93.
I1
A I2
2500
I=
= 11.36 » 12A
220
(Minimum capacity of main fuse)
I1 =
R2
I
3
I2 = I2 = 2
R1
6
2
or I2 = 2I1
Heat flow H = I2 Rt
For Q, H Q = I12 Qt =
I22
´ 4t = I 22 t
4
2
2
2
For S, H S = I 2 St = I2 × 2t = 2I 2 t
\ Greatest amount of heat generated by S.
96. (c) Current capacity of 25 W bulb
W1
25
Am p
=
V1 220
Current capacity of 100 W bulb
W 2 100
I2 = V = 220 Am p
2
The current flowing through the circuit
I1 =
Power
(c) Current in each bulb = Voltage
100
= 0.45A
220
Current through ammeter = 0.45 × 3 = 1.35 A
=
6W
B1
B2
R1
r2
Bulb
(Lead)
94.
B
R=1W S=2W
R1 = P + Q = 2 W + 4 W = 6 W
R2 = R + S = 1W + 2 W = 3 W
I1R1 = I2R2
\ a=
91.
P=2 W Q=4W
95. (b)
(d)
120 V
440V
Power of bulb = 60 W (given)
Resistance of bulb =
Resistance of 25 W bulb,
120 ´ 120
= 240W
60
é
V2 ù
êQ P =
ú
ë
R û
Power of heater = 240W (given)
Resistance of heater =
120 ´120
= 60W
240
V12 (220)2
=
;
P1
25
Resistance of 100 W bulb
R1 =
V22 (220)2
=
P
100
Reff = R1 + R2
Current flowing through circuit
R2 =
P-306
I=
Physics
101. (c) The resistance of the electric bulb is
440
R eff
I=
V 2 (220) 2
=
P
100
The power consumed when operated at 110 V is
R=
440
2
(220)
(220) 2
+
25
100
440
=
;
1 ù
é1
(220) 2 ê +
ë 25 100 úû
I=
40
Am p
220
æ 25 ö
æ 40 ö < I æ = 100 A ö
A÷ < I ç =
A÷
Q I1 ç =
2ç
÷
è 200 ø
è 220 ø
è 220 ø
Thus the bulb rated 25 W–220 will fuse.
97.
(b) Total resistance =
6´3
= 2W
6+3
6
= 3A
2
Therefore current through bulb 1 is 2A and bulb 2 is 1A.
So bulb 1 will glow more
(c) Resistors 4 W, 6 W and 12 W are connected in parallel,
its equivalent resistance (R) is given by
Current in circuit =
98.
1 1 1 1
12
= + +
Þ R=
= 2W
R 4 6 12
6
Again R is connected to 1.5 V battery whose internal
resistance r = 1 W.
Equivalent resistance now,
R¢ = 2W + 1W = 3W
V 1.5 1
=
= A
Current, Itotal =
R'
3
2
1
Itotal = = 3x + 2x + x = 6x
2
1
12
\ Current through 4W resistor = 3x
Þx=
1
1
= A
12 4
Therefore, rate of Joule heating in the 4W resistor
= 3×
2
1
æ1ö
= I2R = ç ÷ ´ 4 = = 0.25W
4
è4ø
99. (c)
100. (b) Let resistance of bulb filament be R0 at 0°C using R =
R0 (1 + a Dt) we have
R1 = R0 [1 + a × 100] = 100 ....(1)
R2 = R0 [1 + a × T] = 200
....(2)
On dividing we get
200
1 + aT
1 + 0.005 T
=
Þ2=
100 1 + 100a
1 + 100 ´ 0.005
Þ T = 400°C
Note : We may use this expression as an approximation
because the difference in the answers is appreciable. For
accurate results one should use R = R0eaDT
P¢ =
V2
R
Þ P=
(110) 2
2
=
(220) /100
102. (b) Heat generated,
100
= 25 W
4
V 2t
R
After cutting equal length of heater coil will become half.
As R µ l
R
Resistance of half the coil =
2
H=
V 2t
= 2H
R
2
\ As R reduces to half, ‘H’ will be doubled.
H¢ =
V2
R
\ Resistance of tungsten filament when in use
103. (b) Power, P = Vi =
V 2 200 ´ 200
=
= 400 W
P
100
Resistance when not in use i.e., cold resistance
Rhot =
400
= 40 W
10
104. (a) Thermistors are usually made of metaloxides with
high temperature coefficient of resistivity.
105. (a) Heat supplied in time t for heating 1L water from
10°C to 40°C
DQ = mCp × DT
= 1 × 4180 × (40 – 10) = 4180 × 30
But DQ = P × t = 836 × t
Rcold =
4180 ´ 30
= 150s
836
106. (c) We know that resistance,
Þt=
2
Vrated
(220) 2
=
= 48.4 W
Prated
1000
When this bulb is connected to 110 volt mains supply we
get
R=
V 2 (110)2
=
= 250W
R
48.4
107. (b) Case 1 Initial power dissipation,
P=
R
P1 =
V2
R
V
P-307
Current Electricity
Case 2
When wire is cut into two equal pieces, the resistance of
R
each piece is . When they are connected in parallel
2
Equivalent resistance, Req =
R/2
R/4
V
æV ö
V
= 4ç
= 4 P1
ç R ÷÷
R/4
è
ø
108. (b) The equivalent resistance of parallel combination of
2W and R is
P2 =
Req =
R 25 1
=
=
S 75 3
New resistance,
l
r
l´2
R' = 2 = r
A
A
4
Þ R¢ = 2R
l2
R'
=
Þ
S 100 – l 2
Þ
=
2
1.02
= 0.02 volt/cm
100 - 49
112. (40) For the given meter bridge
\x =
l1
R
=
Where, l1 = balancing length
S 100 – l1
R/2 R
=
2
4
R/2
V
Power dissipated,
Balancing length from P = 100 – 49
2
2´R
2+ R
Þ
\ Power dissipation P =
V2
Re q
\ 15 0 =
(15)2
Req
Þ
...(i)
lö
æ
çQ R = r ÷
Aø
è
l2
1
2´ =
3 100 – l 2
l2
2R
=
S 100 – l 2
Using (i)
l!
2 = 40 cm
113. (d)
225 ´ ( R + 2)
2R
3
Þ
=
2R
2+ R 2
Þ 4 R = 6 + 3R Þ R = 6W
109. (d) The voltmeter of resistance 10kW is parallel to the
resistance of 400W. So, their equivalent resistance is
Þ 150 =
1
1
1
1
1
=
+
=
+
R ' 10 k W 400W 10000 400
Þ
Let R be the resistance of the whole wire
Potential gradient for the potentiometer wire
1 1 + 25
26
=
=
R ' 10000 10000
' AB ' = -
10000
Þ R' =
W
26
Using Ohm's law, current in the circuit
æ dV ö
60 ´ R
V AP = ç
lAP =
´ 1000mV
1200
è d l AB ÷ø
Þ VAP = 50 R mV
Also, VAP = 5 V (for balance point at P)
Voltage
6
=
Net Resistance 10000 + 800
26
Potential difference measured by voltmeter
I=
V = IR ' =
6
10000
´
10000
26
+ 800
26
150
= 1.95 volt
77
110. (b) Multimeter shows deflection in both cases i.e. before
and after reversing the probes if the chosen component is
capacitor.
Potential drop
111. (a) Potential gradient, x =
length
Here, Potential drop = 1.02
dV
I ´ R é 60 ´ R ù
=
=ê
ú mv / m
dl
l
ë l AB û
\R =
V AP
50 ´ 10
–3
=
5
= 100W
50 ´ 10 –3
114. (10)
ÞV =
As per Wheatstone bridge balance condition
P S
=
Q R
P-308
Physics
Let resistance R’ is connected in parallel with resistance S
of 10W
10 R '
15
10 R '
Þ 5=
\ =
10 + R '
12 10 + R '
4
Þ 50 + 5R’ = 10R’
50
= 10W
5
115. (12) We know that
E µ l where l is the balancing length
\ E = k (560)
....(i)
When the balancing length changes by 60 cm
...(ii)
dl
l
Let R1 and R2 be the resistance of AP and PB respectively.
Using wheatstone bridge principle
\
R ' R1
or R1 = R 2
=
R ' R2
Now, ò dR = k ò
l
r + 10 56
=
Þ 50 r + 500 = 560
10
50
6
N
Þ r = W = W Þ N = 12
5
10
116. (c) For a balanced bridge
R1 l 2
=
R 2 l1
\ R1 = k ò l -1 2dl = k.2. l
R
l
=
X 100 – l
Using the above expression
1
R 2 = k ò l -1 2dl = k.(2 - 2 l )
l
Putting R1 = R2
k2 l = k(2 - 2 l )
\ 2 l =1
l=
R(100 – l )
l
100 ´ 40 2000
=
W
for observation (1) X =
60
3
for observation (2) X =
100 ´ 87 8700
=
W
13
13
for observation (3) X =
10 ´ 98.5 1970
=
W
1.5
3
1 ´ 99
= 99W
1
Clearly we can see that the value of x calculated in
observation (4) is inconsistent than other.
117. (c) The resistance of potentiometer wire
R = 0.01 × 400 = 4 W
Current in the wire
for observation (4) X =
i=
3
1
V
=
= A
RT 4 + 0.5 + 0.57 + 1 2
Now V = iRAJ =
1
× (0.01 × 50) = 0.25 V..
2
1
2
1
i.e., l = m Þ 0.25 m
4
R
4v
So
X=
dl
l
0
Dividing (i) by (ii) we get
Þ
dR
1
dR
1
(where k is constant)
µ
Þ
= k´
dl
dl
l
l
dR = k
\ R' =
E
10 = k (500)
r + 10
118. (c) We have given
119. (c)
i
5W
i
1m
Current flowing through the circuit (I) is given by
æ 4 ö
I=ç
A
è R + 5 ÷ø
Resistance of length 10 cm of wire
10
= 0.5W
100
According to question,
æ 4 ö
5 ´ 10 -3 = ç
.(0.5)
è R + 5 ÷ø
4
\
= 10 -2 or R + 5 = 400 W
R +5
\ R = 395W
120. (b) Given, Emf of cell, e = 0.5 v
Rheostat resistance, Rh = 2W
Potential gradient is
=5´
dv æ 6 ö 4
=
´
dL çè 2 + 4 ÷ø L
Let null point be at l cm when cell of emfe = 0.5 v is used.
æ 6 ö 4
thus e1 = 0.5V = ç
´ ´l
è 2 + 4 ÷ø L
... (i)
P-309
Current Electricity
For resistance R h = 6 W new potential gradient is
æ 6 ö 4 and at null point
çè
÷´
4 + 6ø L
æ 6 ö æ 4ö
çè
÷ ç ÷ ´ l = e2
4 + 6ø è L ø
R2=1000 – R1
... (ii)
Dividing equation (i) by (ii) we get
0.5 10
thus e2 = 0.3v
=
e2
6
121. (d)
R1 2
=
...(i)
R2 3
Finally at null deflection, when null point is shifted
R1 + 10
= 1 Þ R1 + 10 = R2 ...(ii)
R2
Solving equations (i) and (ii) we get
2R 2
+ 10 = R 2
3
R
10 = 2 Þ R 2 = 30W
3
& R1 = 20W
Now if required resistance is R then
30 ´ R
30 + R = 2
30
3
R = 60W
123. (c) Let x be the length AJ at which galvanometer shows
null deflection current,
e
3
æx
ö e
i=
=
or, i ç 12r ÷ =
L
12r + r 13r
è
ø 2
e éx
e éx
ù e
ù e
·12r ú = Þ
·12r ú =
Þ
ê
ê
13r ë L
13r ë L
û 2
û 2
13L
or, x =
24
124. (a) Given, colour code of resistance,
R1 = Orange, Red and Brown
\ R1= 32 × 10 = 320
using balanced wheatstone bridge principle,
122. (c) Initially at null deflection
R1 R 2 320 80
= Þ
=
R3 R 4
R 3 40
\ R3 = 160 i.e. colour code for R3 Brown, Blue and Brown
125. (b) Using formula, internal resistance,
æl -l ö
52 - 40 ö
r = ç 1 2 ÷ s = æç
÷ ´ 5 = 1.5W
l
è 40 ø
è 2 ø
126. (c) R1 + R2 = 1000
Þ R2 = 1000 – R1
R1
On balancing condition
R1(100 – l) = (1000 – R1)l
...(i)
On Interchanging resistance balance point shifts left by 10 cm
R1
G
(l – 10)
(100 – l + 10)
=(110 – l)
On balancing condition
(1000 – R1) (110 – l) = R1 (l – 10)
or, R1 (l – 10) = (1000 – R1) (110 – l)
Dividing eqn (i) by (ii)
100 - l
l
=
l - 10 110 - l
Þ (100 – l) (110 – l) = l(l – 10)
Þ 11000 – 100l – 110l + l2 = l2 – 10l
Þ 11000 = 200l
or, l = 55
Putting the value of ‘l’ in eqn (i)
R1 (100 – 55) = (1000 – R1) 55
Þ R1 (45) = (1000 – R1) 55
Þ R1 (9) = (1000 – R1) 11
Þ 20 R1 = 11000
\ R1 = 550KW
127. (b) For a balanced meter bridge,
X
Y
Þ Y = 39.5 = X × (100 – 39.5)
=
39.5 (100 - 39.5)
12.5 ´ 39.5
or, X =
= 8.16 W
60.5
When X and Y are interchanged l1 and (100 – l1) will also
interchange so, l2 = 60.5 cm
128. (d) There is no change in null point, if the cell and the
galvanometer are exchanged in a balanced wheatstone
bridge.
On balancing condition
After exchange
R1 R2
=
R3 R4
R2 = 1000 – R1
G
(l )
100 – l
...(ii)
On balancing condition
R1 R3
=
R2
R
P-310
Physics
P
l
=
Q (100 - l )
Initially neutral position is 60 cm. from A, so
129. (d) In balance position of bridge,
4
Q
16 8
=
ÞQ= = W
60 40
6 3
Now, when unknown resistance R is connected in series
to P, neutral point is 80 cm from A then,
4+ R Q
=
80
20
4+ R 8
=
80
60
64
64 - 24 40
R=
-4=
=
W
6
6
6
20
Hence, the value of unknown resistance R is =
W
3
130. (d) When key is at point (1)
V1 = iR1 = xl1
When key is at (3)
V2 = i (R1 + R2) = xl2
R1
l
R
l
= 1 Þ 1= 1
R1 + R 2 l2
R 2 l2 - l1
131. (c) As the two cells oppose each other hence, the
effective emf in closed circuit is 15 – 10 = 5 V and net
resistance is 1 + 0.6 = 1.6 W (because in the closed
circuit the internal resistance of two cells are in series.
Current in the circuit,
I=
effective emf
5
=
A
total resistance 1.6
The potential difference across voltmeter will be same as
the terminal voltage of either cell.
Since the current is drawn from the cell of 15 V
\ V1 = E1 – Ir1
5
´ 0.6 = 13.1 V
= 15 –
1.6
132. (c) Balancing length l will give emf of cell
\ E = Kl
Here K is potential gradient.
If the cell is short circuited by resistance 'R'
Let balancing length obtained be l¢ then
V = kl¢
æ E -V ö
R
r= ç
è V ÷ø
Þ V = E – V [Q r = R given]
Þ 2V = E
or, 2Kl¢ = Kl
\
l¢ =
l
2
133. (c) To measure the emf of a cell we prefer potentiometer
rather than voltmeter because
(i) the length of potentiometer which allows greater
precision.
(ii) in case of potentiometer, no current flows through the
cell.
(iii) of high sensitivity.
134. (a) Bridge wire in a sensitive meter bridge wire should be
of high resistivity and low temperature coefficient.
x
40
2
135. (c) From question, y = 100 - 40 = 3
Þx=
2
y
3
Again,
3x
Z
=
y 100 - Z
or
2y
Z
3 =
y
100 - Z
3´
Solving we get Z = 67 cm
Therefore new position of null point @ 67 cm
136. (c) Potential gradient
Þk=
k=
V IR I æ rl ö I r
=
= ç ÷=
l
l lè A ø A
0.2 ´ 4 ´ 10 -7 0.8
=
= 0.1 V/m
8
8 ´ 10-7
137. (b) Given,
Balance point from one end, l1 = 20 cm
From the condition for balance of metre bridge, we have
55
l1
=
R 100 – l1
55 20
=
R 80
Þ R = 220W
138. (b) From balanced wheat stone bridge
S=
P R
where
=
Q S
S1S 2
S1 + S 2
139. (c) Initial balancing length, l1 = 240 cm New balancing
length, l2 = 120 cm.
The internal resistance of the cell,
æl -l ö
240 - 120
´ 2 = 2W
r= ç 1 2 ÷´R =
l
120
è
2 ø
P-311
Current Electricity
140. (c) From the balanced wheat stone bridge
E
R1 l1
=
R2 l 2
i
i
where l2 = 100 – l1
In the first case X = 20
Y 80
Y = 4X
In the second case
4X
l
=
Y
100 - l
Þ
4X
l
=
4 X 100 – l
Þ l = 50
141. (d) From the principle of potentiometer, V µ l
If a cell of emF E is employed in the circuit between the
ends of potentiometer wire of length L, then
V
l
= ;
E L
Þ V=
El 30E
=
L 100
r
E'
Note : In this arrangement, the internal resistance of the
battery E does not play any role as current is not passing
through the battery.
142. (d) ig × G = (i – ig) S
\ S=
ig ´ G
i - ig
=
1 ´ 0.81
= 0.09W
10 - 1
143. (c) To use an ammeter in place of voltmeter, we must
connect a high resistance in series with the ammeter.
Connecting high resistance in series makes its resistance
much higher.
18
P-312
Physics
Moving Charges
and Magnetism
Motion of Charged Particle in
TOPIC 1
Magnetic Field
1.
An electron is moving along + x direction with a velocity
of 6 × 106 ms–1. It enters a region of uniform electric field of
300 V/cm pointing along + y direction. The magnitude and
direction of the magnetic field set up in this region such
that the electron keeps moving along the x direction will
be :
[Sep. 06, 2020 (I)]
4.
5.
(a) 3 × 10–4 T, along + z direction
(b) 5 × 10–3 T, along – z direction
(c) 5 × 10–3 T, along + z direction
2.
(b) 0.88 m
(d) 3 × 10–4 T, along – z direction
A particle of charge q and mass m is moving with a
(c) 0.44 m
velocity – v $i (v ¹ 0) towards a large screen placed in the
Y-Z plane at a distance d. If there is a magnetic field
ur
B = B0 k$ , the minimum value of v for which the particle
will not hit the screen is:
[Sep. 06, 2020 (I)]
6.
qdB0
2qdB0
(b)
3m
m
qdB0
qdB0
(c)
(d)
m
2m
A charged particle carrying charge 1 mC is moving with
velocity (2iˆ + 3 ˆj + 4kˆ) ms–1. If an external magnetic field
7.
(a)
3.
of (5iˆ + 3 ˆj - 6kˆ) ´10-3 T exists in the region where the
particle is moving then the force on the particle is
ur
ur
[Sep. 03, 2020 (I)]
F ´10-9 N. The vector F is :
(a)
- 0.30iˆ + 0.32 ˆj - 0.09kˆ
(b)
- 30iˆ + 32 ˆj - 9kˆ
(c)
- 300iˆ + 320 ˆj - 90kˆ
(d)
- 3.0iˆ + 3.2 ˆj - 0.9kˆ
A beam of protons with speed 4 × 105 ms–1 enters a uniform
magnetic field of 0.3 T at an angle of 60° to the magnetic
field. The pitch of the resulting helical path of protons is
close to : (Mass of the proton = 1.67 × 10–27 kg, charge of
the proton = 1.69 × 10–19 C)
[Sep. 02, 2020 (I)]
(a) 2 cm
(b) 5 cm
(c) 12 cm (d) 4 cm
The figure shows a region of length ‘l’ with a uniform
magnetic field of 0.3 T in it and a proton entering the region
with velocity 4 × 105 ms–1 making an angle 60° with the
field. If the proton completes 10 revolution by the time it
cross the region shown, ‘l’ is close to (mass of proton
= 1.67 × 10–27 kg, charge of the proton = 1.6 × 10–19 C)
[Sep. 02, 2020 (II)]
B
(a) 0.11 m
60°
l
(d) 0.22 m
Proton with kinetic energy of 1 MeV moves from south to
north. It gets an acceleration of 1012 m/s2 by an applied
magnetic field (west to east). The value of magnetic field:
(Rest mass of proton is 1.6 ´ 10–27 kg) [8 Jan 2020, I]
(a) 0.71 mT
(b) 7.1 mT
(c) 0.071 mT
(d) 71 mT
A particle having the same charge as of electron moves
in a circular path of radius 0.5 cm under the influence of
a magnetic field of 0.5T. If an electric field of 100V/m
makes it to move in a straight path then the mass of the
particle is (Given charge of electron = 1.6 × 10–19C)
[12 April 2019, I]
kg
(b) 1.6 × 10–27 kg
(c) 1.6 × 10–19 kg
(d) 2.0 × 10–24 kg
(a) 9.1 ×
8.
10–31
An electron, moving along the x-axis with an initial energy
ur
of 100 eV, enters a region of magnetic field B = (1.5×10–3T) k$
at S (see figure). The field extends between x = 0 and x =
2 cm. The electron is detected at the point Q on a screen
P-313
Moving Charges and Magnetism
placed 8 cm away from the point S. The distance d between
P and Q (on the screen) is :
(Electron’s charge = 1.6 × 10–19 C, mass of electron
= 9.1 × 10–31 kg)
[12 April 2019, II]
(a) 11.65 cm
(c) 1.22 cm
(b) 12.87 cm
(d) 2.25 cm
9.
A proton, an electron, and a Helium nucleus, have the
same energy. They are in circular orbits in a plane due to
magnetic field perpendicular to the plane. Let rp, re and rHe
be their respective radii, then,
[10 April 2019, I]
(a) re > rp = rHe
(b) re < rp = rHe
(c) re < rp < rHe
(d) re > rp > rHe
10. A proton and an α -particle (with their masses in the ratio
of 1 : 4 and charges in the ratio 1 : 2) are accelerated from
rest through a potential difference V. If a uniform magnetic
field (B) is set up perpendicular to their velocities, the ratio
of the radii rp : ra of the circular paths descrfibed by them
will be:
[12 Jan 2019, I]
11.
(a) 1: 2
(b) 1: 2
(c) 1: 3
(d) 1: 3
In an experiment, electrons are accelerated, from rest, by
applying a voltage of 500 V. Calculate the radius of the
path if a magnetic field 100 mT is then applied.
[Charge of the electron = 1.6 × 10–19 C
Mass of the electron = 9.1 × 10–31 kg] [11 Jan 2019, I]
(a) 7.5 × 10–3 m
(b) 7.5 × 10–2 m
(c) 7.5 m
(d) 7.5 ×10–4 m
12. The region between y = 0 and y = d contains a magnetic
r
field B = Bzˆ . A particle of mass m and charge q enters
mv
the region with a velocity vr = viˆ . if d =
, the
2qB
acceleration of the charged particle at the point of its
emergence at the other side is :
[11 Jan 2019, II]
(a)
qv B æ 1 ˆ
3 ˆö
j÷
ç im è2
2 ø
(b)
qv B æ 3 ˆ 1 ˆ ö
i + j÷
ç
m è 2
2 ø
(c)
qvB æ - ˆj + iˆ ö
ç
÷
m è 2 ø
(d)
qvB æ iˆ + ˆj ö
ç
÷
m è 2 ø
13. An electron, a proton and an alpha particle having the same
kinetic energy are moving in circular orbits of radii re,
rp, ra respectively in a uniform magnetic field B. The
relation between re, rp, ra is :
[2018]
(a) re > rp = ra
(b) re < rp = ra
(c) re < rp < ra
(d) re < ra < rp
14. A negative test charge is moving near a long straight wire
carrying a current. The force acting on the test charge is
parallel to the direction of the current. The motion of the
charge is :
[Online April 9, 2017]
(a) away from the wire
(b) towards the wire
(c) parallel to the wire along the current
(d) parallel to the wire opposite to the current
15. In a certain region static electric and magnetic fields exist.
r
ˆ . If a
The magnetic field is given by B = B0 (iˆ + 2ˆj - 4k)
r
ˆ
test charge moving with a velocity v = v 0 (3iˆ - ˆj + 2k)
experiences no force in that region, then the electric field
in the region, in SI units, is :
[Online April 8, 2017]
r
r
ˆ (b) E = -v B (iˆ + ˆj + 7k)
ˆ
E = -v0B0(3iˆ - 2jˆ - 4k)
0 0
r
r
ˆ
ˆ
(c) E = v0 B0 (14jˆ + 7k)
(d) E = -v0 B0 (14jˆ + 7k)
16. Consider a thin metallic sheet perpendicular to the plane
of the paper moving with speed 'v' in a uniform magnetic
field B going into the plane of the paper (See figure). If
charge densities s1 and s2 are induced on the left and
right surfaces, respectively, of the sheet then (ignore fringe
effects):
[Online April 10, 2016]
(a)
v
- Î0 vB
Î vB
, s2 = 0
2
2
(a)
s1 =
(b)
s1 = Î0 vB, s 2 = - Î0 vB
(c)
Î vB
- Î0 vB
s1 = 0
, s2 =
2
2
B
(d) s1 = s 2 = Î0 vB
s1 s2
17. A proton (mass m) accelerated by a potential difference V
flies through a uniform transverse magnetic field B. The
field occupies a region of space by width ‘d’. If a be the
angle of deviation of proton from initial direction of motion
(see figure), the value of sin a will be :
[Online April 10, 2015]
B
a
d
Bd
2m
(a)
qV
(c)
B
q
d 2mV
(b)
B qd
2 mV
(d) Bd
q
2mV
P-314
Physics
18. A positive charge ‘q’ of mass ‘m’ is moving along the + x
axis. We wish to apply a uniform magnetic field B for
time Dt so that the charge reverses its direction crossing
the y axis at a distance d. Then: [Online April 12, 2014]
19.
20.
(a) B =
pd
mv
and Dt =
v
qd
(c) B =
2mv
pd
and Dt =
(d)
qd
2v
22.
pd
2mv
and Dt =
v
qd
ra = rp = rd
(b)
ra = rp < rd
rd > rp
(c) ra > rd > rp
(d) ra =
This question has Statement 1 and Statement 2. Of the
four choices given after the Statements, choose the one
that best describes the two Statements.
Statement 1: A charged particle is moving at right angle
to a static magnetic field. During the motion the kinetic
energy of the charge remains unchanged.
Statement 2: Static magnetic field exert force on a moving
charge in the direction perpendicular to the magnetic field.
[Online May 26, 2012]
(a) Statement 1 is false, Statement 2 is true.
(b) Statement 1 is true, Statement 2 is true, Statement 2 is
not the correct explanation of Statement 1.
(c) Statement 1 is true, Statement 2 is false.
(d) Statement 1 is true, Statement 2 is true, Statement 2 is
the correct explanation of Statement 1.
A proton and a deuteron are both accelerated through the
same potential difference and enter in a magnetic field
perpendicular to the direction of the field. If the deuteron
follows a path of radius R, assuming the neutron and proton
masses are nearly equal, the radius of the proton’s path
will be
[Online May 19, 2012]
R
R
(c)
(d) R
2
2
The magnetic force acting on charged particle of charge 2
mC in magnetic field of 2 T acting in y-direction, when the
(a)
23.
B=
pd
mv
and Dt =
2v
2qd
A particle of charge 16 × 10–16 C moving with velocity
10 ms–1 along x-axis enters a region where magnetic field
ur
of induction B is along the y-axis and an electric field
of magnitude 104 Vm–1 is along the negative z-axis. If
the charged particle continues moving along x-axis, the
ur
magnitude of B is :
[Online April 23, 2013]
(a) 16 × 103 Wb m–2
(b) 2 × 103 Wb m–2
(c) 1 × 103 Wb m–2
(d) 4 × 103 Wb m–2
Proton, deuteron and alpha particle of same kinetic energy
are moving in circular trajectories in a constant magnetic
field. The radii of proton, deuteron and alpha particle are
respectively rp, rd and ra. Which one of the following
relation is correct?
[2012]
(a)
21.
(b) B =
2R
(b)
(
)
6
-1
particle velocity is 2iˆ + 3 ˆj ´ 10 ms is
[Online May 12, 2012]
(a) 8 N in z-direction
(b) 8 N in y-direction
(c) 4 N in y-direction
(d) 4 N in z-direction
24. The velocity of certain ions that pass undeflected through
crossed electric field E = 7.7 k V/m and magnetic field
B = 0.14 T is
[Online May 7, 2012]
(a) 18 km/s
(b) 77 km/s
(c) 55 km/s
(d) 1078 km/s
r
25. An electric charge +q moves with velocity v = 3iˆ + 4 ˆj + kˆ
ur
in an electromagnetic field given by E = 3i$ + $j + 2k$ and
ur
B = iˆ + ˆj - 3kˆ The y - component of the force experienced
by + q is :
[2011 RS]
(a) 11 q
(b) 5 q
(c) 3 q
(d) 2 q
26. A charged particle with charge q enters a region of
ur
constant, uniform and mutually orthogonal fields E and
ur
ur
ur
r
B with a velocity v perpendicular to both E and B ,
and comes out without any change in magnitude or
r
direction of v . Then
[2007]
r ur ur 2
r ur ur 2
(a) v = B ´ E / E
(b) v = E ´ B / B
r ur ur 2
r ur ur
(c) v = B ´ E / B
(d) v = E ´ B / E 2
27. A charged particle moves through a magnetic field
perpendicular to its direction. Then
[2007]
(a) kinetic energy changes but the momentum is
constant
(b) the momentum changes but the kinetic energy is
constant
(c) both momentum and kinetic energy of the particle
are not constant
(d) both momentum and kinetic energy of the particle
are constant
28. In a region, steady and uniform electric and magnetic fields
are present. These two fields are parallel to each other. A
charged particle is released from rest in this region. The
path of the particle will be a
[2006]
(a) helix
(b) straight line
(c) ellipse
(d) circle
29. A charged particle of mass m and charge q travels on a
circular path of radius r that is perpendicular to a magnetic
field B. The time taken by the particle to complete one
revolution is
[2005]
2 pmq
2pm
2pqB
2pq 2 B
(b)
(d)
(c)
B
qB
m
m
30. A uniform electric field and a uniform magnetic field are
acting along the same direction in a certain region. If an
electron is projected along the direction of the fields with
a certain velocity then
[2005]
(a) its velocity will increase
(b) Its velocity will decrease
(c) it will turn towards left of direction of motion
(d) it will turn towards right of direction of motion
(a)
P-315
Moving Charges and Magnetism
31. A particle of mass M and charge Q moving with velocity
r
v describe a circular path of radius R when subjected to
a uniform transverse magnetic field of induction B. The
work done by the field when the particle completes one
full circle is
[2003]
32.
33.
æ Mv 2 ö
(a) ç
(b) zero
÷ 2 pR
è R ø
(d) B Qv 2p R
(c) B Q 2 p R
If an electron and a proton having same momenta enter
perpendicular to a magnetic field, then
[2002]
(a) curved path of electron and proton will be same
(ignoring the sense of revolution)
(b) they will move undeflected
(c) curved path of electron is more curved than that of
the proton
(d) path of proton is more curved.
The time period of a charged particle undergoing a circular
motion in a uniform magnetic field is independent of its
(a) speed
(b) mass
[2002]
(c) charge
(d) magnetic induction
36. Magnitude of magnetic field (in SI units) at the centre of
a hexagonal shape coil of side 10 cm, 50 turns and
m0 I
is :
p
[Sep. 03, 2020 (I)]
carrying current I (Ampere) in units of
(a)
250 3 (b) 50 3 (c) 500 3 (d) 5 3
37. A long, straight wire of radius a carries a current distributed uniformly over its cross-section. The ratio of the
a
magnetic fields due to the wire at distance
and 2a,
3
respectively from the axis of the wire is: [9 Jan 2020, I]
2
1
3
(b) 2
(c)
(d)
3
2
2
An electron gun is placed inside a long solenoid of radius
R on its axis. The solenoid has n turns/length and carries
a current I. The electron gun shoots an electron along the
radius of the solenoid with speed v. If the electron does
not hit the surface of the solenoid, maximum possible value
of v is (all symbols have their standard meaning):
(a)
38.
[9 Jan 2020, II]
Magnetic Field Lines,
TOPIC 2 Biot-Savart's
Law and Ampere's Circuital
34. A charged particle going around in a circle can be considered to be a current loop. A particle of mass m carrying charge q is moving in a plane wit speed v under the
®
influence of magnetic field B . The magnetic moment
of this moving particle :
[Sep. 06, 2020 (II)]
®
(a)
mv 2 B
2 B2
®
(b) -
35.
2 pB 2
mv 2 B
mv 2 B
(d) 2 B2
B2
A wire A, bent in the shape of an arc of a circle, carrying a
current of 2 A and having radius 2 cm and another wire B,
also bent in the shape of arc of a circle, carrying a current
of 3 A and having radius of 4 cm, are placed as shown in
the figure. The ratio of the magnetic fields due to the wires
A and B at the common centre O is :
[Sep. 04, 2020 (I)]
A
O
(a) 4 : 6
(c) 2 : 5
39.
(b)
em0 nIR
2m
em0 nIR
2em 0 nIR
(d)
4m
m
A very long wire ABDMNDC is shown in figure carrying
current I. AB and BC parts are straight, long and at
right angle. At D wire forms a circular turn DMND of
radius R.
AB, BC parts are tangential to circular turn at N and D.
Magnetic field at the centre of circle is:
[8 Jan 2020, II]
(a)
m0 I æ
1 ö
p+
÷
2 pR çè
2ø
(b)
m0 I æ
1 ö
p÷
2 pR çè
2ø
(c)
m0 I
(p + 1)
2pR
(d)
m0 I
2R
B
90°
60°
(b) 6 : 4
(d) 6 : 5
em0 nIR
m
(c)
mv 2 B
®
®
(c) -
(a)
P-316
45. As shown in the figure, two infinitely long, identical wires
are bent by 90º and placed in such a way that the segments
LP and QM are along the x-axis, while segments PS and
QN are parallel to the y-axis. If OP = OQ = 4 cm, and the
magnitude of the magneticf field at O is 10–4 T, and the
two wires carry equal currents (see figure), the magnitude
of the current in each wire and the direction of the magnetic
field at O will be (µ0 = 4p × 10–7 NA–2): [12 Jan 2019, I]
S
These wires carry currents of equal magnitude I, whose
directions are shown in the figure. The net magnetic field
at point P will be :
[12 April 2019, I]
(a) Zero
(b)
–
m0 I
( xˆ + yˆ )
2 pd
+m 0 I
m0 I
( zˆ )
( xˆ + yˆ )
(d)
pd
2 pd
A thin ring of 10 cm radius carries a uniformly distributed
charge. The ring rotates at a constant angular speed of 40
À rad s–1 about its axis, perpendicular to its plane. If the
magnetic field at its centre is 3.8 × 10–9 T, then the charge
carried by the ring is close to (µ0 = 4p × 10–7 N/A2).
[12 April 2019, I]
–6
–5
(a) 2×10 C (b) 3×10 C (c) 4×10–5C (d) 7×10–6C
Find the magnetic field at point P due to a straight line
segment AB of length 6 cm carrying a current of 5 A. (See
figure) (mo=4p×10–7 N-A–2)
[12 April 2019, II]
(c)
41.
OQ
P
L
m
3c
°
Q
2 cm
2 cm
R
P
43.
44.
(a) 2.0×10–5T
(b) 1.5×10–5T
(c) 3.0×10–5T
(d) 2.5×10–5T
The magnitude of the magnetic field at the center of an
equilateral triangular loop of side 1 m which is carrying a
current of 10 A is :
[10 April 2019, II]
[Take mo = 4p×10–7 NA–2]
(a) 18 mT (b) 9 mT
(c) 3 mT
(d) 1 mT
A square loop is carrying a steady current I and the
magnitude of its magnetic dipole moment is m. If this
square loop is changed to a circular loop and it carries the
same current, the magnitude of the magnetic dipole moment
of circular loop will be :
[10 April 2019, II]
(a)
m
p
(b)
3m
p
(c)
2m
p
(d)
4m
p
x
M
N
(a) 20 A, perpendicular out of the page
(b) 40 A, perpendicular out of the page
(c) 20 A, perpendicular into the page
(d) 40 A, perpendicular into the page
46. A current loop, having two circular arcs joined by two
radial lines is shown in the figure. It carries a current of 10
A. The magnetic field at point O will be close to:
[9 Jan. 2019 I]
O
3 cm
42.
y
5
Two very long, straight, and insulated wires are kept at
90° angle from each other in xy-plane as shown in the
figure.
q =4
40.
Physics
S
i = 10A
(a) 1.0 × 10–7 T
(b) 1.5 × 10–7 T
–5
(c) 1.5 × 10 T
(d) 1.0 × 10–5 T
47. One of the two identical conducting wires of length L is
bent in the form of a circular loop and the other one into
a circular coil of N identical turns. If the same current is
passed in both, the ratio of the magnetic field at the
central of the loop (B1) to that at the centre of the coil
B
(BC), i.e., L will be:
[9 Jan 2019, II]
BC
(a) N
(b)
(c) N2
(d)
1
N
1
N2
48. The dipole moment of a circular loop carrying a current I,
is m and the magnetic field at the centre of the loop is B1.
When the dipole moment is doubled by keeping the current
P-317
Moving Charges and Magnetism
current constant, the magnetic field at the centre of the
B
loop is B2. The ratio 1 is:
[2018]
B2
1
2
A Helmholtz coil has pair of loops, each with N turns and
radius R. They are placed coaxially at distance R and the
same current I flows through the loops in the same
direction. The magnitude of magnetic field at P, midway
between the centres A and C, is given by (Refer to figure):
(a) 2
49.
(b)
(c)
3
(d)
2
[Online April 15, 2018]
2R
A
C
P
R
(a)
50.
4 N m0 I
8 N m0 I
8 N m0 I
(b)
(c)
(d)
53/ 2 R
53/ 2 R
51/ 2 R
51/ 2 R
A current of 1A is flowing on the sides of an equilateral
triangle of side 4.5 × 10–2m . The magnetic field at the
centre of the triangle will be:
[Online April 15, 2018]
(a) 4 × 10–5Wb/m2
51.
4 N m0 I
(b) Zero
(c) 2 × 10–5Wb/m2
(d) 8 × 10–5Wb/m2
Two identical wires A and B, each of length 'l', carry the
same current I. Wire A is bent into a circle of radius R
and wire B is bent to form a square of side 'a'. If BA and
BB are the values of magnetic field at the centres of the
B
circle and square respectively, then the ratio A is:
BB
[2016]
p2
p2
p2
p2
(b)
(c)
(d)
16
8
8 2
16 2
Two long current carrying thin wires, both with current I,
are held by insulating threads of length L and are in
equilibrium as shown in the figure, with threads making
an angle 'q' with the vertical. If wires have mass l per unit
length then the value of I is :
(g = gravitational acceleration)
[2015]
(a)
52.
(a)
(b)
(c)
2
pgL
tan q
µ0
plgL
tan q
µ0
sin q
(d) 2sin q
L
q
plgL
µ0 cos q
plgL
µ0 cos q
I
I
53. Consider two thin identical conducting wires covered
with very thin insulating material. One of the wires is
bent into a loop and produces magnetic field B1, at its
centre when a current I passes through it. The ratio B1 :
B2 is:
[Online April 12, 2014]
(a) 1 : 1
(b) 1 : 3
(c) 1 : 9
(d) 9 : 1
54. A parallel plate capacitor of area 60 cm2 and separation 3
mm is charged initially to 90 mC. If the medium between the
plate gets slightly conducting and the plate loses the charge
initially at the rate of 2.5 × 10–8 C/s, then what is the magnetic
field between the plates ?
[Online April 23, 2013]
–8
(a) 2.5 × 10 T
(b) 2.0 × 10–7 T
(c) 1.63 × 10–11 T (d)
Zero
55. A current i is flowing in a straight conductor of length L.
The magnetic induction at a point on its axis at a distance
L
from its centre will be :
4
[Online April 22, 2013]
(a) Zero
(b)
m 0i
(d)
(c)
m0i
2pL
4m0i
2L
5pL
56. Choose the correct sketch of the magnetic field lines of a
circular current loop shown by the dot e and the cross
[Online April 22, 2013]
Ä.
(a)
(b)
(c)
(d)
57. An electric current is flowing through a circular coil of
radius R. The ratio of the magnetic field at the centre of
the coil and that at a distance 2 2R from the centre of
the coil and on its axis is :
[Online April 9, 2013]
(a) 2 2
(b) 27
(c) 36
(d) 8
58. A charge Q is uniformly distributed over the surface of
non-conducting disc of radius R. The disc rotates about
an axis perpendicular to its plane and passing through its
centre with an angular velocity w. As a result of this
rotation a magnetic field of induction B is obtained at
the centre of the disc. If we keep both the amount of charge
placed on the disc and its angular velocity to be constant
and vary the radius of the disc then the variation of the
magnetic induction at the centre of the disc will be
represented by the figure :
[2012]
P-318
Physics
B
B
(a)
(b)
R
R
B
B
(c)
59.
R
R
A current I flows in an infinitely long wire with cross section
in the form of a semi-circular ring of radius R. The magnitude
of the magnetic induction along its axis is:
[2011]
(a)
60.
(d)
m0 I
2p2 R
m0 I
2pR
(b)
(c)
m0 I
4pR
(d)
m0 I
p2R
Two long parallel wires are at a distance 2d apart. They
carry steady equal currents flowing out of the plane of
the paper as shown. The variation of the magnetic field B
along the line XX' is given by
[2010]
B
X¢
(a) X
d
X¢
d
d
X¢
X
d
d
X¢
X
d
61.
i
. The value of the magnetic field at its centre is
3
[2006]
(a) 1.05 × 10–2 Weber/m2 (b) 1.05 × 10–5 Weber/m2
(c) 1.05 × 10–3 Weber/m2 (d) 1.05 × 10–4 Weber/m2
66. Two concentric coils each of radius equal to 2 p cm are
placed at right angles to each other. 3 ampere and 4 ampere are the currents flowing in each coil respectively.
The magnetic induction in Weber/m2 at the centre of the
coils will be (m0 = 4p ´10-7 Wb / A.m)
B
(d)
)
1
a current
B
(c)
(
(b)
m 0 æ I1 + I 2 ö 2
ç
÷
2p è d ø
m0
m0
(d)
I 2 + I 22 2
( I1 + I 2 )
2 pd
2pd 1
65. A long solenoid has 200 turns per cm and carries a current
i. The magnetic field at its centre is 6.28 × 10–2 Weber/m2.
Another long solenoid has 100 turns per cm and it carries
d
X
(a)
1
m0
( I12 + I 2 2 )
2pd
(c)
B
(b)
(a) 2.5 × 10–7 T southward
(b) 5 × 10–6 T northward
(c) 5 × 10–6 T southward
(d) 2.5 × 10–7 T northward
62. A long straight wire of radius a carries a steady current i.
The current is uniformly distributed across its cross
section. The ratio of the magnetic field at a/2 and 2a is
[2007]
(a) 1/2
(b) 1/4
(c) 4
(d) 1
63. A current I flows along the length of an infinitely long,
straight, thin walled pipe. Then
[2007]
(a) the magnetic field at all points inside the pipe is the
same, but not zero
(b) the magnetic field is zero only on the axis of the
pipe
(c) the magnetic field is different at different points
inside the pipe
(d) the magnetic field at any point inside the pipe is zero
64. Two identical conducting wires AOB and COD are placed
at right angles to each other. The wire AOB carries an
electric current I1 and COD carries a current I2. The
magnetic field on a point lying at a distance d from O, in
a direction perpendicular to the plane of the wires AOB
and COD, will be given by
[2007]
d
A horizontal overhead powerline is at height of 4m from
the ground and carries a current of 100A from east to west.
The magnetic field directly below it on the ground is
(m0 = 4p×10 –7 Tm A–1)
[2008]
(a) 10 -5
[2005]
(b) 12 ´ 10 -5
(c) 7 ´ 10 -5
(d) 5 ´ 10 -5
67. A current i ampere flows along an infinitely long straight
thin walled tube, then the magnetic induction at any point
inside the tube is
[2004]
m 0 2i
.
tesla
(a)
(b) zero
4p r
2i
tesla
(c) infinite
(d)
r
P-319
Moving Charges and Magnetism
68. A long wire carries a steady current. It is bent into a circle
of one turn and the magnetic field at the centre of the coil
is B. It is then bent into a circular loop of n turns. The
magnetic field at the centre of the coil will be
[2004]
(a) 2n B
(b) n2 B
(c) nB
(d) 2 n2 B
69. The magnetic field due to a current carrying circular loop
of radius 3 cm at a point on the axis at a distance of 4 cm
from the centre is 54 mT. What will be its value at the
centre of loop?
[2004]
(a) 125 mT (b) 150 mT (c) 250 mT (d) 75 mT
70. If in a circular coil A of radius R, current I is flowing and in
another coil B of radius 2R a current 2I is flowing, then the
ratio of the magnetic fields BA and BB, produced by them
will be
[2002]
(a) 1
(b) 2
(c) 1/2
(d) 4
TOPIC 3
71.
A square loop of side 2a and carrying current I is kept is xz
plane with its centre at origin. A long wire carrying the
same current I is placed parallel to z-axis and passing
through point (0, b, 0), (b >> a). The magnitude of torque
on the loop about z-axis will be : [Sep. 06, 2020 (II)]
(a)
(a)
73.
2m 0 I 2 a 2
pb
(b)
m 0 I2 a 2b
2m 0 I 2 a 2 b
p(a 2 + b2 )
m 0 I2 a 2
2 p (a + b )
2 pb
A square loop of side 2a, and carrying current I, is kept in
XZ plane with its centre at origin. A long wire carrying the
same current I is placed parallel to the z-axis and passing
through the point (0, b, 0), (b >> a). The magnitude of the
torque on the loop about z-axis is given by :
[Sep. 05, 2020 (I)]
(c)
72.
Force and Torque on Current
Carrying Conductor
2
m0 I 2a2
2 pb
2
æ ˆj
kˆ ö
abI , along ç
+
÷
ç 2
2 ÷ø
è
æ ˆj
kˆ ö
(b)
2abI , along ç
+
÷
ç 2
2 ÷ø
è
æ ˆj
2kˆ ö
(c)
2abI , along ç
+
÷
ç 5
5 ÷ø
è
æ ˆj
2kˆ ö
(d) abI , along çç
+
÷
5 ÷ø
è 5
74. A small circular loop of conducting wire has radius a and
carries current I. It is placed in a uniform magnetic field B
perpendicular to its plane such that when rotated slightly
about its diameter and released, it starts performing simple
harmonic motion of time period T. If the mass of the loop is m
then :
[9 Jan 2020, II]
(a)
(a) T =
2m
IB
(b) T =
pm
2 IB
2pm
pm
(c) T =
IB
IB
75. Two wires A & B are carrying currents I1 and I2 as shown
in the figure. The separation between them is d. A third
wire C carrying a current I is to be kept parallel to them
at a distance x from A such that the net force acting on it
is zero. The possible values of x are : [10 April 2019, I]
(c) T =
(d)
(b)
(a)
æ I ö
I2
x = ç 1 ÷ d and x =
d
(I1 + I2 )
è I1 - I2 ø
(b)
æ I2 ö
æ I2 ö
x=ç
d and x = ç
d
÷
è (I1 + I2 ) ø
è (I1 - I2 ) ÷ø
m 0 I 2 a3
2pb2
2m 0 I 2 a 3
2m 0 I 2 a 2
(c)
(d)
pb
pb2
A wire carrying current I is bent in the shape ABCDEFA as
shown, where rectangle ABCDA and ADEFA are
perpendicular to each other. If the sides of the rectangles
are of lengths a and b, then the magnitude and direction of
magnetic moment of the loop ABCDEFA is :
[Sep. 02, 2020 (II)]
Z
E
I
I
F
C
D
Y
O
b
A a
B
X
æ I1 ö
æ I2 ö
(c) x = çè (I + I ) ÷ø d and x = çè (I - I ) ÷ø d
1
2
1
2
(d)
x=±
I1d
(I1 - I2 )
76. A rectangular coil (Dimension 5 cm × 2.5 cm) with 100
turns, carrying a current of 3 A in the clock-wise
direction, is kept centered at the origin and in the X-Z
plane. A magnetic field of 1 T is applied along X-axis. If
the coil is tilted through 45° about Z-axis, then the torque
on the coil is:
[9 April 2019 I]
(a) 0.38 Nm
(b) 0.55 Nm
(c) 0.42 Nm
(d) 0.27 Nm
P-320
77.
Physics
A rigid square of loop of side ‘a’ and carrying current I2 is
lying on a horizontal surface near a long current I1 carrying
wire in the same plane as shown in figure. The net force
on the loop due to the wire will be:
[9 April 2019 I]
I2
I1
cm carries a current 1 of 12 A. Out of the following different orientations which one corresponds to stable equilibrium ?
[Online April 9, 2017]
Z
Z
(a)
a
X
a
78.
m II
(a) Repulsive and equal to o 1 2
2p
mo I1I 2
(b) Attractive and equal to
3p
mo I1I 2
(c) Repulsive and equal to
4p
(d) Zero
A circular coil having N turns and radius r carries a current
I. It is held in the XZ plane in a magnetic field B. The
torque on the coil due to the magnetic field is :
[8 April 2019 I]
(a)
Br 2 I
pN
(b) Bpr2I N
Bpr 2 I
(d) Zero
N
An infinitely long current carrying wire and a small current
carrying loop are in the plane of the paper as shown. The
redius of the loop is a and distance of its centre from the
wire is d (d>>a). If the loop applies a force F on the wire
then:
[9 Jan. 2019 I]
(c)
79.
B
Bd c
I
I
a b
c
I
Y (b)
Z
B
B
(c)
b
a
I d
I c
Y (d)
b
I
I
I
y
I
x
d
z
B
2
æ a2 ö
æ aö
(c) F µ ç 3 ÷
(d) F µ ç ÷
è dø
èd ø
80. A charge q is spread uniformly over an insulated loop of
radius r . If it is rotated with an angular velocity w with
respect to normal axis then the magnetic moment of the
loop is
[Online April 16, 2018]
81.
1
4
3
qwr 2 (b)
qwr 2 (c)
qwr 2 (d) qwr 2
(a)
2
3
2
A uniform magnetic field B of 0.3 T is along the positive Zdirection. A rectangular loop (abcd) of sides 10 cm × 5
I d
I c
Y
83. A rectangular loop of sides 10 cm and 5 cm carrying a
current 1 of 12 A is placed in different orientations as shown
in the figures below :
z
(A)
æ aö
(b) F µ çè ÷ø
d
a
X
X
82. Two coaxial solenoids
of
different
radius
carry current I in
uur
force
on the inner
the same direction. F1 be the magnetic
uur
solenoid due to the outer one and F2 be the magnetic
force on the outer solenoid due to the inner one. Then :
[2015]
uur
uur
(a) F1 is radially inwards and F2 = 0
uur
uur
(b) F1 is radially outwards and F2 = 0
uur uur
(c) F1 = F2 = 0
uur
uur
(d) F1 is radially inwards and F2 is radially outwards
B
(a) F = 0
Y
b
X
Z
d
I
a
(B)
I
I
I
z
x
y
B
I
(C)
I
I
I
y
I
x
z
B
(D)
I
I
x
I
I
y
P-321
Moving Charges and Magnetism
84.
If there is a uniform magnetic field of 0.3 T in the positive
z direction, in which orientations the loop would be in (i)
stable equilibrium and (ii) unstable equilibrium ?
[2015]
(a) (B) and (D), respectively
(b) (B) and (C), respectively
(c) (A) and (B), respectively
(d) (A) and (C), respectively
Two long straight parallel wires, carrying (adjustable)
current I1 and I2, are kept at a distance d apart. If the force
‘F’ between the two wires is taken as ‘positive’ when the
wires repel each other and ‘negative’ when the wires
attract each other, the graph showing the dependence of
‘F’, on the product I1 I2, would be :
[Online April 11, 2015]
F
F
(a)
O
I1I2
(b)
O
85.
O
I1I2
(b)
(c) IBR
(d)
86.
(d)
O
q0
y
B
x
–1.5
(a) 1.57 W (b) 2.97 W (c) 14.85 W (d) 29.7 W
87. Three straight parallel current carrying conductors are
shown in the figure. The force experienced by the middle
conductor of length 25 cm is: [Online April 11, 2014]
I1 = 30 A
I2 = 20 A
5 cm
I1I2
Q
P
I
2.0
3 cm
A wire carrying current I is tied between points P and
Q and is in the shape of a circular arc of radius R due
to a uniform magnetic field B (perpendicular to the plane
of the paper, shown by xxx) in the vicinity of the wire.
If the wire subtends an angle 2q0 at the centre of the
circle (of which it forms an arc) then the tension in the
wire is :
[Online April 11, 2015]
IBR
(a)
B
2sin q0
IBRq0
sin q0
1.5
F
F
(c)
I1I2
z
I = 10 A
10–4
(a) 3 ×
N toward right
–4
(b) 6 × 10 N toward right
(c) 9 × 10–4 N toward right
(d) Zero
88. A rectangular loop of wire, supporting a mass m, hangs
r
with one end in a uniform magnetic field B pointing out
of the plane of the paper. A clockwise current is set up
such that i > mg/Ba, where a is the width of the loop.
Then :
[Online April 23, 2013]
R
IBR
sin q0
A conductor lies along the z-axis at -1.5 £ z < 1.5 m and
carries a fixed current of 10.0 A in -â z direction (see figure).
r
For a field B = 3.0 ´10-4 e-0.2x aˆ y T, find the power required
to move the conductor at constant speed to x = 2.0 m,
y = 0 m in 5 ´10-3 s. Assume parallel motion along the
x-axis.
[2014]
Q
P
y
Ii
S
x
a
mg
R
(a) The weight rises due to a vertical force caused by
the magnetic field and work is done on the system.
(b) The weight do not rise due to vertical for caused by
the magnetic field and work is done on the system.
P-322
89.
90.
Physics
(c) The weight rises due to a vertical force caused by the
magnetic field but no work is done on the system.
(d) The weight rises due to a vertical force caused by
the magnetic field and work is extracted from the
magnetic field.
Currents of a 10 ampere and 2 ampere are passed through
two parallel thin wires A and B respectively in opposite
directions. Wire A is infinitely long and the length of the
wire B is 2 m. The force acting on the conductor B, which
is situated at 10 cm distance from A will be
[Online May 26, 2012]
(a) 8 × 10–5 N
(b) 5 × 10–5 N
(c) 8p × 10–7 N
(d) 4p × 10–7 N
The circuit in figure consists of wires at the top and bottom
and identical springs as the left and right sides. The wire
at the bottom has a mass of 10 g and is 5 cm long. The wire
is hanging as shown in the figure. The springs stretch 0.5
cm under the weight of the wire and the circuit has a total
resistance of 12 W. When the lower wire is subjected to a
static magnetic field, the springs, stretch an additional
0.3 cm. The magnetic field is
[Online May 12, 2012]
mo I é b - a ù
4p êë ab úû
mo I
[2(b - a ) + p / 3(a + b)]
(c)
4p
(d) zero
92. Due to the presence of the current I1 at the origin:
(a) The forces on AD and BC are zero.
(b) The magnitude of the net force on the loop is given
(b)
I1 I
mo [2(b - a ) + p / 3(a + b] .
4p
(c) The magnitude of the net force on the loop is given
by
m o II1
(b - a).
24 ab
(d) The forces on AB and DC are zero.
Two long conductors, separated by a distance d carry
current I1 and I2 in the same direction. They exert a force F
on each other. Now the current in one of them is increased
to two times and its direction is reversed. The distance is
also increased to 3d. The new value of the force between
them is
[2004]
by
93.
24 V
2F
F
F
(b)
(c) –2 F
(d) 3
3
3
If a current is passed through a spring then the spring
will
[2002]
(a) expand
(b) compress
(c) remains same
(d) none of these
Wires 1 and 2 carrying currents i1 and i2 respectively are
inclined at an angle θ to each other. What is the force on
a small element dl of wire 2 at a distance of r from wire 1 (as
shown in figure) due to the magnetic field of wire 1? [2002]
(a) m 0 i1i2 dl tan q
1
2
2 pr
m0
i1
i1i2 dl sin q
(b)
i2
r
2pr
m0
dl
q
(c)
i1i2 dl cos q
2 pr
m0
(d)
i1i2 dl sin q
4 pr
(a)
Magnetic
field region
94.
5 cm
95.
(a) 0.6 T and directed out of page
(b) 1.2 T and directed into the plane of page
(c) 0.6 T and directed into the plane of page
(d) 1.2 T and directed out of page
Directions : Question numbers 91 and 92 are based on the
following paragraph.
A current loop ABCD is held fixed on the plane of the paper as
shown in the figure. The arcs BC (radius = b) and DA (radius =
a) of the loop are joined by two straight wires AB and CD. A
steady current I is flowing in the loop. Angle made by AB and
CD at the origin O is 30°. Another straight thin wire with steady
current I1 flowing out of the plane of the paper is kept at the
origin.
[2009]
B
a
I1
A
30°
O
I
D
b
91.
C
The magnitude of the magnetic field (B) due to the loop
ABCD at the origin (O) is :
(a)
m o I (b - a )
24ab
-
Galvanometer and its
TOPIC 4 Conversion into Ammeter and
Voltmeter
96. A galvanometer of resistance G is converted into a voltmeter of ragne 0 – 1V by connecting a resistance R1 in
series with it. The additional resistance R1 in series with
it. The additional resistance that should be connected in
series with R1 to increase the range of the voltmeter to 0
– 2V will be :
[Sep. 05, 2020 (I)]
(a) G
(b) R1
(c) R1 – G
(d) R1 + G
P-323
Moving Charges and Magnetism
97. A galvanometer is used in laboratory for detecting the
null point in electrical experiments. If, on passing a
current of 6 mA it produces a deflection of 2°, its figure
of merit is close to :
[Sep. 05, 2020 (II)]
(a) 333° A/div. (b)
6 × 10–3 A/div.
(c) 666° A/div. (d)
3 × 10–3 A/div.
98. A galvanometer coil has 500 turns and each turn has an
average area of 3 × 10–4 m2. If a torque of 1.5 Nm is required
to keep this coil parallel to a magnetic field when a current
of 0.5 A is flowing through it, the strength of the field (in T)
is __________.
[NA Sep. 03, 2020 (II)]
99. A galvanometer of resistance 100 W has 50 divisions on
its scale and has sensitivity of 20 µA/division. It is to be
converted to a voltmeter with three ranges, of 0–2V, 0–10
V and 0–20 V. The appropriate circuit to do so is :
[12 April 2019, I]
102.
103.
104.
(a)
(b)
105.
(c)
0 – 5 V. Therefore the value of shunt resistance required
to convert the above galvanometer into an ammeter of range
0 – 10 mA is :
[10 April 2019, I]
(a) 500 W (b) 100 W (c) 200 W (d) 10 W
A moving coil galvanometer has resistance 50 W and it
indicates full deflection at 4 mA current. A voltmeter is
made using this galvanometer and a 5 k W resistance. The
maximum voltage, that can be measured using this
voltmeter, will be close to:
[9 April 2019 I]
(a) 40 V
(b) 15 V
(c) 20 V
(d)
10 V
A moving coil galvanometer has a coil with 175 turns and
area 1 cm2. It uses a torsion band of torsion constant 10–
6 N-m/rad. The coil is placed in a magnetic field B parallel
to its plane. The coil deflects by 1° for a current of 1mA.
The value of B (in Tesla) is approximately:
[9 April 2019, II]
(a) 10–4
(b) 10–2
(c) 10–1
(c) 10–3
The resistance of a galvanometer is 50 ohm and the
maximum current which can be passed through it is 0.002
A. What resistance must be connected to it order to
convert it into an ammeter of range 0 – 0.5 A?
[9 April 2019, II]
(a) 0.5 ohm
(b) 0.002 ohm
(c) 0.02 ohm
(d) 0.2 ohm
The galvanometer deflection, when key K1 is closed but
K2 is open, equals θ 0 (see figure). On closing K2 also
and adjusting R2 to 5W, the deflection in galvanometer
θ0
. The resistance of the galvanometer is, then,
5
given by [Neglect the internal resistance of battery]:
[12 Jan 2019, I]
becomes
(d)
100. A moving coil galvanometer, having a resistance G,
produces full scale deflection when a current Ig flows
through it. This galvanometer can be converted into (i)
an ammeter of range 0 to I0 (I0 > Ig) by connecting a shunt
resistance RA to it and (ii) into a voltmeter of range 0 to V
(V=GI0) by connecting a series resistance Rv to it. Then,
[12 April 2019, II]
ö
RA æ I g
=ç
÷÷ and
RV çè I 0 - I g
ø
(a)
æ I0 - I g
R A RV = G ç
ç Ig
è
(b)
æ Ig
R
RA RV = G and A = çç
RV è I 0 - I g
(c)
æ Ig
R A RV = G ç
ç I0 - I g
è
(d)
2
2
R1 = 220 W
G
2
ö
÷÷
ø
2
K1
ö
÷÷
ø
ö
RA æ I 0 - I g
=ç
÷÷ and
RV çè I g
ø
Ig
RA
=
RA RV = G 2 and
RV ( I0 - I g )
2
R2
K2
2
ö
÷÷
ø
101. A moving coil galvanometer allows a full scale current
of 10– 4 A. A series resistance of 2 MW is required to
convert the above galvanometer into a voltmeter of range
(a) 5W
(b) 22W
(c) 25W
(d) 12W
106. A galvanometer, whose resistance is 50 ohm, has 25
divisions in it. When a current of 4 × 10–4 A passes through
it, its needle (pointer) deflects by one division. To use this
galvanometer as a voltmeter of range 2.5 V, it should be
connected to a resistance of :
[12 Jan 2019, II]
(a) 250 ohm
(b) 200 ohm
(c) 6200 ohm
(d) 6250 ohm
P-324
Physics
107. A galvanometer having a resistance of 20 W and 30
division on both sides has figure of merit 0.005 ampere/
division. The resistance that should be connected in series
such that it can be used as a voltmeter upto 15 volt, is:
[11 Jan 2019, II]
(a) 100 W (b) 120 W (c) 80 W
(d) 125 W
108. A galvanometer having a coil resistance 100 W gives a full
scale deflection when a current of 1 mA is passed through
it. What is the value of the resistance which can convert
this galvanometer into a voltmeter giving full scale
deflection for a potential difference of 10 V?
[8 Jan 2019, II]
(a) 10 kW
(b) 8.9 kW
(c) 7.9 kW
(d) 9.9 kW
109. In a circuit for finding the resistance of a galvanometer by
half deflection method, a 6 V battery and a high resistance
of 11kW are used. The figure of merit of the galvanometer
60mA/division. In the absence of shunt resistance, the
galvanometer produces a deflection of q = 9 divisions when
current flows in the circuit. The value of the shunt
resistance that can cause the deflection of q/2 , is closest
to
[Online April 16, 2018]
(a) 55W
(b) 110W
(c) 220W (d) 550W
110. A galvanometer with its coil resistance 25W requires a
current of 1mA for its full deflection. In order to construct
an ammeter to read up to a current of 2A, the approximate
value of the shunt resistance should be
[Online April 16, 2018]
(a) 2.5 × 10–2W
(b) 1.25 × 10–3W
(c) 2.5 × 10–3W
(d) 1.25 × 10–2W
111. When a current of 5 mA is passed through a galvanometer
having a coil of resistance 15 W , it shows full scale
deflection. The value of the resistance to be put in series
with the galvanometer to convert it into to voltmeter of
range 0 - 10 V is
[2017]
(a)
2.535 ´10 W
3
(b)
114. To know the resistance G of a galvanometer by half
deflection method, a battery of emf VE and resistance R
is used to deflect the galvanometer by angle q. If a shunt
of resistance S is needed to get half deflection then G, R
and S related by the equation:
[Online April 9, 2016]
(a) S (R + G) = RG
(b) 2S (R + G) = RG
(c) 2G = S
(d) 2S = G
115. The AC voltage across a resistance can be measured
using a :
[Online April 11, 2015]
(a) hot wire voltmeter
(b) moving coil galvanometer
(c) potential coil galvanometer
(d) moving magnet galvanometer
116. In the circuit diagrams (A, B, C and D) shown below, R is
a high resistance and S is a resistance of the order of
galvanometer r esistance G. The correct circuit,
corresponding to the half deflection method for finding
the resistance and figure of merit of the galvanometer, is
the circuit labelled as:
[Online April 11, 2014]
K2
R
S
G
(A)
K1
K2
R
G
S
(B)
4.005 ´10 W
3
(c) 1.985 ´ 10 W
(d) 2.045 ´ 103 W
112. A galvanometer having a coil resistance of 100 W gives a
full scale deflection, when a currect of 1 mA is passed
through it. The value of the resistance, which can convert
this galvanometer into ammeter giving a full scale deflection
for a current of 10 A, is :
[2016]
(a) 0.1 W (b) 3W
(c) 0.01W (d) 2W
113. A 50 W resistance is connected to a battery of 5V. A
galvanometer of resistance 100 W is to be used as an
ammeter to measure current through the resistance, for
this a resistance rs is connected to the galvanometer. Which
of the following connections should be employed if the
measured current is within 1% of the current without the
ammeter in the circuit ?
[Online April 9, 2016]
(a) rs = 0.5 W in series with the galvanometer
(b) rs = 1 W in series with galvanometer
(c) rs = 1W in parallel with galvanometer
(d) rs = 0.5 W in parallel with the galvanometer.
3
K1
K2
S
G
R
(C)
K1
S
R
G
(D)
K1
K2
P-325
Moving Charges and Magnetism
(a) Circuit A with G =
RS
Ii
( R - S)
R1
(b) Circuit B with G = S
(c) Circuit C with G = S
(d) Circuit D with G =
R2
G
RS
( R - S)
117. This questions has Statement I and Statement II. Of the
four choices given after the Statements, choose the one
that best describes into two Statements.
Statement-I : Higher the range, greater is the resistance of
ammeter.
Statement-II : To increase the range of ammeter, additional
shunt needs to be used across it.
[2013]
(a) Statement-I is true, Statement-II is true, Statement-II
is the correct explanation of Statement-I.
(b) Statement-I is true, Statement-II is true, Statement-II
is not the correct explanation of Statement-I.
R3
(a) 107 W (b) 137 W (c) 107/2 W (d) 77 W
119. A shunt of resistance 1 W is connected across a
galvanometer of 120 W resistance. A current of 5.5 ampere
gives full scale deflection in the galvanometer. The current
that will give full scale deflection in the absence of the
shunt is nearly :
[Online April 9, 2013]
(a) 5.5 ampere
(b) 0.5 ampere
(c) 0.004 ampere
(d) 0.045 ampere
120. In the circuit , the galvanometer G shows zero deflection.
If the batteries A and B have negligible internal resistance,
the value of the resistor R will be [2005]
500 W
G
(c) Statement-I is true, Statement-II is false.
(d) Statement-I is false, Statement-II is true.
118. To find the resistance of a galvanometer by the half
deflection method the following circuit is used with
resistances R1= 9970 W, R2 = 30 W and R3 = 0. The
deflection in the galvanometer is d. With R3 = 107 W the
deflection changed to
d
. The galvanometer resistance is
2
approximately :
[Online April 22, 2013]
2V
12V
B
R
A
(a) 100 W (b) 200W (c) 1000 W (d) 500 W
121. A moving coil galvanometer has 150 equal divisions. Its
current sensitivity is 10-divisions per milliampere and
voltage sensitivity is 2 divisions per millivolt. In order that
each division reads 1 volt, the resistance in ohms needed
to be connected in series with the coil will be [2005]
(a) 105
(b) 103
(c) 9995
(d) 99995
P-326
1.
Physics
r
(c) E = 300 ˆj V/cm = 3 ´ 104 V/m
r
V = 6 ´ 106 iˆ
E
4.
\ Pitch = (V cos q)
y
E = 300 j
z
5.
r
B must be in +z axis.
r
r r
qE + qV ´ B = 0
2pm
qB
2pm
qB
= (4 ´ 105 cos 60°)
x
V/cm = 3– ´ 104 V/m
e
V
V = 6 ´ 106 i$
2p æ 1.67 ´10-27
ç
0.3 çè 1.69 ´10-19
E 3 ´ 104
=
= 5 ´ 10-3 T
V 6 ´ 106
Hence, magnetic field B = 5 × 10–3 T along +z direction.
(c) In uniform magnetic field particle moves in a circular
path, if the radius of the circular path is 'r', particle will not
hit the screen.
\ B=
(c) Time period of one revolution of proton, T =
\ Length of region, l = 10 ´ (v cos q)T
Þ l = 10 ´ v cos 60° ´
Þl=
2pm
qB
20pmv 20 ´ 3.14 ´ 1.67 ´ 10 -27 ´ 4 ´ 105
=
qB
1.6 ´ 10 -19 ´ 0.3
Þ l = 0.44 m
d
6.
r=
mv
qB0
é mv
ù
= qvB0 ú
êQ
r
ë
û
(a)
2
Hence, minimum value of v for which the particle will not
hit the screen.
qB0 d
m
(a) [Given: q = 1mC = 1 ´ 10-6 C;
r
V = (2iˆ + 3 ˆj + 4kˆ) m/s and
r
B = (5iˆ + 3 ˆj - 6kˆ) ×10 –3 T ]
As we know, magnetic force F = qvB = ma
r æ qvB ö
\ a =ç
è m ÷ø perpendicular to velocity..
v=
3.
iˆ ˆj
r
r r
-6
-3
F = q(V ´ B) = 10 ´ 10 2 3
\ Also v =
5 3 -6
= (-30iˆ + 32 ˆj - 9kˆ) ´ 10-9 N
r
\ F = ( -30iˆ + 32 ˆj - 9kˆ)
2KE
2 ´ e ´ 106
=
m
m
\ a=
qvB eB 2 ´ e ´ 106
=
m
m
m
\ 1012
æ 1.6 ´ 10 –19 ö 2
. 2 ´ 10 3 B
=ç
–27 ÷
´
1.67
10
è
ø
kˆ
4
ö
÷÷ = 4 cm
ø
Here, m = mass of proton
q = charge of proton
B = magnetic field.
Linear distance travelled in one revolution,
p = T(v cos q) (Here, v = velocity of proton)
E = VB
2.
(d) Pitch = (v cos q)T and T =
3
\B;
1
2
´ 10–3 T = 0.71 mT (approx)
2pm
qB
P-327
Moving Charges and Magnetism
7.
(d) As particle is moving along a circular path
mv
\R =
qB
Path is straight line, then
qE = qvB
...(i)
E
B
From equation (i) and (ii)
E = vB Þ v =
–19
12.
....(ii)
= 7.5 ´ 10 -4
(BONUS)
Assuming particle enters from (0, d)
r=
´ ( 0.5) ´ 0.5 ´10 –2
100
8.
qB2 R 1.6 ´ 10
=
E
\ m = 2.0 × 10–24 kg
(b)
9.
(b) As mvr = qvB Þ r =
m=
r=
2
mv
=
qB
2mK.E.
qB
a=
r=
mv
qB
1
2
r=
2mk
eB
(Q p = mv =
=
2meV
eB
(Q k = eV)
2m
V
e
=
B
2mk
2 ´ 9.1 ´ 10 -31
1.6 ´ 10 -19
100 ´ 10-3
)
(500)
For proton, rp =
2Km e
....(i)
eB
2Kmp
....(ii)
eB
2Km a
qa B
=
2K4m p
2eB
=
2Km p
eB
...(iii)
\ r e < rp = ra
(Q me < mp)
14. (b) The force is parallel to the direction
of current in magnetic field,
r
hence F = q(v ´ B)
According to Fleming's left hand rule,
F
2mqDV
m
Þrµ
qB
q
(d) Radius of the path (r) is given by r =
r=
2Km
qB
For a particle, ra =
2mKE
(Q p = mv = 2mKE )
Þ r=
qB
Q KE = qDV
11.
qVB é - 3i - j ù
ê
ú
m ë
2 û
For electron, re =
(a) Radius of the circular path will be r =
ra
V
this option is not given in the all above four choices.
13. (b) As we know, radius of circular path in magnetic field
m
q
=
Fm
C
\ rHe = rp > re
rp
30°
r
Þ mv = 2m K.E.]
\
r
2
r/2
For proton, electron and a-particle,
mHe = 4mp and mp >> me
Also aHe = 2qp and qp = qe
\ As KE of all the particles is same then,
\ r=
d=
(0, 0)
Þ m 2 v2 = 2m K.E.
10.
mv
,
qB
(0, d)
1
[As : mv2 = K.E.
2
ra
9.1
´ 10 -10
3
0.16
= ´ 10 -4
-1
.4
10
mv
qB
I
v
e
BÄ
we have, the direction of motion of charge is towards the
wire.
15. (d) According to question, as the test charge experiences
no net force in that region i.e., sum of electric force
r
r r
(Fe = qE) and magnetic forces [Fm = q(v ´ B] will be zero.
Hence, Fe + Fm = 0
P-328
Physics
r r
Fe = -q(v ´ B)
= - B0 v0 é( 3iˆ - ˆj + 2kˆ ) ´ ( i + 2jˆ - 4kˆ ) ù
ë
û
ˆ
ˆ
= - B v (14 j + 7k )
16.
0 0
(b) Q F = qE and F = qvB
\ E = vB
And Gauss's law in Electrostatics E =
E=
17.
s
= vB Þ s = e vB
0
e0
s
e0
σ1= – σ 2
(d) From figure, sin a = d/R
a
R
a
d
20. (b) The centripetal force is provided by the magnetic force
m
mv
mv 2
\
= qvB Þ r =
\ rµ
q
Bq
R
\ rp : rd : ra =
mp
qp
:
md
qd
:
ma
qa
= 1 : 2 :1
Thus we have, ra = rp < rd
21. (d) When a charged particle enters the magnetic field
in perpendicular direction then it experience a force in
perpendicular direction.
i.e. F = Bqv sinq
Due to which it moves in a circular path.
22. (b) As charge on both proton and deuteron is same i.e. 'e'
Energy acquired by both, E = eV
For Deuteron.
1
Kinetic energy, mV2 = eV
2
[V is the potential difference]
2eV
md
But md = 2m
v=
And we know,
Þ R=
mv
qB
\ sin a =
mv 2
= qvB
R
dqB
mv
q éQ qV = 1 mv 2 ù
ê
ú
2
û
2mV ë
(c) The applied magnetic field provides the required
centripetal force to the charge particle, so it can move in
d
circular path of radius
2
2eV
eV
=
m
2m
mv
Radius of path, R =
eB
Substituting value of 'v' we get
Therefore, v =
18.
\ Bqv =
mv 2
d/2
2mv
qd
Time interval for which a uniform magnetic field is applied
or,
B=
d
Dt = 2
v
(particle reverses its direction after time Dt by covering
semi circle).
p.
pd
2v
(c) Since particle is moving undeflected
Dt =
19.
So, q E = qvB
ÞB=
E 104
=
= 103 wb / m 2
V 10
ev
m
eB
2m
sin a = Bd
R=
ev
m
R
m
=
2
eB
For proton :
1
mV 2 = eV
2
V=
...(i)
2eV
m
mV
Radius of path, R¢ =
=
eB
R
R¢ = 2 ´ [From eq. (i)]
2
R¢ =
m
R
2
®
æ ® ®ö
23. (a) F = q ç V ´ B ÷
è
ø
-6
6
= 2 ´10 éë(2iˆ + 3 ˆj ) ´10 ´ 2 ˆj ùû
= 2 ´ 4kˆ = 8 N in Z-direction.
2eV
m
eB
P-329
Moving Charges and Magnetism
24.
25.
26.
E 7.7 ´ 103
(c) As velocity v = =
= 55 km/s
B
0.14
(a) The charge experiences both electric and magnetic
force.
Electric force, Fe = qE
r r
Magnetic force, Fm = q ( v ´ B )
ur
ur r ur
\ Net force, F = q é E + v ´ B ù
ë
û
é
iˆ ˆj kˆ ù
ê
ú
= q ê 3$i + $j + 2k$ + 3 4 1 ú
ê
1 1 -3 ú
ë
û
$
$
$
ˆ
= q é3i + j + 2k + i ( -12 - 1) - $j ( -9 - 1) + k ( 3 - 4 )ù
ë
û
$
$
$
$
$
ˆ
é
ù
= q ë3i + j + 2k - 13i + 10 j - k û
= q éë -10i$ + 11$j + k$ ùû
Fy = 11qjˆ
Thus, the y component of the force.
(b) As velocity is not changing, charge particle must go
undeflected, then
qE = qvB
E
Þ v=
B
Also,
r r
E´B
E B sin q
=
2
B
B2
E
r
= |v| =v
B
B
(b) When a charged particle enters a magnetic field at a
direction perpendicular to the direction of motion, the path
of the motion is circular. In circular motion the direction of
velocity changes at every point (the magnitude remains
constant).
Therefore, the tangential momentum will change at every
point. But kinetic energy will remain constant as it is given
1
by mv 2 and v2 is the square of the magnitude of velocity
2
which does not change.
(b) The charged particle will move along the lines of
electric field (and magnetic field). Magnetic field will exert
no force. The force by electric field will be along the lines
of uniform electric field. Hence the particle will move in a
straight line.
(c) Equating magnetic force to centripetal force,
=
27.
28.
29.
E B sin 90°
2
30. (b) Due to electric field, it experiences force and
accelerates i.e. its velocity decreases.
31. (b) The workdone, dW = Fds cosq
The angle between force and displacement is 90°.
Therefore work done is zero.
×
×
×
×
×
F
×
×
S
32. (a) When a moving charged particle is subjected to a
perpendicular magnetic field, then it describes a circular
path of radius.
p
r=
qB
where q = Charge of the particle
p = Momentum of the particle
B = Magnetic field
Here p, q and B are constant for electron and proton, therefore
the radius will be same.
33. (a) The time period of a charged particle of charge q and
2 pm
mass m moving in a magnetic field (B) is T =
qB
Clearly time period is independent of speed of the particle.
34. (d)
v
=
mv 2
= qvB sin 90º
r
Bqr
mv
Þ
= Bq Þ v =
r
m
Time to complete one revolution,
2pr 2pm
=
T=
v
qB
×
+
+q
Length of the circular path, l = 2pr
q
qv
=
T 2 pr
Magnetic moment M = Current × Area
Current, i =
= i ´ pr 2 =
M =
qv
´ pr 2
2 pr
1
q×v×r
2
Radius of circular path in magnetic field, r =
mv
qB
1
mv
mv 2
qv ´
ÞM =
2
qB
2B
r
r
Direction of M is opposite of B therefore
r
r - mv 2 B
M =
2B 2
(By multiplying both numerator and denominator by B).
\M =
P-330
35.
Physics
(d) Given : IA = 2 A, RA = 2 cm, q A = 2p -
p 3p
=
2 2
p 5p
=
3
3
m Iq
Using, magnetic field, B = 0
4pR
3p
2´ ´ 4
B A I A q A RB
6
2
= ´
=
=
5p
BB I B q B RA
5
3´ ´ 2
3
IB = 3 A, RB = 4 cm, q B = 2p -
36.
Magnetic field at point B (outside)
m 0i
BB =
2p ( 2 a )
m 0i
6pa = 4 = 2
m 0i
6 3
2p (2 a)
38. (b) Magnetic field inside the solenoid is given by
B = µ0nI
.... (i)
Here, n = number of turns per unit length
BA
=
\ BB
(c)
30°
30°
a
I
3a
2
Magnetic field due to one side of hexagon
m0 I
B=
(sin 30° + sin 30°)
3a
4p
2
The path of charge particle is circular. The maximum
R
possible radius of electron =
2
mVmax R
=
\
qB
2
qBR eRm 0 nI
Þ Vmax =
=
(using (i))
2m
2m
m0 I æ 1 1 ö
m0 I
çè + ÷ø =
2
2
2 3a
2 3ap
Now, magnetic field due to one hexagon coil
ÞB=
B = 6´
m0 I
39. (a)
2 3ap
Again magnetic field at the centre of hexagonal shape coil
of 50 turns,
B = 50 ´ 6 ´
10
é
ù
êëQ a = 100 = 0.1 m úû
2 3ap
m0 I
p
3 ´ 0.1 ´ p
(a) Let a be the radius of the wire
Magnetic field at point A (inside)
a
m 0 ir m 0 i 3 m 0 i a m 0i
=
=
=
BA =
2 pa 2 2 pa 2 pa 2 6 6 pa
or, B =
37.
m0 I
150m 0 I
= 500 3
B0 = B1 + B2 + B3 + B4
m I
m I m I
= 0 [ sin 90° – sin 45° ] + 0 + 0
4 pR
2 R 4pR
[sin45° + sin90°]
m I æ
1 ö m0 I m0I æ
1 ö
= – 0 ç1 –
+
÷+
ç1 +
÷
4 pR è
2 ø 2R 4pR è
2ø
e
uuur m I æ
1 ö
Be0 = 0 ç p +
÷
2 pR è
2ø
®
®
®
40. (a) B = B + B
1
2
µ0 æ i º ˆ i º ˆ ö
=
. ç .k + -k ÷ = 0
ø
d
2p è d
41. (b) If q is the charge on the ring, then
( )
P-331
Moving Charges and Magnetism
q qw
=
T 2p
Magnetic field,
æ 4a2 ö
M1 = ( I )(p) ç 2 ÷
è p ø
i=
B=
m0 i
2R
M1 =
æ qw ö
m0 ç ÷
è 2p ø
=
2R
( )
(b) B =
µ0 i
, (sin a + sin b)
4p r
Here r =
52 - 32 = 4 cm
a = b = 37°
\ B = 10
43.
-7
a
3 æ a ö
´
=
3 2 çè 2 3 ÷ø
ém l
ù
B0 = 3 ê 0 (sin 60° + sin 60°) ú
ë 4pr
û
æ 3 ö 9 æ m 0l ö
3m 0 l
=
´ (2) ç
ç 2 ÷÷ 2 çè pa ÷ø
æ a ö
è
ø
4p ç
÷
è2 3ø
9 ´ 2 ´ 10 -7 ´ 10
= 18 mT
1
(d) Let a be the area of the square and r be the radius of
circular loop.
=
44.
m 0i m 0i
+
4 pd 4 pd
m0i 2 ´ 10 -7 ´ i
+
2 pd
4 ´ 10 -2
\ i = 20 A and the direction of magnetic field is
perpendicular into the plane
or 10 -4 =
46. (d)
æ 1ö
r = ç ÷ (a sin 60)
è 3ø
=
\ Net magnetic field B0 =
5
´ 2sin 37° = 1.5 × 10–5 T
4
(a)
r=
4M
(Q M = Ia2)
p
45. (c) Let I be the current in each wire. (directed inwards)
Magnetic field at ‘O’ due to LP and QM will be zero.
i.e., B0 = BPS + BQN
M1 =
æ m 0 ö qw
-7 q ´ 40p
or 3.8 × 10–9 = çè 4p ÷ø R = 10
0.10
\
q = 3 × 10–5 C.
42.
4Ia 2
p
æ 2a ö
2pr = 4a Þ r = ç ÷
è pø
For square
M = (I) a2
For circular loop
M1 = (I)pr2
There will be no magnetic field at O due to wire
PQ and RS
Magnetic field at ‘O’ due to arc QR
æ pö
.I
m0 çè 4 ÷ø
=
4p r1
Magnetic field at ‘O’ due to are PS
æ pö
.I
m çè 4 ÷ø
= 0
4p r2
\ Net magnetic field at ‘O’
é
m0
1
1
ê
B = 4p ( p / 4 ) ´ 10 ê
-2
5 ´ 10 -2
êë 3 ´ 10
(
) (
r p
Þ| B |= ´ 10 -5 T » 1 ´ 10 -5 T
3
47. (d)
Loop
L = 2p R
R
L = N × 2pr
r
Coil
)
ù
ú
ú
úû
P-332
Physics
R= Nr Þ r =
R
N
51. (b) Case (a) :
m0i
m 0 Ni m 0 Ni m 0 N i
=
=
Bcoil =
2r
2R
2R
æ Rö
2ç ÷
è Nø
1
B
\ L= 2
BC N
2
µ0 I
µ
I
´ 2p = 0
´ 2p (Q 2pR = l)
4p R
4 p l / 2p
µ0 I
´ (2p) 2
=
4p l
BLoop =
48.
BA=
Case (b) :
(c) Magnetic field at the centre of loop, B1 =
m0 I
2R
Dipole moment of circular loop is m = IA
m1 = I.A = I.pR2 {R = Radius of the loop}
If moment is doubled (keeping current constant) R becomes
2R
m 2 = I.p
(
B2 =
2
(
m0 I
2R
B
\ 1 =
B2
2
49.
(
)
2
= 4´
= 2.IpR 2 = 2m1
)
m0 I
2R
m0 I
2R
µ0
I
2 µ0 I 64 m0 I
´
´
=
´
=
32 2 [4a = l]
4p l / 8
2 4p l
2 4pl
BA
p2
=
BB 8 2
(d) Let us consider 'l' length of current carrying wire.
At equilibrium
T cos q = lgl
52.
= 2
q
)
1
2d
10 –2
m
A
l=
B
(ll)g
pl gL
u 0 cos q
m0 nI
2a
where, a is the radius of loop.
m0 I
Then, B1 =
2a
m I 2nA
Now, for coil B = 0 .
4p x 3
at the centre x = radius of loop
2
O
60°
C
m 0i
(cos q1 + cos q 2 )
4pd
Putting value of µ = 4p × 10–7 and q1 and q2 we will get B =
4 × 10–5 Wb/m2
Magnetic field, B =
FB
53. (b) For loop B =
B2 =
4.5
×
æ 4.5 ´ 10-2 ö
=ç
÷m
2 3 è 2 3 ø
Lsin q
T sin q
T cos q
é FB m0 2 I ´ I ù
m0 I ´ Il
=
êQ
4p 2l sin q úû
2 p 2L sin q ë l
Therefore, I = 2sin q
(a)
magnetic field at the centre of the triangle ‘O’ B = ?
l
Lsin q
and T sin q =
Here, side of the triangle, l = 4.5 × 10–2 m, current, I = 1A
From figure, tan 60° = 3 =
L
T
(b) Point P is situated at the mid-point of the line joining
the centres of the circular wires which have same radii (R).
r
The magnetic fields ( B ) at P due to the currents in the
wires are in same direction.
Magnitude of magnetic field at point, P
Þ d=
µ0 I
[sin 45° + sin 45°]
4p a / 2
Þ
ì
ü m0 NIR 2
8m0 NI
m 0 NIR 2
B = 2 ïï
=
=
ï
3/ 2 ï
53/ 2 R
53 / 2
í æ 2 R2 ö ý
ï2ç R +
÷ ï
8
4 ø þï
îï è
50.
a/2
BB = 4 ×
2R
45°
BB
a
\
m0 .3I
m 0 2 ´ 3 ´ ( I / 3) ´ p ( a / 3 )
.
=
3
2a
4p
( a / 3)
B1
m I / 2a
= 0
B2 m0 .3I / 2a
B1 : B2 = 1: 3
54. (d) Magnetic field between the plates in this case is zero.
55. (a) Magnetic field at any point lies on axial position of
current carrying conductor B = 0
P-333
Moving Charges and Magnetism
56.
57.
(a) If magnetic field is perpendicular and into the plane of
the paper, it is represented by cross Ä and if the direction
of the magnetic field is perpendicular out of the plane of
the paper it is represented by dot e .
(b) Given : Radius = R
Distance x = 2 2R
Bcentre æ
x2
= çç1 + 2
Baxis
è R
58.
ö
÷÷
ø
3/2
3/2
æ (2 2R)2 ö
= çç1 +
÷
R 2 ÷ø
è
= (9)3/2 = 27
(a) The magnetic field due to a disc is given as
m 0 wQ
1
i.e., B µ
2 pR
R
(d) Let R be the radius of semicircular ring. Let an
elementary length dl is cut for finding magnetic field. So,
dq
I
dl = Rdq. Current in a small element, dI =
p
Magnetic field due to the element
m 2dI
m I
dB = 0
= 02
4p R
2p R
The component dB cos q, of the field is cancelled by another
opposite component.
Therefore,
B=
59.
dB
Bnet =
60.
61.
ò
dB sin q =
m0 I
2p
2
p
ò
R
sin qd q =
0
m0I
p2 R
(a) The magnetic field varies inversely with the distance
1
for a long conductor. That is, B µ
d
so, graph in option (a) is the correct one.
(c) The magnetic field is
62. (d) Since uniform current is flowing through a straight
wire, current enclosed in the amperean path formed at a
æ aö
distance r1 ç = ÷ is
è 2ø
æ p r2 ö
i = ç 12 ÷ ´ I ,
ç pa ÷
è
ø
where I is total current
Using Ampere circuital law,
uur
Ñò B × dl = m0i
P1
P2
m0 ´ current enclosed
Path
æ p r12 ö
m0 ´ ç 2 ÷ ´ I
ç pa ÷
m ´Ir
è
ø
= 0 21
Þ B1 =
2p r1
2p a
Now, magnetic field induction at point P2,
m I
m
I
= 0 .
B2 = 0 ×
2p (2a ) 4pa
B1 m 0 Ir1 4pa
\
=
´
B2 2pa 2 m 0 I
Þ B1 =
a
2´
B1 2 r1
2 = 1.
Þ
=
=
B2
a
a
63. (d) There is no current inside the pipe. From Ampere’s
r uur
circuital law Ñò B × dl = m0 I
QI=0
\ B=0
64. (c) The direction of magnetic field induction due to
current through AB and CD at P are indicated as B1 and
B2. The magnetic fields at a point P, equidistant from AOB
and COD will have directions perpendicular to each other,
as they are placed normal to each other.
P
A
D
B1
B2
d
I1
m 2I
2 ´ 100
B= 0
= 10-7 ´
= 5 × 10–6 T
4p r
4
W
a/2
I2
O
N
C
100A
4m
E
S
Ground
B
Current flows from east to west. Point is below the power
line, using right hand thumb rule, the magnetic field is
directed towards south.
B
Magnetic field at P due to current through AB,
m I
B1 = 0 1
2 pd
Magnetic field at P due to current through CD,
m I
B2 = 0 2
2 pd
\ Resultant field, B = B12 + B22
æ m ö
\ B= ç 0 ÷
è 2pd ø
2
(I
2
1
+ I 22
)
P-334
Physics
65.
Þ B2 =
66.
(
)
1/ 2
m0 2
I1 + I 22
2 pd
(a) Magnetic field due to long solenoid is given by B = m0nI
In first case B1 = m0n1I1
In second case, B2 = m0n2I2
m ni
B
\ 2 = 0 22
B1 m 0 n1i1
i
100 ´
B2
3
Þ
=
6.28 ´ 10 -2 200 ´ i
or, B =
(d)
6.28 ´ 10 -2
= 1.05 ´ 10 -2 Wb/m 2
6
(1)
(2)
The magnetic field due to circular coil (1) is
µi
m0i1
m ´ 3 ´ 102
= 0
B1 = 0 1 =
2
2r
4p
2(2p ´ 10 )
Magnetic field due to coil (2)
Total magnetic field
B2 =
m0 i2
2(2p ´ 10 -2 )
=
68.
69.
54(53 )
= 250 µT
3´ 3´ 3
(a) Magnetic field induction at the centre of current
carrying circular coil of radius r is
m I
B = 0 ´ 2p
4p R
m I
Here B A = 0 ´ 2p
4p R
m0 2I
´ 2p
and BB =
4p 2 R
BA
I/R
=
=1
Þ
BB 2 I / 2 R
Þ B¢ =
70.
y
71. (b)
(0,b,0)
m 0 ´ 4 ´ 10 2
4p
Total magnetic field, B =
67.
Magnetic field at the centre of loop is
m i
B¢ = 0
2a
B × ( x 2 + a 2 )3/2
\ B¢ =
a3
Put x = 4 & a = 3
I
B12 + B22
m
= 0 × 5 × 102
4p
Þ B = 10–7 × 5 × 102
Þ B = 5 × 10–5 Wb/m2
(b) From Ampere’s circuital law
r uur
B
ò × dl = m0i
Þ B × 2pr = m0i
Here i is zero, for r < R, whereas R is the radius
\ B=0
(b) Magentic field at the centre of a circular coil of radius
R carrying current i is B = m 0 i
2R
The circumference of the first loop = 2pR. If it is bent into
n circular coil of radius r¢.
n × (2pr¢) = 2pR
Þ nr¢ = R
...(1)
n ×m0i
New magnetic field, B¢ =
...(2)
2 r¢
From (1) and (2),
nm i × n
= n2B
B¢ = 0
2 pR
(c) The magnetic field at a point on the axis of a circular
loop at a distance x from centre is,
m 0i a 2
B=
2( x 2 + a 2 )3 / 2
x
z
y
r
F
b
Fcosq
r = b2 + a2
a
F
a
x
Fcosq
Force, F = BI 2 a =
Force, F =
m0 I
I ´ 2a
2 pr
m0 I 2a
p b2 + a2
Torque, t = F1 ´ Perpendicular distance = F cos q ´ 2a
=
m0 I 2 a
p b2 + a2
Þt=
´
b
b2 + a 2
2m 0 I 2 a 2b
p(a 2 + b2 )
´ 2a
P-335
Moving Charges and Magnetism
r µ I
µ0 I2
ÞF= 0 1 +
=0
2px 2p(d - x)
µ0 I1
µ0 I2
=
2px 2p(x - d)
I1x – I1d = I2x
Id
x= 1
I1 - I2
Two cases may be possible if I1 > I2 or I2 > I1
76. (d) t = MB sin45° = N (iA) B sin 45°
2m 0 I 2 a 2
pb
(c) Torque on the loop,
If b >> a then t =
72.
t = M ´ B = MB sin q = MB sin 90°
Magnetic field, B =
m0 I
2pd
æm I ö
\ t = I1 (2a)2 ç 0 2 ÷ sin 90°
è 2pd ø
2m0 I1 I 2
2m I 2 a 2
´ a2 = 0
pd
pd
(b) Magnetic moment of loop ABCD,
M1 = area of loop × current
r
(Here, ab = area of rectangle)
M1 = (abI )( ˆj )
= 100 ´ 3(5 ´ 2.5) ´ 10-4 ´ 1 ´
=
73.
Magnetic moment of loop DEFA,
r
M 2 = (abI )(i$)
1
2
= 0.27 N-m
m 0 æ ii i2 i1i2 ö
m ii
´a = 012
77. (c) F = 2 çè
2 a ÷ø
p a
4p
r r r
78. (b) | t |=| m ´ B | [m = NIA]
=NIA × B sin 90o
[A = pr2]
Þ t =NIpr2B
Net magnetic moment,
r
r
r
r
M = M1 + M 2 Þ M = abI ($i + $j )
r
æ ˆj
kˆ ö
Þ |M | = 2abI ç
+
÷
2ø
è 2
74.
(c) Torque on circular loop, t = MB sin q
where,
M = magnetic moment
B = magnetic field
Now, using t = Ia
\ t = MB sin q = Ia
Þ pR 2 IBq =
mR 2 a
2
(Q m = IA and moment of inertia of circular loop, I =
Þ pR2IB q =
79. (d)
mR 2
)
2
mR 2
wq
2
Þ w=
2pIB
Þ
m
Þ T=
2pm
IB
2p
=
T
2pIB
m
Force on one pole,
m0 I
F= m´
2p d2 + x 2
Total force, Ftotal = 2F sin q
75.
(d)
= 2´
=
As net force on the third wire C is zero.
m 0 Im
2
2p d + a
2
´
x
2
d + a2
m 0 Im x
p (d 2 + a 2 )
Magnetic moment, M = Ipa2 = m × 2
P-336
Physics
or, Total force, Ftotal
=
m 0 Ia 2
2d 2
=
m0 I a 2
2(d 2 + a 2 )
[Q d >> a ]
a2
Clearly Ftotal µ 2
d
80.
(a) Magnetic moment, m = IA =
qv
( pr 2 )
2 pr
qrw 2
1
(pr ) = qr 2 w
2 pr
2
(c) Magnetic moment of current carrying rectangular loop
of area A is given by M = NIA
magnetic moment of current carrying coil is a vector and
its direction is given by right hand thumb rule, for
r
rectangular loop, B at centre due to current in loop and
r
M are always parallel.
or, m =
81.
B, M
B, M
Ä
Inwards
e
Outwards
82.
83.
84.
85.
Hence, (c) corresponds to stable equilibrium.
r r
(c) F1 = F2 = 0
because of action and reaction pair
r r
(a) For stable equilibrium M || B
r
r
For unstable equilibrium M || (–B)
(a) I1 I2 = Positive
(attract) F = Negative
I1 I2= Negative
(repell) F = Positive
Hence, option (a) is the correct answer.
(c) For small arc length
2T sin q = BIR 2 q
(As F = BIL and L = RZq)
T = BIR
Q
P
T
86.
2
2
0
0
2
-3
-0.2 x
dx
= 9 ´ 10 ò e
l=3m
0
9 ´ 10
0.2
P
Q
Also given; length of wire Q
= 25 cm = 0.25 m
Force on wire Q due to wire R
FQR = 10–7 ´
[- e-0.2´ 2 + 1]
R
2 ´ 20 ´ 10
´ 0.25
0.05
= 20 × 10–5 N (Towards left)
Force on wire Q due to wire P
FQP = 10–7 ×
2 ´ 30 ´ 10
´ 0.25
0.03
= 50 × 10–5 N (Towards right)
Hence, Fnet = FQP – FQR
= 50 × 10–5 N – 20 × 10–5 N
= 3 × 10–4 N towards right
88. (c)
89. (a) Force acting on conductor B due to conductor A is
given by relation
F=
m 0 I1 I 2 l
2 pr
\F=
W = ò Fdx = ò 3.0 ´ 10 -4 e -0.2 x ´ 10 ´ 3dx
=
5 cm
Tsinq
q
(b) Work done in moving the conductor is,
-3
3 cm
l-length of conductor B
r-distance between two conductors
T
R
Tsinq
9 ´ 10-3
´ [1 - e -0.4 ]
0.2
2.97×10 –3
9×10 –3 ´ (0.33)
=
=
2
2
Power required to move the conductor is,
W
P=
t
2.97 ´ 10-3
P=
= 2.97 W
(0.2) ´ 5 ´ 10-3
87. (a) I1 = 30 A I = 10 A
I2 = 20 A
=
I = 10 A
z
x
4p´ 10-7 ´ 10 ´ 2 ´ 2
= 8 × 10–5 N
2 ´ p´ 0.1
90. (a)
91. (a) The magnetic field at O due to current in DA is
m I p
B1 = o ´ (directed vertically upwards)
4p a 6
The magnetic field at O due to current in BC is
m I p
B2 = o ´
(directed vertically downwards)
4p b 6
The magnetic field due to current AB and CD at O is zero.
Therefore the net magnetic field is
P-337
Moving Charges and Magnetism
92.
93.
94.
95.
B = B1 – B2 (directed vertically upwards)
m I p mo I p
= o
´
4 p a 6 4p b 6
m I æ 1 1ö m I
= o ç - ÷ = o (b - a)
24 è a b ø 24ab
r
r r
(d) F = I ( l ´ B)
The force on AD and BC due to current I1 is zero. This is
uur
because the directions of current element I d l and
r
magnetic field B are parallel.
(a) Force acting between two long conductor carrying
current,
m 2I I
F = 0 1 2 ´l
...(i)
4p d
Where d = distance between the conductors
l = length of conductor
m 2(2 I1 ) I 2
In second case, F ¢ = - 0
..(ii)
l
4 p 3d
From equation (i) and (ii), we have
F ¢ -2
=
\
F
3
(b) When current is passed through a spring then current
flows parallel in the adjacent turns in the same direction.
As a result the various turn attract each other and spring
get compress.
(c) Magnetic field due to current in wire 1 at point P distant
r from the wire is
m i
q
B = 0 1 [ cos q + cos q ]
i2
4p r
i
1
B=
m 0 i1 cos q
2p r
r
P
dl
q
This magnetic field is directed perpendicular to the plane of
paper, inwards.
The force exerted due to this magnetic field on current
element i2 dl is
dF = i2 dl B sin 90°
\ dF = i2 dlB
æ m i cos q ö
Þ dF = i2 dl ç 0 1
÷
è 4p r ø
m0
i1 i2 dl cos q
2 pr
96. (d) Galvanometer of resistance (G) converted into a
voltmeter of range 0-1 V.
ig
R1
V = 1 = ig (G + R1 )
...(i)
To increase the range of voltmeter 0-2 V
G
R1
R2
...(ii)
Dividing eq. (i) by (ii),
Þ
G + R1
1
=
2 G + R1 + R2
Þ G + R1 + R2 = 2G + 2 R1
\ R2 = G + R1
97. (d) Given,
Current passing through galvanometer, I = 6 mA
Deflection, q = 2°
Figure of merit of galvanometer
I 6 ´10-3
=
= 3 ´10-3 A/div
2
q
98. (20)
Given,
Area of galvanometer coil, A = 3 × 10–4 m2
Number of turns in the coil, N = 500
Current in the coil, I = 0.5 A
r r
Torque t =| M ´ B |= NiAB sin(90°) = NiAB
=
ÞB=
t
1.5
=
= 20 T
NiA 500 ´ 0.5 ´ 3 ´10 -4
99. (c) ig = 20 × 50 = 1000 µA = 1 mA
Using, V = ig (G + R), we have
2 = 10–3 (100 + R1)
R1 = 1900 W
when, V = 10 volt
10 = 10–3 (100 + R2 + R1)
10000 = (100 + R2 + 1900)
\ R2 = 8000 W
100. (b) In an ammeter,
ig = i0
=
G
2 = ig ( R1 + R2 + G)
RA
RA + G
and for voltmeter,
V = ig (G + RV) = Gi0
On solving above equations, we get
RARV = G2
R A æ ig ö
=ç
and
÷
RV è i0 - ig ø
2
101. (Bonus) v = ig (R + G)
Þ 5 = 10–4 (2 × 106 + x)
x = – 195 × 104W
P-338
Physics
102. (c) V = ig (G + R) = 4 × 10–3 (50 + 5000) = 20V
103. (d) Cq = NBiA sin 90°
æ p ö
or 10-6 ç
= 175B(10-3 ) ´ 10-4
è 180 ÷ø
\ B = 10–3 T
104. (d) Using, ig = i
S
S+G
S
S + 50
On solving, we get
0.002 = 0.5
100
; 0.2 W
498
105. (b) When key K1 is closed and key K2 is open
S=
ig =
E
= Cq 0
220 + R g
... (i)
When both the keys are closed
Cq0
E
5
æ
ö
´
=
ig =
ç
÷
5R g
(R g + 5)
5
ç 220 +
÷
5 + Rg ø
è
Þ
S
... (ii)
E
= Cq0
220 + R g
... (i)
G
G
I
R
Þ 5500 + 25Rg = 225Rg + 1100
200Rg = 4400
Rg = 22W
106. (b) Galvanometer has 25 divisions Ig = 4 × 10–4 × 25 = 10–2 A
ig
G
R
V = Ig Rnet
2.5V
v = Ig (G + R)
2.5 = (50 + R) 10–2 \ R = 200W
107. (c) Deflection current
= Igmax = nxk =0.005 × 30
Where, n = Number of divisions = 30 and k = 0.005 amp/
division
= 15 × 10–2 = 0.15
v = Ig[20 + R]
15 = 0.15 [20 + R]
100 = 20 + R
R = 80 W
I/2
R
E
Dividing (i) by (ii), we get
225R g + 1100
Þ
=5
1100 + 5R g
v
1
2
S=
(R + G)I
e2
RG ´
Cq 0
5E
=
225R g + 1100
5
50W
108. (d) Given,
Resistance of galvanometer, G = 100W
Current, ig = 1 mA
A galvanometer can be converted into voltmeter by
connecting a large resistance R in series with it.
Total resistance of the combination = G + R
According to Ohm’s law, V = ig (G + R)
\ 10 = 1 × 10–3 (100 + R0)
Þ 10000 – 100 = 9900 W = R0
Þ R0 = 9.9 kW
109. (b) Figure of merit of a galvanometer is the correct required
to produce a deflection of one division in the galvanometer
I
i.e., figure of merit =
q
e
1
I=
G = KW
R+G
9
1
S
1
e
eS
=
´
Þ =
2 R + GS
S+ G
2 R(S + G) + GS
G+S
E
1
´ 102 ´ 270 ´ 10 -6
2
S=
= 110 W
æ 6ö
6-ç ÷
è 2ø
110. (d) According to question, current through galvanometer,
Ig = 1 mA
Current through shunt (I – Ig) = 2 A
Galvanometer resistance Rg = 25W
Resistance of shunt, S = ?
Ig
G
I0R0 = (I – Ig)S
11 ´ 103 ´
10-3 ´ 25
I – Ig
S
2
S ; 1.25 × 10–2W
111. (c) Given : Current through the galvanometer,
ÞS=
ig = 5 × 10–3 A
Galvanometer resistance, G = 15W
Let resistance R to be put in series with the galvanometer
to convert it into a voltmeter.
V = ig (R + G)
10 = 5 × 10–3 (R + 15)
\ R = 2000 – 15 = 1985 = 1.985 × 103 W
112. (c) Ig G = ( I – Ig)s
\ 10–3 × 100 = (10 – 10–3) × S
\ S » 0.01W
P-339
Moving Charges and Magnetism
113. (d) As we know, I =
V 5
=
= 0.1
R 50
I' = 0.099
When Galvanometer is connected
100S
V
R eq = 50 +
=
100 + S I
100S
5
Þ
=
- 50
100 + S 0.099
100S
100S
= 50.50 - 50 Þ
= 0.5
100 + S
100 + S
Þ 100S = 50 + 0.55 Þ 99.5S = 50
50
S=
= 0.5 W
99.05
So, shunt of resistance = 0.5W is connected in parallel with
the galvanometer.
V
114. (a) According to Ohm's Law, I =
R
V
Ig =
R +G
where, Ig-Galvanometer current, G-Galvonometer resistance
IG
R
G
Þ
Þ R(G + S) + GS = 2S(R + G)
Þ RG + RS + GS = 2S(R + G)
Þ RG = 2S(R + G) - S(R + G)
\ RG = S(R + G)
115. (b) To measure AC voltage across a resistance a
moving coil galvanometer is used.
116. (d) The correct circuit diagram is D with galvanometer
resistance
RS
R-S
117. (d) Statements I is false and Statement II is true
G=
IgG
I – Ig
Therefore for I to increase, S should decrease, So additional
S can be connected across it.
118. (d)
119. (d) The current that will given full scale deflection in the
absence of the shunt is nearly equal to the current through
the galvanometer when shunt is connected i.e. Ig
For ammeter, shunt resistance, S =
=
V
When shunt of resistance S is connected parallel to the
GS
Galvanometer then G =
G +S
V
\ I=
GS
R+
G +S
R
IG
2
G
Equal potential difference is given by
I'g G = (I - I'g )S
I 'g (G + S) = IS
Ig
2
=
IS
G +S
V
V
S
=
´
2(R + G ) R + GS
G+S
G +S
1
S
Þ
=
2(R + G) R(G + S) + GS
Þ
5.5 ´ 1
= 0.045 ampere.
120 + 1
500 W
120. (a)
A
i
12V
Again, i =
2V
R
12 – 2 = (500W)i Þ i =
S
Þ
IS
G+S
As Ig =
10
1
=
500 50
12
1
=
500 + R 50
Þ 500 + R = 600
Þ R = 100 W
121. (c) Resistance of Galvanometer,
Current sensitivity
10
= 5W
ÞG=
G=
Voltage sensitivity
2
Here ig = Full scale deflection current =
V = voltage to be measured = 150 volts
(such that each division reads 1 volt)
150
ÞR=
- 5 = 9995W
15 ´ 10 -3
150
= 15 mA
10
19
P-340
Physics
Magnetism and
Matter
Magnetism, Gauss’s Law,
TOPIC 1 Magnetic Moment, Properties
of Magnet
1.
2.
3.
A small bar magnet placed with its axis at 30° with an external
field of 0.06 T experiences a torque of 0.018 Nm. The
minimum work required to rotate it from its stable to
unstable equilibrium position is :
[Sep. 04, 2020 (I)]
(a) 6.4 × 10–2 J
(b) 9.2 × 10–3 J
(c) 7.2 × 10–2 J
(d) 11.7 × 10–3 J
A circular coil has moment of inertia 0.8 kg m2 around any
diameter and is carrying current to produce a magnetic
moment of 20 Am2. The coil is kept initially in a vertical
position and it can rotate freely around a horizontal
diameter. When a uniform magnetic field of 4 T is applied
along the vertical, it starts rotating around its horizontal
diameter. The angular speed the coil acquires after rotating
by 60° will be :
[Sep. 04, 2020 (II)]
(a) 10 rad s–1
(b) 10p rad s–1
(c) 20p rad s–1
(d) 20 rad s–1
Two magnetic dipoles X and Y are placed at a separation
d, with their axes perpendicular to each other. The dipole
moment of Y is twice that of X. A particle of charge q is
passing through their midpoint P, at angle q = 45° with the
horizontal line, as shown in figure. What would be the
magnitude of force on the particle at that instant? (d is
much larger than the dimensions of the dipole)
[8 April 2019 II]
(c)
4.
5.
6.
7.
8.
9.
æ m0 ö M
ç ÷
(a) è 4p ø d
2
( )
3
´ qv
(b) 0
æm ö M
2ç 0÷
è 4p ø d
2
( )
3
´ qv
æ m0 ö 2M
´ qv
ç ÷
(d) è 4p ø d 3
2
( )
A magnet of total magnetic moment 10 -2 iˆ A-m2 is
placed in a time varying magnetic field,
B iˆ (coswt)where B = 1 Tesla and w = 0.125 rad/s. The
work done for reversing the direction of the magnetic
moment at t = 1 second, is:
[10 Jan. 2019 I]
(a) 0.01 J
(b) 0.007 J
(c) 0.028 J
(d) 0.014 J
A magnetic dipole in a constant magnetic field has :
[Online April 8, 2017]
(a) maximum potential energy when the torque is maximum
(b) zero potential energy when the torque is minimum.
(c) zero potential energy when the torque is maximum.
(d) minimum potential energy when the torque is maximum.
A magnetic dipole is acted upon by two magnetic fields
which are inclined to each other at an angle of 75°. One of
the fields has a magnitude of 15 mT. The dipole attains
stable equilibrium at an angle of 30° with this field. The
magntidue of the other field (in mT) is close to :
[Online April 9, 2016]
(a) 1
(b) 11
(c) 36
(d) 1060
A 25 cm long solenoid has radius 2 cm and 500 total number
of turns. It carries a current of 15 A. If it is equivalent to a
r
magnet of the same size and magnetization M (magnetic
uur
moment/volume), then M is : [Online April 10, 2015]
(a) 30000p Am–1
(b) 3pAm–1
–1
(c) 30000 Am
(d) 300 Am–1
A bar magnet of length 6 cm has a magnetic moment of
4 J T–1. Find the strength of magnetic field at a distance of
200 cm from the centre of the magnet along its equatorial
line.
[Online May 7, 2012]
(a) 4 × 10–8 tesla
(b) 3.5 × 10–8 tesla
(c) 5 × 10–8 tesla
(d) 3 × 10–8 tesla
A thin circular disc of radius R is uniformly charged with
density s > 0 per unit area. The disc rotates about its axis
with a uniform angular speed w.The magnetic moment of
the disc is
[2011 RS]
P-341
Magnetism and Matter
pR 4
(b)
sw
2
4
(a) pR sw
4
pR
sw
(d) 2pR 4 sw
4
10. A magnetic needle is kept in a non-uniform magnetic field.
It experiences
[2005]
(a) neither a force nor a torque
(b) a torque but not a force
(c) a force but not a torque
(d) a force and a torque
11. The length of a magnet is large compared to its width and
breadth. The time period of its oscillation in a vibration
magnetometer is 2s. The magnet is cut along its length
into three equal parts and these parts are then placed on
each other with their like poles together. The time period
of this combination will be
[2004]
(c)
r
(a) B
(b) zero
r
r
(c) much large than | B | and parallel to B
r
r
(d) much large than | B | but opposite to B
17. Magnetic materials used for making permanent magnets
(P) and magnets in a transformer (T) have different
properties of the following, which property best matches
for the type of magnet required? [Sep. 02, 2020 (I)]
(a) T : Large retentivity, small coercivity
(b) P : Small retentivity, large coercivity
(c) T : Large retentivity, large coercivity
(d) P : Large retentivity, large coercivity
18.
2
2
s
s
(c) 2 s
(d)
3
3
12. A magnetic needle lying parallel to a magnetic field requiers
(a) 2 3 s
(b)
W units of work to turn it through 600 . The torque needed
to maintain the needle in this position will be
[2003]
3
(d) 2 W
W
2
13. The magnetic lines of force inside a bar magnet [2003]
(a) are from north-pole to south-pole of the magnet
(b) do not exist
(c) depend upon the area of cross-section of the bar
magnet
(d) are from south-pole to north-pole of the Magnet
(a)
3W
(b) W
(c)
The Earth Magnetism, Magnetic
TOPIC 2
Materials and their properties
14. An iron rod of volume 10–3 m3 and relative permeability
1000 is placed as core in a solenoid with 10 turns/cm. If a
current of 0.5 A is passed through the solenoid, then the
magnetic moment of the rod will be : [Sep. 05, 2020 (II)]
(a) 50 × 102 Am2
(b) 5 × 102 Am2
2
2
(c) 500 × 10 Am
(d) 0.5 × 102 Am2
15. A paramagnetic sample shows a net magnetisation of 6 A/m
when it is placed in an external magnetic field of 0.4 T at a
temperature of 4 K. When the sample is placed in an external
magnetic field of 0.3 T at a temperature of 24 K, then the
magnetisation will be :
[Sep. 04, 2020 (II)]
(a) 1 A/m
(b) 4 A/m
(c) 2.25 A/m
(d) 0.75 A/m
16. A perfectly diamagnetic sphere has a small spherical cavity
at its centre, which is filled with a paramagnetic substance.
r
The whole system is placed in a uniform magnetic field B.
Then the field inside the paramagnetic substance is :
[Sep. 03, 2020 (II)]
P
19.
20.
21.
22.
The figure gives experimentally measured B vs. H variation
in a ferromagnetic material. The retentivity, co-ercivity
and saturation, respectively, of the material are:
[7 Jan. 2020 II]
(a) 1.5 T, 50 A/m and 1.0 T
(b) 1.5 T, 50 A/m and 1.0 T
(c) 150 A/m, 1.0 T and 1.5 T
(d) 1.0 T, 50 A/m and 1.5 T
A paramagnetic material has 1028 atoms/m3. Its magnetic
susceptibility at temperature 350 K is 2.8 × 10–4 . Its
susceptibility at 300 K is:
[12 Jan. 2019 II]
(a) 3.267 × 10–4
(b) 3.672 × 10–4
(c) 3.726 × 10–4
(d) 2.672 × 10 –4
A paramagnetic substance in the form of a cube with sides
1 cm has a magnetic dipole moment of 20 × 10–6 J/T when
a magnetic intensity of 60 × 103 A/m is applied. Its magnetic
susceptibility is:
[11 Jan. 2019 II]
(a) 3.3 × 10–2
(b) 4.3 × 10 –2
(c) 2.3 × 10–2
(d) 3.3 × 10–4
At some location on earth the horizontal component of
earth’s magnetic field is 18 × 10–6 T. At this location,
magnetic needle of length 0.12 m and pole strength 1.8
Am is suspended from its mid-point using a thread, it
makes 45° angle with horizontal in equilibrium. To keep
this needle horizontal, the vertical force that should be
applied at one of its ends is:
[10 Jan. 2019 II]
(a) 3.6 × 10–5 N
(b) 1.8 × 10–5 N
(c) 1.3 × 10–5 N
(d) 6.5 × 10–5 N
A bar magnet is demagnetized by inserthing it inside a
solenoid of length 0.2 m, 100 turns, and carrying a current
of 5.2 A. The coercivity of the bar magnet is:
[9 Jan. 2019 I]
(a) 285 A/m
(b) 2600 A/m
(c) 520 A/m
(d) 1200 A/m
P-342
Physics
23. The B-H curve for a ferromagnet is shown in the figure.
The ferromagnet is placed inside a long solenoid with
1000 turns/cm.. The current that should be passed in the
solenoid to demagnetise the ferromagnet completely is:
[Online April 15, 2018]
B(T)
2, 0
1, 0
H(A/m)
–200 –100 100 200
–1, 0
–2, 0
(a) 2 mA
(b) 1 mA (c) 40 µA
(d) 20 µA
24. Hysteresis loops for two magnetic materials A and B are
given below :
D
B
H
(A)
H
(B)
These materials are used to make magnets for elecric
generators, transformer core and electromagnet core.
Then it is proper to use :
[2016]
(a) A for transformers and B for electric generators.
(b) B for electromagnets and transformers.
(c) A for electric generators and trasformers.
(d) A for electromagnets and B for electric generators.
25. A fighter plane of length 20 m, wing span (distance from
tip of one wing to the tip of the other wing) of 15m and
height 5m is lying towards east over Delhi. Its speed is
240 ms–1. The earth's magnetic field over Delhi is 5 ×
10–5T with the declination angle ~0° and dip of q such
2
. If the voltage developed is VB between
3
the lower and upper side of the plane and VW between the
tips of the wings then VB and VW are close to :
[Online April 10, 2016]
(a) VB = 40 mV; VW = 135 mV with left side of pilot at
higher voltage
(b) VB = 45 mV; VW = 120 mV with right side of pilot
at higher voltage
(c) VB = 40 mV; VW = 135 mV with right side of pilot
at higher voltage
(d) VB = 45 mV; VW = 120 mV with left side of pilot at
higher voltage
26. A short bar magnet is placed in the magnetic meridian
of the earth with north pole pointing north. Neutral
points are found at a distance of 30 cm from the magnet
on the East – West line, drawn through the middle point
of the magnet. The magnetic moment of the magnet in
that sin q =
Am2 is close to: (Given
=Horizontal component of earth’s magnetic field = 3.6
× 10–5 tesla)
[Online April 11, 2015]
(a) 14.6
(b) 19.4
(c) 9.7
(d) 4.9
27. The coercivity of a small magnet where the ferromagnet
gets demagnetized is 3 ´ 103 Am -1. The current required
to be passed in a solenoid of length 10 cm and number of
turns 100, so that the magnet gets demagnetized when
inside the solenoid, is:
[2014]
(a) 30 mA (b) 60 mA (c) 3 A
(d) 6 A
28. An example of a perfect diamagnet is a superconductor.
This implies that when a superconductor is put in a
magnetic field of intensity B, the magnetic field Bs inside
the superconductor will be such that:
[Online April 19, 2014]
(a) Bs = – B
(b) Bs = 0
(c) Bs = B
(d) Bs < B but Bs ¹ 0
29. Three identical bars A, B and C are made of different
magnetic materials. When kept in a uniform magnetic field,
the field lines around them look as follows:
m0
= 10–7 in SI units and BH
4p
A
C
B
Make the correspondence of these bars with their
material being diamagnetic (D), ferromagnetic (F) and
paramagnetic (P):
[Online April 11, 2014]
(a) A « D, B « P, C « F
(b) A « F, B « D, C « P
(c) A « P, B « F, C « D
(d) A « F, B « P, C « D
30. The magnetic field of earth at the equator is approximately
4 × 10–5 T. The radius of earth is 6.4 × 106 m. Then the
dipole moment of the earth will be nearly of the order of:
[Online April 9, 2014]
(a) 1023 A m2
(b) 1020 A m2
(c) 1016 A m2
(d) 1010 A m2
31. The mid points of two small magnetic dipoles of length d
in end-on positions, are separated by a distance x, (x > >
d). The force between them is proportional to x–n where n
is:
[Online April 9, 2014]
N
S
N
x
S
(a) 1
(b) 2
(c) 3
(d) 4
32. Two short bar magnets of length 1 cm each have magnetic
moments 1.20 Am2 and 1.00 Am2 respectively. They are
placed on a horizontal table parallel to each other with
their N poles pointing towards the South. They have a
common magnetic equator and are separated by a distance
of 20.0 cm. The value of the resultand horizontal magnetic
induction at the mid-point O of the line joining their
centres is close to (Horizontal component of earth.s
magnetic induction is 3.6× 10.5Wb/m2)
[2013]
(a) 3.6 × 10.5 Wb/m2
(b) 2.56 × 10.4 Wb/m2
(c) 3.50 × 10.4 Wb/m2
(d) 5.80 × 10.4 Wb/m2
P-343
Magnetism and Matter
33. The earth’s magnetic field lines resemble
that of a dipole at the centre of the earth. If the magnetic
moment of this dipole is close to 8 × 1022 Am2, the
value of earth’s magnetic field near the equator is close
to (radius of the earth = 6.4 × 106 m)
[Online April 25, 2013]
(a) 0.6 Gauss
(b) 1.2 Gauss
(c) 1.8 Gauss
(d) 0.32 Gauss
34. Relative permittivity and permeability of a material e r and
m r , respectively. Which of the following values of these
quantities are allowed for a diamagnetic material? [2008]
(a) e r = 0.5, m r = 1.5
(b) e r = 1.5, mr = 0.5
(c) e r = 0.5, m r = 0.5
(d) e r = 1.5, mr = 1.5
35. Needles N1, N2 and N3 are made of a ferromagnetic, a
paramagnetic and a diamagnetic substance respectively.
A magnet when brought close to them will
[2006]
(a) attract N1 and N2 strongly but repel N3
(b) attract N1 strongly, N2 weakly and repel N3 weakly
(c) attract N1 strongly, but repel N2 and N3 weakly
(d) attract all three of them
36. The materials suitable for making electromagnets should
have
[2004]
(a) high retentivity and low coercivity
(b) low retentivity and low coercivity
(c) high retentivity and high coercivity
(d) low retentivity and high coercivity
37. A thin rectangular magnet suspended freely has a period
of oscillation equal to T. Now it is broken into two equal
halves (each having half of the original length) and one
piece is made to oscillate freely in the same field. If its
period of oscillation is T¢, the ratio
(a)
1
2 2
(c) 2
(b)
1
2
(d)
1
4
T'
is
T
[2003]
38. Curie temperature is the temperature above which
[2003]
(a) a ferromagnetic material becomes paramagnetic
(b) a paramagnetic material becomes diamagnetic
(c) a ferromagnetic material becomes diamagnetic
(d) a paramagnetic material becomes ferromagnetic
TOPIC 3 Magnetic Equipment
39. A ring is hung on a nail. It can oscillate, without slipping
or sliding (i) in its plane with a time period T1 and, (ii) back
and forth in a direction perpendicular to its plane, with a
period T2. The ratio
(a)
2
3
3
T1
will be :
T2
(b)
[Sep. 05, 2020 (II)]
2
3
2
3
2
40. A magnetic compass needle oscillates 30 times per minute
at a place where the dip is 45o, and 40 times per minute
where the dip is 30o. If B1 and B2 are respectively the total
magnetic field due to the earth and the two places, then
the ratio B1/B2 is best given by :
(c)
(d)
[12 April 2019 I]
(a) 1.8
(b) 0.7
(c) 3.6
(d) 2.2
41. A hoop and a solid cylinder of same mass and radius are
made of a permanent magnetic material with their
magnetic moment parallel to their respective axes. But
the magnetic moment of hoop is twice of solid cylinder.
They are placed in a uniform magnetic field in such a
manner that their magnetic moments make a small angle
with the field. If the oscillation periods of hoop and
cylinder are Th and Tc respectively, then:
[10 Jan. 2019 II]
(a) Th = Tc
(b) Th = 2 Tc
(c) Th = 1.5 Tc
(d) Th = 0.5 Tc
42. A magnetic needle of magnetic moment 6.7 × 10–2 Am2
and moment of inertia 7.5 × 10–6 kg m2 is performing
simple harmonic oscillations in a magnetic field of 0.01
T. Time taken for 10 complete oscillations is : [2017]
(a) 6.98 s
(b) 8.76 s
(c) 6.65 s
(d) 8. 89 s
P-344
1.
Physics
(c) Here, q = 30°, t = 0.018 N-m, B = 0.06 T
Torque on a bar magnet :
and B2 =
t = MB sin q
µ0 2M
4p ( d / 2 ) 3
µ0 2M
4p ( d / 2) 3
0.018 = M ´ 0.06 ´ sin 30°
1
Þ M = 0.6 A-m 2
2
Position of stable equilibrium (q = 0°)
Position of unstable equilibrium (q = 180°)
Minimum work required to rotate bar magnet from stable
to unstable equilibrium
Þ 0.018 = M ´ 0.06 ´
DU = U f - U i = -MB cos180° - (- MB cos0°)
W = 2MB = 2 ´ 0.6 ´ 0.06
2.
\ W = 7.2 ´ 10-2 J
(a) Given,
Moment of inertia of circular coil, I = 0.8 kg m2
Magnetic moment of circular coil, M = 20 Am2
Rotational kinetic energy of circular coil,
1 2
Iw
2
Here, w = angular speed of coil
Potential energy of bar magnet = – MB cos f
From energy conservation
KE =
\
or q = 45º
4.
5.
6.
1 2
I w = U in - U f = - MB cos 60° - (- MB )
2
Þ
MB 1 2
= Iw
2
2
Þ
20 ´ 4 1
= (0.8)w 2
2
2
Þ 100 = w 2 Þ w = 10 rad
3.
(b)
µ
2K
B1 = 0
4 p ( d / 2 )3
B
tan q = 2 =
B1 µ0 2M = 1
4p ( d / 2) 3
7.
8.
uur
The resultant field is 45º from B1. The angle between B
uur
and v zero, so force on the particle is zero.
(c)
Work done, W = 2 m·B
= 2 × 10–2 × 1 cos (0.125)
= 0.02 J
(c) Potential energy of dipole,
U = – pE cos q
Torque experienced by dipole
t = pE sin q
Torque will be maximum (tmax) when q = 90° then potential
energy U = 0
(b) We know that, magnetic dipole moment
M = NiA cosθ i.e., M µ cosθ
When two magnetic fields are inclined at an angle of 75°
the equilibrium will be at 30°, so
1
cos θ < cos(75°, 30°) < cos 45° <
2
x
15
< [ x » 11
2
2
r
NiA
(c) M (mag. moment/volume) =
Al
Ni
(500)15
=
=
= 30000 Am–1
l
25 ´ 10 –2
(c) Along the equatorial line, magnetic field strength
m
M
B= 0
3/2
4p 2
r + l2
(
)
Given: M = 4JT–1
r = 200 cm = 2 m
6cm
l=
= 3 cm = 3 × 10–2 m
2
\B=
4p ´ 10 -7
´
4p
4
(
)
3
é 2
-2 2 ù 2
ê 2 + 3 ´ 10
ú
ë
û
Solving we get, B = 5 × 10–8 tesla
P-345
Magnetism and Matter
9.
(c)
14. (b) Given,
Volume of iron rod, V = 10-3 m3
Relative permeability, m r = 1000
Number of turns per unit length, n = 10
Magnetic moment of an iron core solenoid,
M = (m r - 1) ´ NiA
q
Magnetic dipole moment
=
2m
Angular momentum
a
Þ M = (mr -1) ´ Ni
\ Magnetic dipole moment (M)
q æ mR 2 ö
1
4
.ç
M =
÷ .w = s.pR w.
2m è 2 ø
4
10
´ 0.5 ´ 10 -3 = 499.5 » 500.
10 -2
15. (d) For paramagnetic material. According to curies law
Þ M = 999 ´
10. (d) A magnetic needle kept in non uniform magnetic field
experience a force and torque due to unequal forces acting
on poles.
11. (b) Initially, time period of magnet
1
2
I
= 25 where I = ml
12
MB
When the magnet is cut into three pieces the pole strength
will remain the same and
Moment of inertia of each part,
1 æ mö æ l ö
I
(I¢) =
çè ÷ø çè ÷ø ´ 3 =
12 3 3
9
We have, Magnetic moment (M)
= Pole strength (m) × l
\ New magnetic moment,
æ lö
M ' = m ´ ç ÷ ´ 3 = ml = M
è 3ø
1
T
For two temperatures T1 and T2
cµ
c1T1 = c2T2
T = 2p
New time period, T¢ = 2p
V
N
Þ M = (mr - 1) ´ iV
l
l
I¢
M ¢B
I
2
T
Þ T¢ =
= s.
9 MB
9 3
12. (a) Workdone to turn a magnetic needle from angle q1 to
q2 is given by
W = MB (cos q1 - cos q2 )
= 2p
\ W = MB (cos 0° - cos 60°)
But c =
\
I
B
I1
I
T1 = 2 T2
B1
B2
I
6
0.3
´ 4 = 2 ´ 24 Þ I 2 =
= 0.75 A/m
0.4
0.3
0.4
16. (b) When magnetic field is applied to a diamagnetic
substance, it produces magnetic field in opposite direction
so net magnetic field inside the cavity of sphere will be
zero. So, field inside the paramagnetic substance kept
inside the cavity is zero.
17. (d) Permanent magnets (P) are made of materials with large
retentivity and large coercivity. Transformer cores (T) are
made of materials with low retentivity and low coercivity.
Þ
18. (d)
æ 1 ö MB
= MB ç 1 - ÷ =
è 2ø
2
MB
= 3W
2
13. (d) The magnetif field lines of bar magnet form closed
lines. As shown in the figure, the magnetic lines of force
are directed from south to north inside a bar magnet.
Outside the bar magnet magnetic field lines directed from
north to south pole.
\ Torque, t = MB sin q = MB sin 60° = 3
19. (a) According to Curie law for paramagnetic substance,
1
c
TC2
cµ T Þ 1 = T
c2
C1
C
N
S
2.8 ´ 10 –4
300
=
c2
350
c2 =
2.8 ´ 350 ´10 –4
= 3.266 × 10–4
300
P-346
Physics
20. (d) Magnetic susceptibility,
c=
I
H
where, I =
Now, c =
Magnetic moment 20 ´10 –6
= 20 N/m2
=
Volume
10 –6
20
1
= ´10 –3 = 3.3 ´10–4
60´10 3
3
21. (d) using, MB sinq = F l Sinq (t)
28. (b) Magnetic field inside the superconductor is zero.
Diamagnetic substances are repelled in external magnetic
field.
29. (b) Diamagnetic materials are repelled in an external
magnetic field.
Bar B represents diamagnetic materials.
30. (a) Given, B = 4 × 10–5 T
RE = 6.4 × 106 m
Dipole moment of the earth M = ?
m M
B= 0 3
4p d
4 ´10-5 =
F
B
45°
m
l
MB sin 45° = F sin 45°
2
F = 2MB = 2 × 1.8 × 18 × 10–6 = 6.5 × 10–5N
4p´10-7 ´ M
24.
25.
26.
3.6 ´10–5
(0.3)3
10 –7
Hence, M = 9.7 Am2
27. (c) Magnetic field in solenoid B = m0ni
Þ M =
Þ
B
= ni
m0
(Where n = number of turns per unit length)
B Ni
100i
=
Þ 3 ´ 103 =
m0 L
10 ´ 10-2
Þ i = 3A
Þ
3
B1
B2
O
S
N
or, H =
23.
)
\ M @ 1023 Am2
31. (d) In magnetic dipole
1
Force µ
r4
In the given question,
Force µ x– n
Hence, n = 4
32. (b) Given : M1 = 1.20 Am2
N
BH
B
Nö
æ
22. (b) Corecivity, H = m and B = m 0 ni çè n = ÷ø
l
0
N
100
i=
× 5.2 = 2600 A/m
l
0.2
(b) Given Number of turns,
n = 1000 turns/cm = 1000 × 100 turns/m
Coercivity of ferromagnet, H = 100 A/m
Current to demagnetise the ferromagnet, I = ?
Using,
H = nI
or, 100 = 105 × I
100
\ I = 5 = 1 mA
10
(b) Graph [A] is for material used for making permanent
magnets (high coercivity)
Graph [B] is for making electromagnets and transformers.
(d) VB = VBHl = 240 × 5 × 10–5 cos(q) × 5 = 44.7 mv
By right hand rule, the charge moves to the left of pilot.
(c) Here, r = 30cm = 0.3cm
m0 M
= BH = 3.6 ´ 10 –5
we know
4 pr 3
(
4p´ 6.4 ´ 106
r
S
r
N
S
20
cm = 0.1m
M2 = 1.00 Am2 ; r =
2
Bnet = B1 + B2 + BH
m ( M1 + M 2 )
Bnet = 0
+ BH
4p
r3
=
10 -7 (1.2 + 1)
(0.1)3
+ 3.6 ´ 10 -5 = 2.56 ´ 10 -4 wb/m2
33. (a) Given M = 8 × 1022 Am2
d = Re = 6.4 × 106m
m0 2M
.
Earth’s magnetic field, B =
4p d3
4 p ´ 10 -7 2 ´ 8 ´ 1022
=
´
@ 0.6 Gauss
4p
(6.4 ´ 106 )3
34. (b) For a diamagnetic material, the value of µr is slightly
less than one. For any material, the value of Îr is always
greater than 1.
35. (b) Ferromagnetic substance has magnetic domains
whereas paramagnetic substances have magnetic dipoles
which get attracted to a magnetic field. Ferromagnetic
material magnetised strongly in the direction of magnetism
field, Hence, N1 will be attracted paramagnetic substance
attract weekly in the direction of field. Hence, N2 will weakly
attracted. Diamagnetic substances do not have magnetic
dipole but in the presence of external magnetic field due to
their orbital motion of electrons these substances are
repelled. Hence, N3 will be repelled.
P-347
Magnetism and Matter
36. (b) Electromagnet should be amenable to magnetisation
& demagnetization.
\ Materials suitable for making electromagnets should
have low retentivity and low coercivity should be low.
37. (b) The time period of a rectangular magnet oscillating in
earth’s magnetic field is given by T = 2p
I
MBH
where I = Moment of inertia of the rectangular magnet
M = Magnetic moment
BH = Horizontal component of the earth’s magnetic field
Initially, the time period of the magnet
T = 2p
I
1
where I = M l2
MBH
12
I¢
M ¢BH
\
T12
T22
I
We have, T = 2p MB
x
=
Bx 2
Bx1
B2 cos 45°
æ 2ö
=
çè ÷ø =
1.5
B1 cos 30°
2
2
B2
2
æ 4ö
´
çè ÷ø =
3
B1
6
B1
9
=
B2 8 6
T = 2p
T¢
I¢ M
I /8 M
1 1
´
=
´
=
=
\ =
M /2 I
T
M I
4 2
38. (a) The temperature above which a ferromagnetic
substance becomes paramagnetic is called Curie’s
temperature.
39. (a) Let I1 and I2 be the moment of inertia in first and second
case respectively.
I1 = 2 MR 2
I
MB
Where, M = magnetic moment, I moment of inertia and B =
magnetic field
mR 2
Th = 2p
( 2MB)
Tc = 2p
1 / 2mR 2
MB
X
Using, T = 2p
= 2p
TµI
T2
I
mgd
= 0.46
Clearly, Th = Tc
42. (c) Given : Magnetic moment, M = 6.7 × 10–2 Am2
Magnetic field, B = 0.01 T
Moment of inertia, I = 7.5 × 10–6 Kgm2
MR 2 3
I 2 = MR +
= MR 2
2
2
Axis of rotation
2
Time period, T = 2p
B2 ´ 2
2 ´ B1 ´ 3
41. (a) Using, time /oscillation period,
I
M
1 æ M öæ l ö
ç ÷ ç ÷ = and M ¢ =
12 è 2 ø è 2 ø
8
2
T1
2MR 2
2
=
3
2
3
MR
2
I1
=
I2
2
\
Moment of inertia of each part
I¢ =
T1
=
T2
40. (B onus)
or
Case 2
Magnet is cut into two identical pieces such that each
piece has half the original length.
Then T ¢ = 2p
\
I
MB
7.5 ´ 10-6
-2
=
2p
´ 1.06 s
10
6.7 ´ 10 ´ 0.01
Time taken for 10 complete oscillations
t = 10T = 2p × 1.06
= 6.6568 » 6.65 s
20
P-348
Physics
Electromagnetic
Induction
Magnetic Flux, Faraday's
TOPIC 1
and Lenz's Law
1.
3.
Two concentric circular coils, C1 and C2, are placed in the
XY plane. C1 has 500 turns, and a radius of 1 cm. C2 has 200
turns and radius current 20 cm. C2 carries a time dependent
current I(t) = (5t2 – 2t + 3) A Where t is in s. The emf induced
An elliptical loop having resistance R, of semi major axis a,
and semi minor axis b is placed in a magnetic field as shown
in the figure. If the loop is rotated about the x-axis with
angular frequency w, the average power loss in the loop
due to Joule heating is :
[Sep. 03, 2020 (I)]
B
z
2.
y
(a)
a
magnet
c
b
Three positions shown describe : (1) the magnet’s entry
(2) magnet is completely inside and (3) magnet’s exit.
(1)
(2)
(3)
®
(1)
(2)
®
(b)
(1)
(2)
(1)
(d)
(3)
®
®
(c)
(3)
6.
®
(2)
®
5.
®
(3)
®
p 2 a 2 b 2 B 2 w2
2R
a
x
y
(b) zero
pabBw
p 2 a 2 b 2 B 2 w2
(d)
R
R
A uniform magnetic field B exists in a direction
perpendicular to the plane of a square loop made of a
metal wire. The wire has a diameter of 4 mm and a total
length of 30 cm. The magnetic field changes with time at a
steady rate dB/dt = 0.032 Ts–1. The induced current in
the loop is close to (Resistivity of the metal wire is
1.23 × 10–8 W m)
[Sep. 03, 2020 (II)]
(c)
4.
(a)
b
x
4
. The value of x is
x
______.
[NA Sep. 05, 2020 (I)]
A small bar magnet is moved through a coil at constant
speed from one end to the other. Which of the following
series of observations will be seen on the galvanometer G
attached across the coil ?
[Sep. 04, 2020 (I)]
G
in C1 (in mV), at the instant t = 1 s is
7.
(a) 0.43 A
(b) 0.61 A (c) 0.34 A
(d) 0.53 A
A circular coil of radius 10 cm is placed in a uniform magnetic
field of 3.0 × 10–5 T with its plane perpendicular to the field
initially. It is rotated at constant angular speed about an
axis along the diameter of coil and perpendicular to
magnetic field so that it undergoes half of rotation in 0.2 s.
The maximum value of EMF induced (in mV) in the coil will
be close to the integer _____. [NA Sep. 02, 2020 (I)]
In a fluorescent lamp choke (a small transformer) 100 V of
reverse voltage is produced when the choke current
changes uniformly from 0.25 A to 0 in a duration of 0.025
ms. The self-inductance of the choke (in mH) is estimated
to be ______.
[NA 9 Jan. 2020 I]
At time t = 0 magnetic field of 1000 Gauss is passing
perpendicularly through the area defined by the closed
loop shown in the figure. If the magnetic field reduces
linearly to 500 Gauss, in the next 5 s, then induced EMF
in the loop is:
[NA 8 Jan. 2020 I]
P-349
Electromagnetic Induction
(c)
(d)
(a) 56 mV (b) 28 mV (c) 48 mV (d) 36 mV
Consider a circular coil of wire carrying constant current
I, forming a magnetic dipole. The magnetic flux through
an infinite plane that contains the circular coil and
excluding the circular coil area is given by fi .The
magnetic flux through the area of the circular coil area is
given by f0. Which of the following option is correct?
[7 Jan. 2020 I]
(a) fi = f0 (b) fi > f0 (c) fi < f0 (d) fi = – f0
9. A long solenoid of radius R carries a time (t) - dependent
current I(t) = I0t(l – t). A ring of radius 2R is placed coaxially
near its middle. During the time interval 0 £ t £ 1, the
induced current (IR) and the induced EMF(VR) in the ring
change as:
[7 Jan. 2020 I]
(a) Direction of IR remains unchanged and VR is maximum
at t = 0.5
(b) At t = 0.25 direction of IR reverses and VR is maximum
(c) Direction of IR remains unchanged and VR is zero at t = 0.25
(d) At t = 0.5 direction of IR reverses and VR is zero
10. A loop ABCDEFA of straight edges has six corner points
A(0, 0, 0), B{5, 0, 0), C(5, 5, 0), D(0, 5, 0), E(0, 5, 5) and
F(0, 0, 5). The magnetic field in this region is
r
B = ( 3iˆ + 4kˆ )T. The quantity of flux through the loop
ABCDEFA (in Wb) is _________ . [NA 7 Jan. 2020 I]
11. A planar loop of wire rotates in a uniform magnetic field.
Initially, at t = 0, the plane of the loop is perpendicular to
the magnetic field. If it rotates with a period of 10 s about
an axis in its plane then the magnitude of induced emf will
be maximum and minimum, respectively at:[7 Jan. 2020 II]
(a) 2.5 s and 7.5 s
(b) 2.5 s and 5.0 s
(c) 5.0 s and 7.5 s
(d) 5.0 s and 10.0 s
12. A very long solenoid of radius R is carrying current
I(t) = kte–at (k >0), as a function of time (t >0). Counter
clockwise current is taken to be positive. A circular
conducting coil of radius 2R is placed in the equatorial
plane of the solenoid and concentric with the solenoid.
The current induced in the outer coil is correctly depicted,
as a function of time, by:
[9 Apr. 2019 II]
8.
(a)
(b)
13. Two coils ‘P’ and ‘Q’ are separated by some distance. When
a current of 3A flows through coil ‘P’, a magnetic flux of
10–3 Wb passes through ‘Q’. No current is passed through
‘Q’. When no current passes through ‘P’ and a current of
2A passes through ‘Q’, the flux through ‘P’ is:
[9 Apr. 2019 II]
(a) 6.67 × 10–4 Wb
(b) 3.67 × 10–3 Wb
(c) 6.67 × 10–3 Wb
(d) 3.67 × 10–4 Wb
14. The self induced emf of a coil is 25 volts. When the
current in it is changed at uniiform rate from 10 A to 25
A in 1s, the change in the energy of the inductance is:
[9 Jan. 2019 II]
(a) 740 J
(b) 437.5 J
(c) 540 J
(d) 637.5 J
15. A conducting circular loop made of a thin wire, has area
3.5 × 10 –3m2 and resistance 10W. It is placed perpendicular
to a time dependent magnetic field B (t) = (0.4T) sin (50pt).
The the net charge flowing through the loop during t = 0
s and t = 10 ms is close to:
[9 Jan. 2019 I]
(a) 14 mC
(b) 7 mC (c) 21 mC
(d) 6 mC
16. In a coil of resistance 100 W , a current is induced by
changing the magnetic flux through it as shown in the
figure. The magnitude of change in flux through the coil is
[2017]
(a)
250 Wb
(b) 275 Wb
(c) 200 Wb
(d) 225 Wb
17. A conducting metal circular–wire–loop of radius r is placed
perpendicular to a magnetic field which varies with time as
-t
B = B0 e t , where B0 and t are constants, at time t = 0. If
the resistance of the loop is R then the heat generated in
the loop after a long time (t ® ¥) is ;
[Online April 10, 2016]
(a)
p 2 r 4 B04
2tR
(b)
p 2 r 4 B02
2tR
(c)
p 2 r 4 B02 R
t
(d)
p 2 r 4 B02
tR
P-350
Physics
18. When current in a coil changes from 5 A to 2 A in 0.1 s,
average voltage of 50 V is produced. The self - inductance
of the coil is :
[Online April 10, 2015]
(a) 6 H
(b) 0.67 H
(c) 3 H
(d) 1.67 H
19. Figure shows a circular area of radius
V (t )
t
®
R where a uniform magnetic field B is
going into the plane of paper and
increasing in magnitude at a constant
rate.
R
(A)
(B)
I (t )
I (t )
t
t
In that case, which of the following graphs, drawn
schematically, correctly shows the variation of the induced
electric field E(r)?
[Online April 19, 2014]
(C)
(D)
I (t )
I (t )
t
E
t
E
(a)
(b)
R
r
R
r
R
r
E
E
(c)
(d)
R
r
20. A coil of circular cross-section having 1000 turns and 4
cm2 face area is placed with its axis parallel to a magnetic
field which decreases by 10–2 Wb m–2 in 0.01 s. The e.m.f.
induced in the coil is:
[Online April 11, 2014]
(a) 400 mV
(b) 200 mV
(c) 4 mV
(d) 0.4 mV
21. A circular loop of radius 0.3 cm lies parallel to amuch bigger
circular loop of radius 20 cm. The centre of the small loop
is on the axis of the bigger loop. The distance between
their centres is 15 cm. If a current of 2.0 A flows through
the smaller loop, then the flux linked with bigger loop is
[2013]
(a) 9.1 × 10–11 weber
(b) 6 × 10–11 weber
(c) 3.3 × 10–11 weber
(d) 6.6 × 10–9 weber
22. Two coils, X and Y, are kept in close vicinity of each other.
When a varying current, I(t), flows through coil X, the
induced emf (V(t)) in coil Y, varies in the manner shown
here. The variation of I(t), with time, can then be
represented by the graph labelled as graph :
[Online April 9, 2013]
(a) A
(b) C
(c) B
(d) D
23. A coil is suspended in a uniform magnetic field, with the
plane of the coil parallel to the magnetic lines of force.
When a current is passed through the coil it starts
oscillating; It is very difficult to stop. But if an aluminium
plate is placed near to the coil, it stops. This is due to :
[2012]
(a) developement of air current when the plate is placed
(b) induction of electrical charge on the plate
(c) shielding of magnetic lines of force as aluminium is a
paramagnetic material.
(d) electromagnetic induction in the aluminium plate
giving rise to electromagnetic damping.
24. Magnetic flux through a coil of resistance 10 W is changed
by Df in 0.1 s. The resulting current in the coil varies with
time as shown in the figure. Then |Df| is equal to (in weber)
[Online May 12, 2012]
i(A)
4
0.1
t(s)
(a) 6
(b) 4
(c) 2
(d) 8
25. The flux linked with a coil at any instant 't' is given by
f = 10t2 – 50t + 250. The induced emf at t = 3s is [2006]
(a) –190 V
(b) –10 V
(c) 10 V
(d) 190 V
P-351
Electromagnetic Induction
Motional and Static EMI and
TOPIC 2 Application of EMI
26. An infinitely long straight wire carrying current I, one side
opened rectangular loop and a conductor C with a sliding
connector are located in the same plane, as shown in the
figure. The connector has length l and resistance R. It
slides to the right with a velocity v. The resistance of the
conductor and the self inductance of the loop are
negligible. The induced current in the loop, as a function
of separation r, between the connector and the straight
wire is :
[Sep. 05, 2020 (II)]
one side opened long
conducting wire loop
C
I
R
v
l
r
(a)
m0 Ivl
4p Rr
(b)
m 0 Ivl
p Rr
m0 Ivl
2m 0 Ivl
(d)
p Rr
2p Rr
27. The figure shows a square loop L of side 5 cm which is
connected to a network of resistances. The whole setup is
moving towards right with a constant speed of 1 cm s–1. At
some instant, a part of L is in a uniform magnetic field of 1
T, perpendicular to the plane of the loop. If the resistance
of L is 1.7 &!, the current in the loop at that instant will be
close to :
[12 Apr. 2019 I]
(c)
(a) 60µA
(b) 170 µA
(c) 150 µA
(d) 115 µA
28. The total number of turns and cross-section area in a
solenoid is fixed. However, its length L is varied by
adjusting the separation between windings. The
inductance of solenoid will be proportional to:
[9 April 2019 I]
(a) L
(b) L2
(c) 1/ L2
(d)
1/L
29. A thin strip 10 cm long is on a U shaped wire of negligible
resistance and it is connected to a spring of spring constant
0.5 Nm–1 (see figure). The assembly is kept in a uniform
magnetic field of 0.1 T. If the strip is pulled from its
equilibrium position and released, the number of
oscillations it performs before its amplitude decreases by
a factor of e is N. If the mass of strip is 50 grams, its
resistance 10W and air drag negligible, N will be close to :
[8 April 2019 I]
(a) 1000
(b) 50000 (c) 5000
(d) 10000
30. A 10 m long horizontal wire extends from North East to
South West. It is falling with a speed of 5.0 ms–1, at right
angles to the horizontal component of the earth’s magnetic
field, of 0.3 × 10–4 Wb/m2. The value of the induced emf in
wire is :
[12 Jan. 2019 II]
(a) 1.5 × 10–3 V
(b) 1.1 × 10–3 V
(c) 2.5 × 10–3V
(d) 0.3 × 10–3 V
31. There are two long co-axial solenoids of same length l.
The inner and outer coils have radii r1 and r2 and number
of turns per unit length n1 and n2, respectively. The ratio of
mutual inductance to the self-inductance of the inner-coil
is :
[11 Jan. 2019 I]
(a)
n1
n2
(b)
n2 r1
×
n1 r2
(c)
n2 r22
×
n1 r12
(d)
n2
n1
32. A copper wire is wound on a wooden frame, whose shape
is that of an equilateral triangle. If the linear dimension of
each side of the frame is increased by a factor of 3, keeping
the number of turns of the coil per unit length of the frame
the same, then the self inductance of the coil:
[11 Jan. 2019 II]
(a) decreases by a factor of 9
(b) increases by a factor of 27
(c) increases by a factor of 3
(d) decreases by a factor of 9 3
33. A solid metal cube of edge length 2 cm is moving in a
positive y-direction at a constant speed of 6 m/s. There
is a uniform magnetic field of 0.1 T in the positive
z-direction. The potential difference between the two
faces of the cube perpendicular to the x-axis, is:
[10 Jan. 2019 I]
(a) 12 mV
(b) 6 mV
(c) 1 mV
(d) 2 mV
P-352
Physics
34. An insulating thin rod of length l has a linear charge
x
on it. The rod is rotated about an
l
axis passing through the origin (x = 0) and perpendicular
to the rod. If the rod makes n rotations per second, then
the time averaged magnetic moment of the rod is:
[10 Jan. 2019 I]
p
n r l3
(a) p n r l3
(b)
3
p
n r l3
(c)
(d) n r l3
4
35. A coil of cross-sectional area A having n turns is placed in
a uniform magnetic field B. When it is rotated with an
angular velocity w, the maximum e.m.f. induced in the coil
will be
[Online April 16, 2018]
3
nBAw
(a) nBAw
(b)
2
1
nBAw
(c) 3nBAw
(d)
2
36. An ideal capacitor of capacitance 0.2 mF is charged to a
potential difference of 10V. The charging battery is then
disconnected. The capacitor is then connected to an ideal
inductor of self inductance 0.5mH. The current at a time
when the potential difference across the capacitor is 5V, is:
[Online April 15, 2018]
(a) 0.17A (b) 0.15A (c) 0.34A (d) 0.25A
37. A copper rod of mass m slides under gravity on two smooth
parallel rails, with separation 1 and set at an angle of q
with the horizontal. At the bottom, rails are joined by a
resistance R.There is a uniform magnetic field B normal to
the plane of the rails, as shown in the figure. The terminal
speed of the copper rod is:
[Online April 15, 2018]
density r(x) = r0
39. A square frame of side 10 cm and a long straight wire
carrying current 1 A are in the plate of the paper. Starting
from close to the wire, the frame moves towards the right
with a constant speed of 10 ms–1 (see figure).
I = 1A
x
v
10 cm
The e.m.f induced at the time the left arm of the frame is at
x = 10 cm from the wire is:
[Online April 19, 2014]
(a) 2 mV
(b) 1 mV
(c) 0.75 mV
(d) 0.5 mV
40. A metallic rod of length ‘l’ is tied to a string of length 2l
and made to rotate with angular speed w on a horizontal
table with one end of the string fixed. If there is a vertical
magnetic field ‘B’ in the region, the e.m.f. induced across
the ends of the rod is
[2013]
®
B
l
R
(a)
(c)
mgR cos q
B2l 2
mgR tan q
(a)
q
(b)
mgR sin q
B2l 2
mgR cot q
(d)
B2l 2
B2l 2
38. At the centre of a fixed large circular coil of radius R, a much
smaller circular coil of radius r is placed. The two coils are
concentric and are in the same plane. The larger coil carries
a current I. The smaller coil is set to rotate with a constant
angular velocity w about an axis along their common
diameter. Calculate the emf induced in the smaller coil after a
time t of its start of rotation.
[Online April 15, 2018]
(a)
m0 I 2
wr sin wt
2R
(c)
m0 I
wpr 2 sin wt
2R
(b)
m0 I
wpr 2 sin wt
4R
(d)
m0 I 2
wr sin wt
4R
2 Bwl2
2
(b)
3Bwl 2
2
4 Bwl2
5Bwl 2
(d)
2
2
41. A coil of self inductance L is connected at one end of two
rails as shown in figure. A connector of length l, mass m
can slide freely over the two parallel rails. The entire set up
is placed in a magnetic field of induction B going into the
page. At an instant t = 0 an initial velocity v0 is imparted to
it and as a result of that it starts moving along x-axis. The
displacement of the connector is represented by the figure.
[Online May 19, 2012]
(c)
B
L
x-axis
P-353
(b)
Time
Displacement
(a)
Displacement
Electromagnetic Induction
RW
Time
Displacement
Displacement
(d)
m0 NI
in the middle of the
L
solenoid but becomes less as we move towards its ends.
[Online May 19, 2012]
(a) Statement 1 is true, Statement 2 is false.
(b) Statement 1 is true, Statement 2 is true, Statement 2 is
the correct explanation of Statement 1.
(c) Statement 1 is false, Statement 2 is true.
(d) Statement 1 is true, Statement 2 is true, Statement 2 is
not the correct explanation of Statement 1.
43. A boat is moving due east in a region where the earth's
magnetic field is 5.0 × 10–5 NA–1 m–1 due north and
horizontal. The boat carries a vertical aerial 2 m long. If the
speed of the boat is 1.50 ms–1, the magnitude of the induced
emf in the wire of aerial is:
[2011]
(a) 0.75 mV
(b) 0.50 mV
(c) 0.15 mV
(d) 1mV
44. A horizontal straight wire 20 m long extending from east to
west falling with a speed of 5.0 m/s, at right angles
to the horizontal component of the earth’s magnetic field
0.30 × 10–4 Wb/m2. The instantaneous value of the e.m.f.
induced in the wire will be
[2011 RS]
(a) 3 mV
(b) 4.5 mV (c) 1.5 mV (d) 6.0 mV
45. A rectangular loop has a sliding connector PQ of length l
and resistance R W and it is moving with a speed v as
shown. The set-up is placed in a uniform magnetic field
going into the plane of the paper. The three currents I1, I2
and I are
[2010]
I2
Q
Blv
2 Blv
, I=
6R
6R
(a)
I1 = - I 2 =
(b)
I1 = I 2 =
(c)
I1 = I 2 = I =
Time
pm0 N 2 r 2
.
L
Statement 2: The magnetic induction in the solenoid in
RW
I
Time
42. This question has Statement 1 and Statement 2. Of the
four choices given after the Statements, choose the one
that best describes the two Statements.
Statement 1: Self inductance of a long solenoid of length
L, total number of turns N and radius r is less than
Statement 1 carrying current I is
v
RW
I1
(c)
l
P
Blv
2 Blv
,I =
3R
3R
Blv
R
Bl n
Bl n
, I=
6R
3R
46. Two coaxial solenoids are made by winding thin insulated
wire over a pipe of cross-sectional area A = 10 cm2 and
length = 20 cm. If one of the solenoid has 300 turns and the
other 400 turns, their mutual inductance is
[2008]
(m0 = 4p × 10 –7 Tm A–1)
(a) 2.4p × 10–5 H
(b) 4.8p × 10–4 H
(c) 4.8p × 10–5 H
(d) 2.4p × 10–4 H
47. One conducting U tube can slide inside another as shown
in figure, maintaining electrical contacts between the tubes.
The magnetic field B is perpendicular to the plane of the
figure . If each tube moves towards the other at a constant
speed v, then the emf induced in the circuit in terms of B, l
and v where l is the width of each tube, will be [2005]
(d)
I1 = I 2 =
A
B
v
v
X
C
(a) – Blv
(b) Blv
(c) 2 Blv
(d) zero
48. A metal conductor of length 1 m rotates vertically about one
of its ends at angular velocity 5 radians per second. If the
horizontal component of earth’s magnetic field is 0.2×10–4T,
then the e.m.f. developed between the two ends of the
conductor is
[2004]
(a) 5 mV
(b) 50 mV
(c) 5 mV
(d) 50mV
49. A coil having n turns and resistance RW is connected with
a galvanometer of resistance 4RW. This combination is
moved in time t seconds from a magnetic field W1 weber to
W2 weber. The induced current in the circuit is [2004]
P-354
(a)
Physics
-
(W2 - W1 )
Rnt
(b)
(W2 - W1 )
5 Rnt
(d)
-
n (W 2 - W1 )
5 Rt
n(W2 - W1 )
Rt
50. Two coils are placed close to each other. The mutual
inductance of the pair of coils depends upon
[2003]
(a) the rates at which currents are changing in the two
coils
(b) relative position and orientation of the two coils
(c) the materials of the wires of the coils
(d) the currents in the two coils
51. When the current changes from +2 A to –2A in 0.05 second,
an e.m.f. of 8 V is induced in a coil. The coefficient of self
-induction of the coil is
[2003]
(a) 0.2 H
(b) 0.4 H
(c) 0.8 H
(d) 0.1 H
(c)
-
-
52. A conducting square loop of side L and resistance R moves
in its plane with a uniform velocity v perpendicular to one
of its sides. A magnetic induction B constant in time and
space, pointing perpendicular and into the plane at the
loop exists everywhere with half the loop outside the field,
as shown in figure. The induced emf is
[2002]
L
(a) zero
(b) RvB
v
(c) vBL/R
(d) vBL
P-355
Electromagnetic Induction
1.
(5)
For coil C1, No. of turns N1 = 500 and radius, r = 1 cm.
For coil C2, No. of turns N2 = 200 and radius, R = 20 cm
dI
I = (5t - 2t + 3) Þ
= (10t - 2)
dt
æ m IN ö
fsmall = BA = ç 0 2 ÷ (pr 2 )
è 2R ø
Induced emf in small coil,
Current, i =
But, resistance of wire, R = r
2
e=
\i =
5.
d f æ m 0 N 2 ö 2 di æ m 0 N1 N 2 pr 2 ö
=
pr N1 = ç
÷ (10t - 2)
dt çè 2r ÷ø
dt è
2R
ø
-d f
= ABw sin(wt )
dt
Max. value of Emf = ABw = pR2Bw
4(4p)10-7 ´ 200
10 -4
´ 500 ´ -2 p
20
10
= 3.14 ´ 0.1 ´ 0.1 ´ 3 ´ 10 -5 ´
4
Þ x = 5.
x
(b) Case (a) : When bar magnet is entering with constant
speed, flux (f) will change and an e.m.f. is induced, so
galvanometer will deflect in positive direction.
Case (b) : When magnet is completely inside, flux (f) will
not change, so galvanometer will show null deflection.
Case (c) : When bar magnet is making on exit, again flux
(f) will change and an e.m.f. is induced in opposite direction
so galvanometer will deflect in negative direction i.e.
reverse direction.
(a) As we know, emf e = NABw cos wt , Here N = 1
Average power,
A2 B 2 w 2 æ 1 ö
e
A B w cos wt
<P> = <
>=<
>=
çè ÷ø
R
2
R
R
Therefore average power loss in the loop due to Joule
heating
2
2
2
p
0.2
= 15 ´ 10 -6 V = 15 mV
= 8 ´ 10-4 volt = 0.8 mV=
2
2p
p
=
2T 0.2
Emf induced, e =
= 80 ´ p 2 ´ 10-7 ´ 10 ´ 102 ´ 10-2
3.
dB ( A)2 0.032 ´ {p´ 2 ´ 10-3 }2
=
= 0.61 A.
dt rl
1.23 ´10-8 ´ 0.3
r r
Flux as a function of time f = B × A = AB cos(wt )
æ m N N pr 2 ö
m N N pr 2
e=ç 0 1 2
8=4 0 1 2
÷
2R
R
è
ø
2.
l
A
(15)
Here, B = 3.0 × 10–5 T, R = 10 cm = 0.1 m
w=
At t = 1 s
=
e
R
2
p 2 a 2 b2 B 2 2
(w )
2R
(b) Given,
Length of wire, l = 30 cm
Radius of wire, r = 2 mm = 2 × 10–3 m
6.
(10) Given dI = 0.25 – 0 = 0.25 A
dt = 0.025 ms
Induced voltage
Eind = 100 v
Self-inductance, L = ?
L (0.25 – 0)
Df
Þ 100 =
Dt
.025 ´10-3
–3
Þ L = 10 H = 10 mH
(a) According to question, dB = 1000 – 500 = 500 gauss
= 500 × 10–4T
Time dt = 5 s
Using faraday law
Using, Eind =
7.
Induced EMF , e = –
df
dB
= A
dt
dt
dB 1000 – 500
=
´ 10 –4 = 10 –2 T/sec
dt
5
<P>=
4.
Resistivity of metal wire, r = 1.23 ´ 10 - 8 W m
d f dB
Emf generated, | e | =
=
( A)
dt
dt
(Q f = B.A.)
Area, A = ar of X –2 ar of D = (16 × 4 – 2 × Area of triangle) cm2
1
æ
ö
= ç 64 – 2 ´ ´ 2 ´ 4 ÷ cm 2
2
è
ø
= 56 × 10–4 m2
P-356
Physics
\ einduced = A
8.
dB
= 56 ´ 10 –4 ´ 10–2 = 56 ´ 10 –6 V = 56mV
dt
(d) As magnetic field lines form close loop, hence every
magnetic field line creating magnetic flux through the inner
region (fi) must be passing through the outer region.
Since flux in two regions are in opposite region.
\ fi = –f0
wt = p
= - m0 nAk [e -at (1 - t )]
e - m0 nAk -at
=
[e (1 - t )]
R
R
At t = 0, i Þ –ve
i=
(Q B = m0 nI and A = pR 2 )
13. (a) Qcoil = ( NQ) µ i
–df
dt
So,
VR = m0 npR 2 ( I0 – 2I 0t )
Þ VR = 0 at t =
d
dQ
= -m 0 nAk (te -a t )
dt
dt
= -m0 nAk[t (-1)e- at + e -at ´1]
(d) According to question,
I(t) = I0t(1 – t)
\ I = I0t – I0t2
f = B.A
f = (m0 nI) × (pR2)
VR =
p
=5s
p
5
\ Induced emf is zero at t = 5 s
12. (a) Q = BA
= (m0 ni)A
= m0 n (kt e–at)A
e=-
9.
Þ t=
Q1 i1
3
=
=
Q2 i2 2
2
2
-3
or Q2 = Q1 = ´ 10 = 6.67 × 10–4 Wb
3
3
14. (b) According to faraday’s law of electromagnetic induc-
1
s
2
10. (175.00)
tion, e =
-df
dt
di
15
5
= 25 Þ L ´ = 25 or L = H
dt
1
3
Change in the energy of the inductance,
1
1 5
DE = L i12 – i 22 = ´ ´ (252 –10 2 )
2
2 3
5
= ´ 525 = 437.5J
6
15. [B onus]
L´
(
Flux through the loop ABCDEFA,
r r
ˆ
ˆ + 25k)
ˆ
f = B.A = (3iˆ + 4k).(25i
Þ f = (3 × 25) + (4 × 25) = 175 weber
11. (b) We have given, time period, T = 10s
2p p
=
10 5
Magnetic flux, f(t)= BA cos wt
\
Angular velocity, w =
Emf induced, E =
Þ t=
p
= 2.5 s
2
p
5
For induced emf to be minimum i.e zero
Df 1
1
= A(Bf - Bi ) = ´ 3.5 ´10 -3
R 10
10
p
æ
ö
ç 0.4sin - 0 ÷
2
è
ø
–d f
= BAw sin wt = BAw sin ( wt )
dt
Induced emf, | e | is maximum when wt =
Net charge Q =
)
p
2
1
(3.5 ´ 10-3 )(0.4 - 0)
10
= 1.4 × 10– 4
No option matches, So it should be a bonus.
16. (a) According to Faraday's law of electromagnetic
df
induction, e =
dt
Also, e = iR
df
\ iR =
Þ ò d f = R ò idt
dt
=
P-357
Electromagnetic Induction
Magnitude of change in flux (df) = R × area under current
vs time graph
1 1
´ ´ 10 = 250 Wb
2 2
17. (b) Electric flux is given by
f = B.A
or, df = 100 ´
(Q B = B0e-t/ t )
f = B0 pr 2 e - t / t
Induced E.m.f. e =
Heat =
df B0 pr 2 - t / t
e
=
dt
t2
¥ 2
p 2 r 4 B02
e
=
ò R 2tR
18. (d) According to Faraday’s law of electromagnetic
induction,
Ldi
dt
æ 5–2 ö
50 = L ç
÷
è 0.1sec ø
50 ´ 0.1 5
= = 1.67 H
Þ L=
3
3
19. (a) Inside the sphere field varies linearly i.e., E µ r with
1
distance and outside varies according to E µ
r2
Hence the variation is shown by curve (a)
20. (a) Given: No. of turns N = 1000
Face area, A = 4 cm2 = 4 × 10–4 m2
Change in magnetic field,
DB = 10–2 wbm–2
Time taken, t = 0.01s = 10–2 sec
Emf induced in the coil e = ?
Applying formula,
Induced emf, e =
=
-d f
DB ö
= N æç
÷ A cos q
dt
è Dt ø
1000 ´ 10-2 ´ 4 ´ 10 -4
10-2
2
2
f = B × A = Bldr
Þf=
´ p(0.3 ´ 10 -2 ) 2
2[(0.2) + (0.15) ]
On solving
= 9.216 × 10–11 = 9.2 × 10–11 weber
22. (a) Induced emf
- di
eµ
dt
23. (d) Because of the Lenz’s law of conservation of energy.
Length of straignt wire, l = 20m Earth’s Magneti field,
B = 0.30 × 10–4 Wb/m2.
[Q A = l dr and B.A = BA cos 0°]
m0 I
l dr
2pr
V
I
r
Emf, e =
dr
m Ivl
d f m 0 Il dr
=
× Þe= 0 ×
dt
2 pr dt
2p r
Induce current in the loop, i =
e m 0 Ivl
=
×
R 2p Rr
27. (b) Induced emf,
e = Bvl= 1 × 10–2 × 0.05 = 5 × 10–4 V
Equivalent resistance,
R=
4´ 2
4
+ 1.7 = + 1.7 ; 3 W
3
4+2
Current, i =
= 400 mV
21. (a) As we know, Magnetic flux, f = B. A
m0 (2)(20 ´ 10 -2 ) 2
m0 I
2 pr
Magnetic flux for small displacement dr,
B=
0
Induced emf, e =
Df
or Ri = Df
(Q e = Ri )
Dt
Dt
Þ Df = R(i.Dt)
= R × area under i – t graph
1
= 10 × × 4 × 0.1 = 2 weber
2
25. (b) Electric flux, f = 10t2 – 50t + 250
df
= - (20t - 50)
Induced emf, e = dt
et = 3 = –10 V
26. (d) Magnetic field at a distance r from the wire
24. (c) As e =
e 5 ´10-4
=
; 170 m A
R
3
28. (d) Inductance =
m0 N 2 A
L
29. (c) Force on the strip when it is at stretched position x
from mean position is
F = -kx - iIB = -kx -
BIv
´ IB
R
B2 I 2
´v
R
Above expression shows that it is case of damped
oscillation, so its amplitude can be given by
F = -kx -
P-358
Physics
Þ A = A0 e
-
33. (a) Potential difference
perpendicular to x-axis
bt
2m
bt
A
Þ 0 = A0 e 2 m
e
Þ t=
2m
æ B2I 2
çç
è R
ö
÷÷
ø
=
A0
[as per question A =
]
e
2 ´ 50 ´ 10-3 ´ 10
0.01 ´ 0.01
Given, m = 50 × 10–3 kg
B = 0.1 T
l = 0.1 m
R = 10 W
k = 0.5 N
m
; 2s
k
so, required number of oscillations,
Time period, T = 2p
10000
= 5000
2
30. (a) Induced emf, e = Bvl
= 0.3 × 10–4 × 5 × 10
= 1.5 × 10–3 V
31. (d) The rate of mutual inductance is given by
N=
M = m0n1n2 pr12
...(i)
The rate of self inductance is given by
L = m 0 n12 pr12
...(ii)
Dividing (i) by (ii) Þ
M n2
=
L n1
32. (c) As total length L of the wire will remain constant
L = (3a) N
(N = total turns )
and length of winding = (d) N
a
a
(d = diameter of wire)
Magnetic moment, M = NIA
dQ = r dx
dQ
.w
2p
dM = dI × A
dI =
l
r0
w r0
n p ò x 3 dx
. . x p x 2 dx Þ M =
=
l
2p l
0
p
3
= . nrl
4
df
= NBAsin wt
35. (a) Induced emf in a coil, e = dt
Also, e = e0 sin wt
\ Maximum emf induced, e0 = nBAw
36. (a) Given: Capacitance, C = 0.2 mF = 0.2 × 10–6 F
Inductance L = 0.5 mH = 0.5 × 10–3 H
Current I = ?
Using energy conservation
1
1
1
CV 2 = CV12 + LI 2
2
2
2
1
´ 0.2 ´10 –6 ´10 2 + 0
2
1
–6
2 1
–3 2
= ´ 0.2 ´10 ´ 5 + ´ 0.5 ´10 I
2
2
\ I = 3 ´ 10 –1 A = 0.17 A
37. (b) From Faraday’s law of electomagnetic induction,
e=
d f d ( BA) d ( Bll )
=
=
dt
lt
dt
Bdl ´ l
= BVl
dt
B 2l 2V
æ BV ö 2
(
)
l
B
=
Also, F = ilB = ç
è R ÷ø
R
At equilibrium
®
B
l
æ 3a 2 ö
= m0n2 ç
÷ dN
è 4 ø
So self inductance will become 3 times
faces
34. (c)
self inductance = m0n2Al
1
µ a2 N µ a [as N = L/3a Þ Nµ ]
a
Now ‘a’ increased to ‘3a’
two
= l V.B = 2 ´10 –2 (6 ´ 0.1) =12mV
=
a
between
R
mg sin q =
q
B 2lV
mgR sin q
ÞV =
R
B 2l 2
P-359
Electromagnetic Induction
38. (c) According to Faraday’s law of electromagnetic
induction,
df
e= and f = BA cos wt = Bpr 2 cos wt
dt
d
2
2
Þ e = - (pr B cos wt ) = pr B sin wt (w )
dt
m0 I
m0 I ö
æ
e
pwr 2 sin wt çQ=
B
\=
÷
è
2R
2R ø
39. (b) In the given question,
Current flowing through the wire, I = 1A
Speed of the frame, v = 10 ms–1
Side of square loop, l = 10 cm
Distance of square frame from current carrying wires
x = 10 cm.
We have to find, e.m.f induced e = ?
According to Biot-Savart’s law
m Idlsin q
B= 0
4p x 2
=
4p´10-7 1´ 10-1
´
2
4p
10-1
( )
= 10–6
Induced e.m.f. e = Blv
= 10–6 × 10–1 × 10 = 1 mv
40. (d) Here, induced e.m.f.
w
l
2l
x
e=
3l
dx
ò (wx) Bdx = Bw
2l
45. (b) Due to the movement of resistor R, an emf equal to
Blv will be induced in it as shown in figure clearly,
P
l Blv
RW
5 Bl 2 w
=
2
41. (d)
42. (b) Self inductance of a long solenoid is given by
m0 N 2 A
l
Magnetic field at the centre of solenoid
m0 NI
B=
l
So both the statements are correct and statement 2 is
correct explanation of statement 1
43. (d) As magnetic field lines form close loop,
hence every magnetic field line creating magnetic flux
through the inner region (fi) must be passing through
the outer region. Since flux in two regions are in opposite
region.
L=
\ fi = –f0
44. (a) Induced, emF, e = Bvl
= 0.3 × 10–4 × 5 × 20
= 3 × 10–3 V = 3 mV.
RW
I
I2
I1
I = I1 + I 2
Q
Also, I1 = I2
Solving the circuit,
we get I1 = I 2 =
and I = 2 I1 =
Blv
3R
2 Blv
3R
46. (d) Given, Area of cross-section of pipe,
A = 10 cm2
Length of pipe, l = 20 cm
M=
=
m0 N1 N2 A
l
4p ´ 10-7 ´ 300 ´ 400 ´ 100 ´ 10 -4
0.2
M=
[(3l)2 – (2l)2 ]
2
v
RW
m0 N1 N2 A
l
= 2.4p × 10–4 H
47. (c) Relative velocity of the tube of width l,
= v – (–v) v = 2v
\ Induced emf. = B.l (2v)
48. (b) Given, length of conductor l = 1m,
Angular speed, w = 5 rad/s,
Magnetic field, B = 0.2 × 10–4 T
EmF generated between two ends of conductor
e=
49. (b)
Bwl 2 0.2 ´ 10-4 ´ 5 ´ 1
=
= 50mV
2
2
Df (W2 - W1 )
=
Dt
t
Rtot = ( R + 4 R )W = 5R W
i=
- n(W2 - W1 )
nd f
=
Rtot dt
5Rt
(Q W2 & W1 are magnetic flux)
50. (b) Mutual inductance depends on the relative position
and orientation of the two coils.
P-360
51. (d) Induced emf,
Df -D ( LI )
DI
=
= -L
Dt
Dt
Dt
DI
\ | e |= L
Dt
[2 – (–2)]
Þ 8= L ´
0.05
8 ´ 0.05
Þ L=
= 0.1H
4
52. (d) As the side BC is outside the field, no emf is induced
across BC. Further, sides AB and CD are not cutting any
flux. So, they will not centribute in flux.
Only side AD is cutting the flux 50 emf will be induced due
to AD only.
The induced emf is
e=-
Physics
r r
-d f
d ( B × A) -d ( BA cos0º )
e=
==
dt
dt
dt
×
×
×
×
×
×
×
×
×
× l
×
×
×
×
×
A
D
×
×
C
×
dA
d (l ´ x )
= -B
dt
dt
dx
\ e = - Bl = - Blv
dt
\ e = –B
V
×
X
×
B
21
Alternating Current
Alternating Current,
TOPIC 1
Voltage and Power
1.
2.
An alternating voltage v(t) = 220 sin 100Àt volt is applied
to a purely resistive load of 50W. The time taken for the
current to rise from half of the peak value to the peak
value is :
[8 April 2019 I]
(a) 5 ms
(b) 2.2 ms (c) 7.2 ms (d) 3.3 ms
A small circular loop of wire of radius a is located at the
centre of a much larger circular wire loop of radius b. The
two loops are in the same plane. The outer loop of radius
b carries an alternating current I = Io cos (wt). The emf
induced in the smaller inner loop is nearly :
[Online April 8, 2017]
(a)
2
pm0 Io a 2
. w sin (wt) (b) pm0 Io . a w cos (wt)
2
b
2
b
a2
pm0 I o b 2
w sin (wt) (d)
w cos ( wt)
b
a
A sinusoidal voltage V(t) = 100 sin (500t) is applied across
a pure inductance of L = 0.02 H. The current through the
coil is:
[Online April 12, 2014]
(a) 10 cos (500 t)
(b) – 10 cos (500t)
(c) 10 sin (500t)
(d) – 10 sin (500t)
In an a.c. circuit the voltage applied is E = E0 sin wt. The
(a)
4.
pö
æ
resulting current in the circuit is I = I 0 sin ç wt - ÷ . The
è
2ø
power consumption in the circuit is given by
[2007]
E I
(a) P = 2 E0 I0
(b) P = 0 0
2
E0 I 0
(d) P =
2
In a uniform magnetic field of induction B a wire in the
form of a semicircle of radius r rotates about the diameter
of the circle with an angular frequency w. The axis of
rotation is perpendicular to the field. If the total resistance
of the circuit is R, the mean power generated per period of
rotation is
[2004]
(c) P = zero
5.
(b)
( B pr 2 w ) 2
8R
B pr 2 w
( B pr w 2 ) 2
(d)
2R
8R
Alternating current can not be measured by D.C. ammeter
because
[2004]
(a) Average value of current for complete cycle is zero
(b) A.C. Changes direction
(c) A.C. can not pass through D.C. Ammeter
(d) D.C. Ammeter will get damaged.
(c)
6.
AC Circuit, LCR Circuit,
TOPIC 2
Quality and Power Factor
7.
(c) pm0 Io
3.
( B pr w )2
2R
A part of a complete circuit is shown in the figure. At some
instant, the value of current I is 1 A and it is decreasing at a
rate of 102A s–1. The value of the potential difference VP –VQ,
(in volts) at that instant, is ______.
[NA Sep. 06, 2020 (I)]
L=50 mH I
R=2W
P
8.
9.
Q
30 V
An AC circuit has R = 100 W, C = 2 mF and L = 80 mH,
connected in series. The quality factor of the circuit is :
[Sep. 06, 2020 (I)]
(a) 2
(b) 0.5
(c) 20
(d) 400
In a series LR circuit, power of 400 W is dissipated from a
source of 250 V, 50 Hz. The power factor of the circuit is 0.8. In
order to bring the power factor to unity, a capacitor of value C
is added in series to the L and R. Taking the value C as
æ nö
çè ÷ø mF , then value of n is ______. [NA Sep. 06, 2020 (II)]
3p
10. A series L-R circuit is connected to a battery of emf V. If
the circuit is switched on at t = 0, then the time at which the
æ 1ö
energy stored in the inductor reaches ç ÷ times of its
è nø
maximum value, is :
[Sep. 04, 2020 (II)]
P-362
L æ n ö
÷
(a) R ln ç
è n -1 ø
(b)
L æ n ö
ln ç
÷
R è n +1ø
(d)
L æ n +1ö
ln ç
÷
R è n -1 ø
L æ n -1 ö
ln ç
÷
R è n ø
A 750 Hz, 20 V (rms) source is connected to a resistance of
100 W, an inductance of 0.1803 H and a capacitance of 10
mF all in series. The time in which the resistance (heat
capacity 2 J/°C) will get heated by 10°C. (assume no loss
of heat to the surroundings) is close to :
(c)
11.
Physics
2
1
(a) ln 2
(b)
ln 2
2
(c) 2 ln 2
(d) ln 2
16. A circuit connected to an ac source of emf e = e0sin(100t)
p
between
4
the emf e and current i. Which of the following circuits
will exhibit this ?
[8 April 2019 II]
with t in seconds, gives a phase difference of
[Sep. 03, 2020 (I)]
12.
(a) 418 s
(b) 245 s
(c) 365 s
(d) 348 s
(a) 1.1 × 10–2 H
(b) 1.1 × 10–1 H
5.5 × 10–5 H
6.7 × 10–7 H
(c)
13.
(a) RL circuit with R = 1 kW and L = 10 mH
An inductance coil has a reactance of 100 W. When an AC
signal of frequency 1000 Hz is applied to the coil, the applied
voltage leads the current by 45°. The self-inductance of
the coil is :
[Sep. 02, 2020 (II)]
(d)
Consider the LR circuit shown in the figure. If the switch
S is closed at t = 0 then the amount of charge that passes
through the battery between t = 0 and t =
L
is :
R
(b) RL circuit with R = 1 kW and L = 1 mH
(d) RC circuit with R = 1 kW and C = 1 mF
(d) RC circuit with R = 1 kW and C = 10 mF.
17. In the figure shown, a circuit contains two identical
resistors with resistance R = 5 W and an inductance with L
= 2 mH. An ideal battery of 15 V is connected in the circuit.
What will be the current through the battery long after the
switch is closed?
[12 Jan. 2019 I]
[12 April 2019 II]
S
L
R
R
(a)
(c)
14.
2.7EL
R
2
7.3EL
R2
(d)
EL
(a) 5.5 A
(c) 3 A
2.7 R 2
EL
18.
7.3R 2
A coil of self inductance 10 mH and resistance 0.1 W is
connected through a switch to a battery of internal
resistance 0.9 W. After the switch is closed, the time taken
for the current to attain 80% of the saturation value is
[take ln 5 = 1.6]
15.
(b)
(b) 7.5 A
(d) 6 A
I2
R2
R1
C
L
I1
~
[10 April 2019 II]
(a) 0.324 s
(b) 0.103 s
(c) 0.002 s
(d) 0.016 s
A 20 Henry inductor coil is connected to a 10 ohm
resistance in series as shown in figure. The time at which
rate of dissipation of energy (Joule’s heat) across
resistance is equal to the rate at which magnetic energy is
stored in the inductor, is :
[8 April 2019 I]
3
3
µF, R2 = 20 W, L =
H and
10
2
R1 = 10 W. Current in L-R1 path is I1 and in C-R2 path it is
In the above circuit, C =
I2. The voltage of A.C source is given by, V = 200 2 sin
(100 t) volts. The phase difference between I1 and I2 is :
[12 Jan. 2019 II]
(a) 60°
(b) 30°
(c) 90°
(d) 0
P-363
Alternating Current
19.
In the circuit shown,
R
(a)
L
(c)
23.
S2
S1
e
the switch S1 is closed at time t = 0 and the switch S2 is
kept open. At some later time (t0), the switch S1 is opened
and S2 is closed. the behaviour of the current I as a function
of time ‘t’ is given by:
[11 Jan. 2019 II]
I
I
(a)
(b)
tO
tO
t
tO
21.
t
t
tO
t
A series AC circuit containing an inductor (20 mH), a
capacitor (120 mF) and a resistor (60 W) is driven by an
AC source of 24 V/50 Hz. The energy dissipated in the
circuit in 60 s is:
[9 Jan. 2019 I]
(a) 5.65 × 102 J
(b) 2.26 × 103 J
(c) 5.17 × 102 J
(d) 3.39 × 103 J
In LC circuit the inductance L = 40 mH and capacitance C
= 100 mF. If a voltage V(t) = 10 sin(314 t) is applied to the
circuit, the current in the circuit is given as:
[9 Jan. 2019 II]
(a) 0.52 cos 314 t
(c) 5.2 cos 314 t
(b)
2
(d)
eL
ÎL
Î L æ 1ö
ç1 - ÷
R2 è e ø
ÎR
R
eL2
A LCR circuit behaves like a damped harmonic oscillator.
Comparing it with a physical spring-mass damped
oscillator having damping constant ‘b’, the correct
equivalence would be:
[7 Jan. 2020 I]
(a) L « m, C « k, R « b
1
1
1
(b) L « , C « , R «
b
m
k
(c) L « k, C « b, R « m
1
(d) L « m, C « , R « b
k
An emf of 20 V is applied at time t = 0 to a circuit containing
in series 10 mH inductor and 5 W resistor. The ratio of the
currents at time t = ¥ and at t = 40 s is close to:
[7 Jan. 2020 II]
(b) 1.15
(c) 1.46
(d) 0.84
25. In an a.c. circuit, the instantaneous e.m.f. and current are
given by
e = 100 sin 30 t
(d)
(c)
2
(Take e2 = 7.389)
(a) 1.06
I
I
20.
24.
ÎR
(b) 10 cos 314 t
(d) 0.52 sin 314 t
22.
As shown in the figure, a battery of emf Î is connected to an
inductor L and resistance R in series. The switch is closed at
t = 0. The total charge that flows from the battery, between t
= 0 and t = tc (tc is the time constant of the circuit) is:
[8 Jan. 2020 II]
pö
æ
i = 20 sin ç 30 t - ÷
4ø
è
In one cycle of a.c., the average power consumed by the
circuit and the wattless current are, respectively: [2018]
1000
(a) 50W, 10A
(b)
W, 10A
2
50
W, 0
(c)
(d) 50W, 0
2
26. For an RLC circuit driven with voltage of amplitude vm and
frequency w0 =
1
the current exhibits resonance. The
LC
quality factor, Q is given by:
[2018]
CR
w0 L
w0 R
R
(b)
(c)
(d)
w0
R
L
(w0C)
27. A sinusoidal voltage of peak value 283 V and angular
frequency 320/s is applied to a series LCR circuit. Given
that R = 5 W, L= 25 mH and C = 1000 mF. The total impedance,
and phase difference between the voltage across the
source and the current will respectively be :
[Online April 9, 2017]
(a)
(a) 10 W and tan–1 æç 5 ö÷ (b) 7 W and 45°
è 3ø
æ8ö
5
(c) 10 W and tan –1 ç ÷ (d) 7 W and tan–1 æç ö÷
è3ø
è 3ø
P-364
28.
29.
Physics
An arc lamp requires a direct current of 10 A at 80 V to
function. If it is connected to a 220 V (rms), 50 Hz AC
supply, the series inductor needed for it to work is close to :
[2016]
(a) 0.044 H
(b) 0.065 H
(c) 80 H
(d) 0.08 H
A series LR circuit is connected to a voltage source with
V(t) = V0 sinwt. After very large time, current l(t) behaves
Lö
æ
as çè t 0 >> ÷ø :
R
(a) 6.7 mA
(b) 0.67 mA
(c) 100 mA
(d) 67 mA
31. An LCR circuit is equivalent to a damped pendulum. In an
LCR circuit the capacitor is charged to Q0 and then
connected to the L and R as shown below :
[Online April 9, 2016]
C
If a student plots graphs of the square of maximum charge
Q 2Max on the capacitor with time(t) for two different
values L1 and L2 (L1 > L2) of L then which of the following
represents this graph correctly ? (plots are schematic and
not drawn to scale)
[2015]
I(t)
(
(a)
)
t
t = t0
L1
2
I(t)
(a)
QMax
L2
2
(b)
QMax
Q0 (For both L1 and L2)
t
t
t
(b) t = t
0
2
(c)
QMax
L1
2
Q
(d) Max L1
L2
L2
t
I(t)
t
t = t0
t
32. For the LCR circuit, shown here, the current is observed
to lead the applied voltage. An additional capacitor C’,
when joined with the capacitor C present in the circuit,
makes the power factor of the circuit unity. The
capacitor C’, must have been connected in :
[Online April 11, 2015]
(c)
L
I(t)
(d)
L
R
C
R
t
t0
~
30.
An inductor (L = 0.03 H) and a resistor (R = 0.15 kW) are
connected in series to a battery of 15V emf in a circuit
shown below. The key K1 has been kept closed for a long
time. Then at t = 0, K1 is opened and key K2 is closed
simultaneously. At t = l ms, the current in the circuit will
be : ( e5 @ 150 )
[2015]
0.03 H
0.15 kW
K2
15V
K1
V = V0sintw
(a) series with C and has a magnitude
(b) series with C and has a magnitude
2
C
(w LC –1)
1 - w2 LC
w2 L
(c) parallel with C and has a magnitude
1 - w2 LC
w2 L
(d) parallel with C and has a magnitude
C
(w LC - 1)
2
P-365
Alternating Current
33.
V
In the circuits (a) and (b) switches S1 and S2 are closed
at t = 0 and are kept closed for a long time. The variation of
current in the two circuits for t ³ 0 are roughly shown by
figure (figures are schematic and not drawn to scale) :
[Online April 10, 2015]
C
L
R
R
S1
C
R
L
S1
E
E
R
E
(a)
(b)
(a)
(a) i
E
R
(b)
(a)
(b) i
(b)
t
t
E
R
(a)
(b)
(c) i
E
R
(b)
(d) i
(a) Work done by the battery is half of the energy
dissipated in the resistor
(b) At, t = t, q = CV/2
(c) At, t = 2t, q = CV (1 – e–2)
(d) At, t = 2 t, q = CV (1 – e–1)
37. A series LR circuit is connected to an ac source of
frequency w and the inductive reactance is equal to 2R. A
capacitance of capacitive reactance equal to R is added in
series with L and R. The ratio of the new power factor to
the old one is :
[Online April 25, 2013]
(a)
(a)
In the circuit shown here, the point ‘C’ is kept connected
to point ‘A’ till the current flowing through the circuit
becomes constant. Afterward, suddenly, point ‘C’ is
disconnected from point ‘A’ and connected to point ‘B’ at
time t = 0. Ratio of the voltage across resistance and the
inductor at t = L/R will be equal to:
[2014]
A
C
R
L
B
(a)
35.
36.
e
1- e
(b) 1
(c) –1
(d)
1- e
e
When the rms voltages VL, VC and VR are measured
respectively across the inductor L, the capacitor C and the
resistor R in a series LCR circuit connected to an AC source,
it is found that the ratio VL : VC : VR = 1 : 2 : 3. If the rms
voltage of the AC sources is 100 V, the VR is close to:
[Online April 9, 2014]
(a) 50 V
(b) 70 V (c) 90 V
(d) 100 V
In an LCR circuit as shown below both switches are open
initially. Now switch S1 is closed, S2 kept open. (q is charge
on the capacitor and t = RC is Capacitive time constant).
Which of the following statement is correct ?
[2013]
2
3
2
5
(b)
(c)
3
2
(d)
5
2
38. When resonance is produced in a series LCR circuit, then
which of the following is not correct ?
[Online April 25, 2013]
(a) Current in the circuit is in phase with the applied
voltage.
(b) Inductive and capacitive reactances are equal.
(c) If R is reduced, the voltage across capacitor will
increase.
(d) Impedance of the circuit is maximum.
39. The plot given below is of the average power delivered to
an LRC circuit versus frequency. The quality factor of the
circuit is :
[Online April 23, 2013]
average power (microwatts)
t
t
34.
S2
1.0
0.5
0.0
3
4
5
6
7
frequency (kHz)
(a) 5.0
(b) 2.0
(c) 2.5
(d) 0.4
–11
40. In a series L-C-R circuit, C = 10 Farad, L = 10–5 Henry
and R = 100 Ohm, when a constant D.C. voltage E is applied
to the circuit, the capacitor acquires a charge 10 –9 C. The
D.C. source is replaced by a sinusoidal voltage source in
P-366
41.
Physics
which the peak voltage E0 is equal to the constant D.C.
voltage E. At resonance the peak value of the charge
acquired by the capacitor will be : [Online April 22, 2013]
(a) 10–15 C (b) 10–6 C (c) 10–10 C (d) 10–8 C
An LCR circuit as shown in the figure is connected to a
voltage source Vac whose frequency can be varied.
V
24 H
2 µF
~
42.
15 W
Vac = V0 sin wt
The frequency, at which the voltage across the resistor is
maximum, is :
[Online April 22, 2013]
(a) 902 Hz (b) 143 Hz (c) 23 Hz
(d) 345 Hz
In the circuit shown here, the voltage across E and C are
respectively 300 V and 400 V. The voltage E of the ac source
is :
[Online April 9, 2013]
L
C
~
E
(a) 400 Volt (b) 500 Volt(c) 100 Volt (d) 700 Volt
43.
(a)
p
LC
4
(a) 1.7 × 105 W
(b) 2.7 × 106 W
(c) 3.3 × 107 W
(d) 1.3 × 104 W
47. Combination of two identical capacitors, a resistor R and
a dc voltage source of voltage 6V is used in an experiment
on a (C-R) circuit. It is found that for a parallel combination
of the capacitor the time in which the voltage of the fully
charged combination reduces to half its original voltage is
10 second. For series combination the time for needed for
reducing the voltage of the fully charged series
combination by half is
[2011 RS]
(a) 10 second
(b) 5 second
(c) 2.5 second
(d) 20 second
48. In the circuit shown below, the key K is closed at t = 0. The
current through the battery is
[2010]
(c) 2 RC ln 7
C
R1
R2
In an LCR circuit shown in the following figure, what will
be the readings of the voltmeter across the resistor and
ammeter if an a.c. source of 220V and 100 Hz is connected
to it as shown?
[Online May 7, 2012]
L
L
3 RC ln 7
(d)
K
V
2 RC ln 2
(b)
(b) 2p LC
(c) LC
(d) p LC
46. A resistor ‘R’ and 2µF capacitor in series is connected
through a switch to 200 V direct supply. Across the
capacitor is a neon bulb that lights up at 120 V. Calculate
the value of R to make the bulb light up 5 s after the switch
has been closed. (log10 2.5 = 0.4)
[2011]
A resistance R and a capacitance C are connected in series
to a battery of negligible internal resistance through a key.
The key is closed at t = 0. If after t sec the voltage across
the capacitance was seven times the voltage across R, the
value of t is
[Online May 12, 2012]
(a) 3 RC ln 2
44.
45. A fully charged capacitor C with initial charge q0 is
connected to a coil of self inductance L at t = 0. The time at
which the energy is stored equally between the electric
and the magnetic fields is:
[2011]
100 W
(a)
VR1R2
R12 + R22
V
at t = 0 and R at t = ¥
2
V ( R1 + R2 )
V
at t = 0 and
at t = ¥
R1 R2
R2
VR1R2
V
(c)
at t = 0 and
at t = ¥
R2
R 2 + R2
(b)
1
V
V
V
300V 300 V
A
VR
220 V, 100 Hz
(a) 800 V, 8 A
(c) 300 V, 3 A
(b) 110 V, 1.1 A
(d) 220V, 2.2 A
(d)
2
V ( R1 + R2 )
V
at t = 0 and
at t = ¥
R1 R2
R2
49. In a series LCR circuit R = 200W and the voltage and the
frequency of the main supply is 220V and 50 Hz
respectively. On taking out the capacitance from the circuit
the current lags behind the voltage by 30°. On taking out
the inductor from the circuit the current leads the voltage
by 30°. The power dissipated in the LCR circuit is [2010]
(a) 305 W (b) 210 W (c) Zero W (d) 242 W
P-367
Alternating Current
50.
E
L
R1
R2
S
An inductor of inductance L = 400 mH and resistors of
resistance R1 = 2W and R2 = 2W are connected to a battery
of emf 12 V as shown in the figure. The internal resistance
of the battery is negligible. The switch S is closed at t = 0.
The potential drop across L as a function of time is [2009]
(a)
51.
52.
(
12 -3t
e V
t
)
-t / 0.2
V
(b) 6 1 - e
(c) 12e–5t V
(d) 6e–5t V
In a series resonant LCR circuit, the voltage across R is
100 volts and R = 1 kW with C = 2mF. The resonant
frequency w is 200 rad/s. At resonance the voltage across
L is
[2006]
–2
(a) 2.5 × 10 V
(b) 40 V
(c) 250 V
(d) 4 × 10–3 V
An inductor (L = 100 mH), a resistor (R = 100 W) and a
battery (E = 100 V) are initially connected in series as
shown in the figure. After a long time the battery is
disconnected after short circuiting the points A and B.
The current in the circuit 1 ms after the short circuit is
[2006]
L
B
E
53.
(a) 1/eA
(b) eA
(c) 0.1 A
(d) 1 A
In an AC generator, a coil with N turns, all of the same area
A and total resistance R, rotates with frequency w in a
magnetic field B. The maximum value of emf generated in
the coil is
[2006]
(a) N.A.B.R.w
(b) N.A.B
(c) N.A.B.R.
(d) N.A.B.w
The phase difference between the alternating current and
p
emf is . Which of the following cannot be the constituent
2
of the circuit?
[2005]
(a) R, L
(b) C alone(c) L alone (d) L, C
A circuit has a resistance of 12 ohm and an impedance of
15 ohm. The power factor of the circuit will be
[2005]
(a) 0.4
(b) 0.8
(c) 0.125
(d) 1.25
(c) 2 mF
(d) 1 mF
58. In an LCR series a.c. circuit, the voltage across each of the
components, L, C and R is 50V. The voltage across the LC
combination will be
[2004]
(b) 50 2 V
(a) 100 V
(c) 50 V
(d) 0 V (zero)
59. In a LCR circuit capacitance is changed from C to 2 C. For
the resonant frequency to remain unchanged, the
inductance should be changed from L to
[2004]
(a) L/2
(b) 2 L
(c) 4 L
(d) L/4
60. The power factor of an AC circuit having resistance (R)
and inductance (L) connected in series and an angular
velocity w is
[2002]
(a) R/ w L
(c)
w 2L2)1/2
(d) R/(R2 – w 2L2)1/2
(b) R/(R2 +
w L/R
61. The inductance between A and D is
(a) 3.66 H
(c) 0.66 H
TOPIC 3
3H
3H
3H
[2002]
D
(b) 9 H
(d) 1 H
Transformers and LC
Oscillations
62. For the given input voltage waveform Vin(t), the output
voltage waveform Vo(t), across the capacitor is correctly
depicted by :
[Sep. 06, 2020 (I)]
1kW
+5V 5m s
0V
10nF
t
m
VO(t)
s
55.
(b) 4 mF
5
54.
(a) 8 mF
A
R
A
56. A coil of inductance 300 mH and resistance 2 W is
connected to a source of voltage 2V. The current reaches
half of its steady state value in
[2005]
(a) 0.1 s
(b) 0.05 s (c) 0.3 s
(d) 0.15 s
57. The self inductance of the motor of an electric fan is 10 H.
In order to impart maximum power at 50 Hz, it should be
connected to a capacitance of
[2005]
0
Vo(t)
3V
(a) 2V
5ms
10ms 15ms
t
P-368
Physics
Vo(t)
(b)
having 4000 turns. The output power is delivered at 230
V by the transformer. If the current in the primary of the
transformer is 5A and its efficiency is 90%, the output
current would be:
[9 Jan. 2019 II]
2V
5ms
10ms 15ms
t
Vo(t)
(c)
2V
5ms
Vo(t)
(d)
63.
t
(b) 45 A
(c) 35 A
(d) 25 A
65. A power transmission line feeds input power at 2300 V to a
step down transformer with its primary windings having
4000 turns, giving the output power at 230 V. If the current in
the primary of the transformer is 5 A, and its efficiency is
90%, the output current would be: [Online April 16, 2018]
(a) 20 A
(b) 40 A (c) 45 A
(d) 25 A
66. In an oscillating LC circuit the maximum charge on the
capacitor is Q. The charge on the capacitor when the
energy is stored equally between the electric and magnetic
field is
[2003]
2V
t
5ms 10ms 15ms
A transformer consisting of 300 turns in the primary and
150 turns in the secondary gives output power of 2.2kW.
If the current in the secondary coil is 10 A, then the input
voltage and current in the primary coil are :
(a) 220 V and 20 A
(c) 440 V and 5 A
64.
10ms 15ms
(a) 50 A
[10 April 2019 I]
(b) 440 V and 20 A
(d) 220 V and 10 A
A power transmission line feeds input power at 2300 V
to a step down transformer with its primary windings
(a)
Q
2
(b)
Q
3
(c)
Q
2
(d) Q
67. The core of any transformer is laminated so as to [2003]
(a) reduce the energy loss due to eddy currents
(b) make it light weight
(c) make it robust and strong
(d) increase the secondary voltage
68. In a transformer, number of turns in the primary coil are 140
and that in the secondary coil are 280. If current in primary
coil is 4 A, then that in the secondary coil is
[2002]
(a) 4 A
(b) 2 A
(c) 6 A
(d) 10 A.
P-369
Alternating Current
1.
(d) As V(t) = 220 sin 100 pt
pö
æ
then I = I0 sin ç wt - ÷
è
2ø
Now, given v(t) = 100 sin (500 t)
220
sin 100 pt
50
i.e., I = Im = sin (100 pt)
For I = Im
so, I(t) =
t1 =
E0
100
=
[Q L = 0.02H ]
wL 500 ´ 0.02
pö
æ
I0 = 10sin ç 500t - ÷
2ø
è
and I0 =
p
1
1
´
=
sec.
2 100 p 200
and for I =
I0 = -10cos ( 500t )
Im
2
I
p
Þ m = I m sin(100 pt 2 ) Þ
= 100 pt 2
2
6
4.
1
s
600
Þ t2 =
1
1
2
1
=
=
s = 3.3 ms
200 600 600 300
(a) For two concentric circular coil,
\ treq =
2.
Mutual Inductance M =
m 0 pN1N 2 a 2
2b
here, N1 = N2 = 1
5.
2
m 0 pa
..... (i)
2b
and given I = I0 cos wt
..... (ii)
Now according to Faraday's second law induced emf
Hence, M =
e = -M
(c) We know that power consumed in a.c. circuit is given
by,
P = Erms.Irms cos f
Here, E = E0 sin wt
pö
æ
I = I0 sin ç wt - ÷
è
2ø
p
This means the phase difference, is f =
2
p
Q cos f = cos = 2
2
p
\ P = Erms .I rms .cos = 0
2
r r
(b) f = B. A ; f = BA cos wt
e=-
wBA
df
= wBA sin wt ; i =
sin wt
dt
R
2
æ wBA ö
Pinst = i 2 R = ç
´ R sin 2 wt
è R ÷ø
dI
dt
ò Pinst ´ dt
Pavg =
From eq. (ii),
-m 0 pa 2 d
(I0 cos w t)
2b dt
e=
m 0 pa 2
I0 sin w t (w )
2b
3.
p
the emf by .
2
If v ( t ) = v0 sin wt
T
(wBA)
=
R
2
ò sin
b
6.
7.
2
wtdt
0
0
pm 0 I0 a 2
. w sin w t
2
b
(b) In a pure inductive circuit current always lags behind
e=
0
ò dt
a
\ Pavg
e=
T
T
(w B pr 2 ) 2
=
8R
T
=
ò dt
1 ( wBA) 2
2
R
0
é
pr 2 ù
êA =
ú
2 ûú
ëê
(a) D.C. ammeter measure average value of current. In AC
current, average value of current in complete cycle is zero.
Hence reading will be zero.
(33)
Here, L = 50 mH = 50 × 10–3 H; I = 1 A, R = 2W
VP - L
dl
- 30 + RI = VQ
dt
Þ VP - VQ = 50 ´ 10 -3 ´ 10 2 + 30 - 1 ´ 2
= 5 + 30 – 2 = 33 V.
P-370
8.
9.
Physics
(a) Quality factor,
Þ-
1 L
1 80 ´ 10 -3
=
Q=
R C 100 2 ´ 10-6
1
200
=
40 ´ 103 =
=2
100
100
(400)
Given: Power P = 400 W, Voltage V = 250 V
æ n - 1ö
Rt
= ln ç
÷
L
n ø
è
Þt =
11.
L æ
n ö
ln ç
R è n - 1÷ø
(d) Here, R = 100, XL = Lw = 0.1803 × 750 × 2p = 850W,
P = Vm × I rms × cos f
XC =
Þ 400 = 250 ´ I rms ´ 0.8 Þ I rms = 2 A
1
1
=
= 21.23W
C w 10 -5 ´ 2p ´ 750
Impedance Z =
2
Using P = I rms
R
R 2 + ( X L - X C )2
= 1002 + (850 - 21.23) 2 = 834.77 ; 835
( I rms )2 × R = P Þ 4 ´ R = 400
Þ R = 100W
Power factor is,
100W
R
cos f =
2
R + X L2
Þ 0.8 =
100
100 2 + X L2
æ 100 ö
Þ 100 2 + X L2 = ç
è 0.8 ÷ø
20V/750 Hz
2
2
æV ö
2
H = irms
Rt = ç rms ÷ RT = (ms )Dt
è |Z |ø
2
æ 100 ö
Þ X L = -1002 + ç
Þ X L = 75W
è 0.8 ÷ø
Þ
20 20
´
´ 100t = (2) ´ 10
835 835
QVrms = 20 V and Dt = 10°C
When power factor is unity,
X C = X L = 75 Þ
ÞC=
1
= 75
wC
1
1
=
F
75 ´ 2p ´ 50 7500p
æ 106
1ö
400
=ç
´ ÷ mF =
mF
3p
è 2500 3p ø
10.
N = 400
(a) Potential energy stored in the inductor
1 2
LI
2
During growth of current,
U=
(
i = I max 1 - e - Rt / L
)
I max
n
= I max (1 - e - Rt / L )
Þ e - Rt / L = 1 -
1
n
=
\ Time, t = 348.61 s.
12. (a) Given,
Reactance of inductance coil, Z = 100W
Frequency of AC signal, v = 1000 Hz
Phase angle, f = 45°
XL
= tan 45° = 1
R
Þ XL = R
tan f =
Reactance, Z = 100 =
Þ 100 =
X L2 + R 2
R2 + R2
Þ 2 R = 100 Þ R = 50 2
\ X L = 50 2
I
U
For U to be max ; i has to be max
n
n
\
10mF
0.1803 H
n -1
n
Þ Lw = 50 2
ÞL=
50 2
2p ´ 1000
25 2
mH
p
= 1.1 × 10–2 H
=
(Q X L = wL)
(Q w = 2pv)
P-371
Alternating Current
13.
(b) We have, i = i0 (1 – e–t/c) =
e
(1 - e - t / c )
R
t
Charge, q = ò idt
XC
tan q1 = R =
2
t
14.
Rt
æ
– ö
(d) I = I0 ç1 - e L ÷ Here R = RL + r = 1W
çè
÷ø
t
æ
– ö
.01
÷
0.8I 0 = I0 ç1 - e
çè
÷ø
Þ 0.8 = 1 - e -100t
æ 1ö
Þ e -100t = 0.2 = ç ÷
è 5ø
Þ 100t = ln5 Þ t =
15.
16.
æ di ö
2
(c) i R = çè t ÷ø i
dt
di i
Þ =
dt t
L 20
é
ù
= 2ú
Þ t = t ln2 = 2ln2 ê ast = =
R 10
ë
û
(d) w = 100 rad/s
We know that
tan f =
XC
1
=
wCR
R
or tan45º =
17.
1
ln 5 = 0.016 s
100
i
q1 is close to 90°
For L-R circuit
0
e
EL
-t /t
)dt = E t = E ´ ( L / R ) =
= R ò (1 - e
Re
R
e
2.7 R 2
0
103
3
XL = wL = 100 ´
q1
3
= 10 3
10
R1 = 10
XL
q2
tan q2 = R
1
or wCR = 1
LHS: wCR = 10 × 10 × 10–6 × 103 = 1
(d) Long time after switch is closed, the inductor will be
idle so, the equivalent diagram will be as below
i
q2 =60°
So, phase difference comes out 90° + 60° = 150°
If R2 is 20 KW
then phase difference comes out to be 60 + 30 = 90°.
Therefore Ans. is Bonus
19. (b)
I
t
t0
The current will grow for the time t = 0 to
t = t0 and after that decay of current takes place.
20. (c) Given: R = 60W, f = 50 Hz, w = 2 pf = 100 p and v = 24v
C = 120 mf = 120 × 10–6f
1
1
=
= 26.52W
wC 100p ´ 120 ´ 10 -6
xL = wL = 100 p × 20 × 10–3 = 2 pW
xC – xL= 20.24 » 20
R = 60W
f
Z
e
R
R
z = R 2 + ( xC – x L )
2
z =20 10W
I=
18.
e
2e 2 ´ 15
=
=
= 6A
R
5
æ R ´Rö
çè
÷
R +Rø
(Bonus)
Capacitive reactance,
Xc =
4
1
2 ´ 104
= –6
=
wC
3
10 ´ 3 ´ 100
v
tan q2 = 3 Þ q2 = tan –1( 3)
xC =
1
wCR
v
cos f=
R
60
3
=
=
z 20 10
10
v v2
= cos f = 8.64 watt
z
z
Energy dissipated (Q) in time t = 60s is
Q = P.t = 8.64 × 60 = 5.17 × 102J
21. (a) Given, Inductance, L= 40 mH
Capacitance, C = 100 mF
Impedance, Z = XC – XL
Pavg = VI cos f, I =
P-372
Physics
1
1
æ
ö
– wL
and X L = wL÷
çèQ X c =
ø
wC
wC
1
– 314 ´ 40 ´ 10 –3
=
–6
314 ´ 100 ´ 10
= 19.28W
V0
sin(wt + p / 2)
Current, i =
Z
10
Þ i=
cos wt = 0.52 cos (314 t)
19.28
(a) For series connection of a resistor and inductor, time
Þ Z=
22.
variation of current is I = I0 (1– e – t /Tc )
d2
dq q
+ = 0 ...(ii)
dt C
dt
Comparing equations (i) & (ii)
1
L « m, C « , R « b
k
24. (a) The current (I) in LR series circuit is given by
tR
– ö
Væ
I = ç1 – e L ÷
R çè
÷ø
At t = ¥,
–¥ ö
æ
20
I¥ = ç I – e L/ R ÷ = 4
...(i)
5 çè
÷ø
At t = 40s,
L
2
+R
–40 ´ 5 ö
æ
–20,000
)
çè1 – e
÷ = 4(1 – e
10 ´ 10 –3 ø
Dividing (i) by (ii) we get
Tc
ò idt
0
Þ ò dq =
ò R (1 – e
E
– t / tc
) dt
tc
Îé
t + tC e – t / tc ù
ë
û
0
R
tC
Îé
ù
Þ q = êtC +
– tC ú
Rë
e
û
Þq=
Þq=
\q =
23.
Î L
R Re
ÎL
R 2e
(d) In damped harmonic oscillation,
md 2 x
dt 2
Þ
I¥
1
=
,
–20,000
I 40 1– e
25. (b) As we know, average power Pavg = Vrms Irms cosq
æ V öæ I ö
æ 100 ö æ 20 ö
= ç 0 ÷ ç 0 ÷ cos q = ç
֍
÷ cos 45° (Q q = 45°)
è 2 øè 2 ø
è 2 øè 2 ø
1000
watt
Pavg =
2
Wattless current I = Irms sin q
I
20
= 0 sin q =
sin 45° = 10A
2
2
w0
w L
= 0
26. (a) Quality factor Q =
2Dw
R
27. (b) Given,
V0 = 283 volt, w = 320, R = 5 W, L = 25 mH, C = 1000 µF
xL = wL = 320 × 25 × 10–3 = 8 W
1
1
xC =
=
= 3.1 W
wC 320 ´ 1000 ´ 10 -6
Total impedance of the circuit :
Þ
L
Here, TC =
R
q=
= – kx – bv
md 2 x
dt
2
+b
...(ii)
dx
+ kx = 0
dt
....(i)
28.
Z = R 2 + (X L - X C ) 2 = 25 + (4.9) 2 = 7 W
Phase difference between the voltage and current
X - XC
tan f = L
R
4.9
tan f =
» 1 Þ f = 45°
5
(b) Here
i=
e
2
R +
10 =
–q
Ldi
– iR –
=0
In LCR circuit,
C
dt
X L2
=
220
e
2
R +w L
64 + 4p2 (50) 2 L
V 80
=
= 8]
I 10
On solving we get
L = 0.065 H
[Q R =
2 2
=
e
2
R + 4p 2v 2 L2
P-373
Alternating Current
29.
(d)
30.
(b) I (0) =
I(¥) = 0
15 ´ 100
0.15 ´ 103
I(t) = [I (0) – I (¥)]
I(t) = 0.1
–t
e L/ R
34. (c) Applying Kirchhoff's law of voltage in closed loop
V
–VR –VC = 0 Þ R = -1
VC
= 0.1A
–t
L
e /R
A
R
+ i (¥)
R
= 0.1 e L
L VL
B
0.15´1000
e 0.03
31.
VR
C
I(t) = 0.1
= 0.67mA
(c) From KVL at any time t
R
+
L
di
dt –
i
35. (c) Given, VL : VC : VR = 1 : 2 : 3
V = 100 V
VR = ?
As we know,
V = VR2 + ( VL - VC )
+ –
q c
q
di
- iR - L = 0
c
dt
i=-
dq
q dq
Ld 2 q
Þ + R+
=0
dt
c dt
dt 2
d 2q
R dq q
+
+
=0
2
L dt Lc
dt
From damped harmonic oscillator, the amplitude is given
dt
by A = Ao e 2m
Double differential equation
d 2x
b dx k
+
+ x =0
2
m dt m
dt
Qmax = Q oe
32.
Rt
2L
Solving we get, VR ; 90V
36. (c) Charge on he capacitor at any time t is given
by q = CV (1– et/t)
at t = 2t
q = CV (1 – e–2)
37. (d) Power factor (old)
=
R
R 2 + XL2
=
R
R
=
R 2 + (2R)2
5R
Power factor(new)
=
R
2
R + (X L - XC )
2
R
=
2
R + (2R - R)
Rt
2
Þ Q max = Qo e L
2
1– w2 LC
w2 L
Hence option (c) is the correct answer.
(c) For capacitor circuit, i = i0e–t/RC
Rt
æ
– ö
ç
For inductor circuit, i = i0 1– e L ÷
ç
÷
è
ø
Hence graph (c) correctly depicts i versus t graph.
2
=
R
2R = 5
R
2
5R
38. (d) Impedance (Z) of the series LCR circuit is
New power factor
\ Old power factor =
Hence damping will be faster for lesser self inductance.
(c) Power factor
R
cos f =
=1
2
é
ù
1
R 2 + ê wL –
ú
w(C + C ') û
ë
On solving we get,
1
wL =
w(C + C ')
C'=
33.
-
2
Z = R 2 + (X L - X C ) 2
At resonance, X L = XC
Therefore, Zminimum = R
39. (b)
Pmax
1.0
P
0.5
0.0
Pmax
=P
2
4 5 6
w1 w0 w2
Quality factor of the circuit
w0
5
= 2.0
= w -w =
2.5
2
1
3
7
R
2R
P-374
40.
41.
42.
43.
44.
Physics
(d)
(c) Frequency f =
1
2p LC
=
V = V0 e
1
2 ´ 3.4 24 ´ 2 ´ 10 -6
; 23Hz
(c) Voltage E of the ac source
E = VC – VL = 400 V – 300 V = 100 V
(a) t = 3 RC ln 2
(d) In case of series RLC circuit,
Equation of voltage is given by
48.
2
\ VR = V 2 = 220V
45.
V 220
=
= 2.2A
R 100
(a) Energy stored in magnetic field =
1 2
Li
2
49.
1 q2
Energy stored in electric field =
2 C
Energy will be equal when
\
1 2 1 q2
Li =
2
2 C
tan wt = 1
q = q0 cos wt
1
( q cos wt )2
Þ L(wq0 sin wt)2 = 0
2
2C
p
1
Þw=
Þ wt =
4
LC
Þ t=
46.
....(2)
tan f =
1
w CR
Þ
(b) We have, V = V0(1 – e–t/RC)
æ 1
ö
Z = R2 + ç
- wL ÷
w
C
è
ø
200 – 120
80
=
200
200
t = loge(2.5)
e–t/r =
Þ t = RC in (2.5)
[Q r = RC]
=
= Vrms ×
=
...(1)
In series grouping, equivalent capacitance =
æ 200 200 ö
(200) 2 + ç
÷ = 200 W
3ø
è 3
Power dissipated in the circuit = VrmsIrms cos f
)
V
V = V0
= 0
2
In second case :
2
2
Þ R = 2.71 × 106 W
(c) Time constant for parallel combination
= 2RC
RC
Time constant for series combination =
2
In first case :
t
–
t
V0
–
CR
= V0 – V0e
V = V0
Þ
1 – e CR
2
t
- 1
RC
2
e
V0
2
1
200
1
=
= R tan f = 200 ´
3
3
wC
Impedance of the circuit,
p
LC
4
(
=
t 10
Þ t2 = 1 = = 2.5 sec.
4 4
(c) At t = 0, no current will flow through L and R1 as
inductor will offer infinite resistance.
V
\ Current through battery, i =
R2
At t = ¥, inductor behave as conducting wire
RR
Effective resistance, Reff = 1 2
R1 + R2
V
V ( R1 + R2 )
\ Current through battery = R =
R1R2
eff
(d) When only the capacitance is removed phase
difference between current and voltage is
XL
tan f =
R
wL
Þ tan f =
R
1
200
=
Þ wL = R tan f = 200 ´
3
3
When only inductor is removed, phase difference between
current and voltage is
\
Þ 120 = 200(1 – e–t/RC)
47.
t2
( RC / 2)
From (1) and (2)
t1
t2
=
2 RC ( RC / 2 )
V 2 = VR2 + (VL - VC )
Here, V = 220 V; VL = VC = 300 V
Current i =
-
C
2
Vrms R
×
Z Z
(220)2 ´ 200
(200)2
R ö V 2rms R
æ
çQ cos f = ÷ =
Zø
è
Z2
=
220 ´ 220
= 242 W
200
50. (c) Growth in current in branch containing L and R2 when
switch is closed is given by
E
[1 - e - R2t / L ]
i=
R2
Þ
di E R2 - R2t / L
E =
× ×e
= e
dt R2 L
L
R2t
L
P-375
Alternating Current
Hence, potential drop across L
Ldi æ E - R2t / L ö
=ç e
VL =
÷L
dt è L
ø
= Ee
51.
- R2t / L
= 12e
-
2t
400 ´10 -3
(c) Across resistor, I =
At resonance,
= 12e–5tV
V
100
=
= 0.1 A
R 1000
1
1
=
= 2500
wC 200 ´ 2 ´ 10-6
Voltage across L is
I X L = 0.1 ´ 2500 = 250 V
(a) Initially, when steady state is achieved,
X L = XC =
52.
E
i=
R
Let E is short circuited at t = 0. Then
At t = 0
E 100
= 1A
=
R 100
Let during decay of current at any time the current flowing
di
is - L - iR = 0
dt
di
R
Þ = - dt
i
L
Maximum current, i0 =
i
t
i0
0
i
R
=- t
i0
L
Þ i = i0
54.
55.
56.
1
2
(2 pn ) L
i = i0 1 - e
-
4p ´ 50 ´ 50 ´ 10
\ Net voltage difference across
LC = 50 – 50 = 0
1
2p LC
For resonant frequency to remain same
LC = constant
\ LC = L' C'
Þ LC = L¢ × 2C
59. (a) Resonant frequency, Fr =
L
2
60. (b) Resistance of the inductor, XL = wL
The impedance triangle for resistance (R) and inductor (L)
connected in series is shown in the figure.
R
-100´10 -3
)
Rt
2
58. (d) In a series LCR circuit voltage across the inductor
and capacitor are in opposite phase
2
Rt
L
1
=
\ C = 0.1 ´ 10-5 F = 1mF
2
-3
1
E - t
Þ i = e L = 1 ´ e 100´10 =
R
e
ur ur
df
d ( N B. A)
=(d) e = dt
dt
d
= - N ( BA cos wt ) = NBAw sin wt
dt
Þ e max = NBAw
(a) Phase difference for R–L circuit lies between
æ pö
ç 0, ÷ but 0 or p/2
è 2ø
(b) Given, Resistance of circuit, R = 12 W
Imedance of circuit, Z = 15 W
R
12 4
=
= = 0.8
Power factor = cos f =
Z
15 5
(a) Current in inductor circuit is given by,
(
C=
R
- t
e L
R
53.
L
300 ´ 10 -3
´ 0.69
log 2 =
R
2
Þ t = 0.1 sec.
57. (d) For maximum power, XL = XC, which yields
Þ t=
Þ L' =
di
R
Þ ò = ò - dt
i
L
Þ log e
Taking log on both the sides,
Rt
= log1 - log 2
L
Rt
i0
1
= i0 (1 - e L ) Þ e L =
2
2
+w
2
L
XL= w L
f
R
Net impedance of circuit Z =
Power factor, cos f =
Þ cos f =
X L2 + R 2
R
Z
R
2
R + w 2 L2
61. (d) All three inductors are connected in parallel. The
equivalent inductance Lp is given by
1
1
1
1
1 1 1 3
= +
+
= + + = =1
L p L1 L2 L3 3 3 3 3
\ Lp = 1
62. (a) When first pulse is applied, the potential across
capacitor
1 ö
æ
V0 (t ) = Vin ç1 - e RC ÷
è
ø
At t = 5ms = 5 ´ 10-6 s
P-376
Physics
Þ Is = 0.9 ´ 50 = 45A
10m F
Vin
V0(t)
Efficiency n = 0.9 =
5´10-6
æ
ö
3
-9
10
V0 (t ) = 5 ç 1 - e ´10 ´10 ÷ = 5(1 - e -0.5 ) = 2V
ç
÷
è
ø
When no pulse is applied, capacitor will discharge.
Now, Vin = 0 means discharging.
1
V0 (t ) = 2e RC = 2e -0.5 = 1.21 V
Now for next 5 ms
1
V0 (t ) = 5 - 3.79e RC
After 5 ms again, V0 (t ) = 2.79 Volt » 3 V
Hence, graph (a) correctly depicts.
63.
(c) Power output (V2I2) = 2.2 kW
2.2kW
= 220 volts
(10A )
\ V2 =
\ Input voltage for step-down transformer
V1 N1
=
=2
V2 N 2
Vinput = 2 × Voutput = 2 × 220
= 440 V
I1 N 2
Also I = N
2
1
\
64.
I1 =
1
´ 10 = 5A
2
P
VI
(b) Efficiency, h= out = s s
Pin VpIp
Þ 0.9 =
Output current = 45A
65. (c) Given : VP = 2300 V, Vs = 230 V, IP = 5 A, n = 90% = 0.9
230 ´ Is
2300 ´ 5
Ps
Þ Ps = 0.9 Pp
PP
VsIs = 0.9 ×VPIP (Q P = VI)
Is =
0.9 ´ 2300 ´ 5
= 45A
230
66. (c) When the capacitor is completely charged, the total
energy in the LC circuit is with the capacitor and that
energy is given by
Umax =
1 Q2
2 C
When half energy is with the capacitor in the form of electric
field between the plates of the capacitor we get
U max 1 q¢2
=
2
2 C
Here q' is the charge on the plate of capacitor when energy
is shared equally.
\
Q
1 1 Q 2 1 q¢2
Þ q¢ =
´
=
2 2 C
2 C
2
67. (a) Laminated core provide less area of cross-section for
the current to flow. Because of this, resistance of the core
increases and current decreases there by decreasing the
energy loss due to eddy current.
68. (b) Number of turns in primary
Np = 140
Number of turns in secondary Ns = 280, Ip = 4A, Is = ?
Using transformation ratio for a transformer
I s 140
=
4 280
Þ Is = 2 A
Þ
Is N p
=
Ip
Ns
22
Electromagnetic
Waves
Electromagnetic Waves,
TOPIC 1 Conduction
and Displacement Current
1.
For a plane electromagnetic wave, the magnetic field at a
point x and time t is
4.
®
B( x, t ) = [1.2 ´ 10-7 sin(0.5 ´ 103®x + 1.5 ´ 1011t )k$ ]T ®
The instantaneous electric field E corresponding to B
is: (speed of light c = 3 × 108 ms–1) [Sep. 06, 2020 (II)]
where c = 3 × 108 ms–1 is the speed of light.
The corresponding electric field is : [Sep. 03, 2020 (I)]
ur
(a) E = 9sin[200p ( y + ct )]kˆ V/m
ur
(b) E = -10-6 sin[200p ( y + ct )]kˆ V/m
ur
(c) E = 3 ´ 10-8 sin[200p ( y + ct )]kˆ V/m
ur
(d) E = -9sin[200p ( y + ct )]kˆ V/m
®
V
3
11
(a) E ( x, t ) = [ -36 sin(0.5 ´ 10 x + 1.5 ´ 10 t ) $j ]
m
®
V
(b) E( x, t ) = [36 sin(1 ´ 103 x + 0.5 ´ 1011 t ) $j ]
m
®
V
(c) E ( x, t ) = [36 sin(0.5 ´ 103 x + 1.5 ´ 1011 t ) k$ ]
m
5.
®
2.
V
(d) E ( x, t ) = [36 sin(1 ´ 103 x + 1.5 ´ 1011 t )$i]
m
An electron is constrained to move along the y-axis with a
speed of 0.1 c (c is the speed of light) in the presence
of
®
electromagnetic wave, whose electric field is E = 30j$
sin(1.5 × 107t – 5 × 10–2x) V/m. The maximum magnetic force
experienced by the electron will be :
(given c = 3 × 108 ms–1 & electron charge = 1.6 × 10–19C)
[Sep. 05, 2020 (I)]
(a)
3.
3.2 × 10–18 N
(b)
E0
(- xˆ + yˆ ) sin(kz - wt )
c
E
(b) 0 ( xˆ + yˆ )sin(kz - wt )
c
(a)
The electric field of a plane electromagnetic wave
propagating along the x direction in vacuum is
r
r
E = E0 ˆj cos(wt - kx ) . The magnetic field B, at the
moment t = 0 is :
[Sep. 03, 2020 (II)]
r
E0
(a) B =
cos(kx)kˆ
m0 e 0
r
(b) B = E0 m0 e0 cos(kx) ˆj
r
(c) B = E0 m0 e0 cos(kx)kˆ
r
(d) B =
2.4 × 10–18 N
(c) 4.8 × 10–19 N
(d) 1.6 × 10–19 N
The electric field of a plane electromagnetic wave is given
r
by E = E0 ( xˆ + yˆ ) sin(kz - wt )
Its magnetic field will be given by : [Sep. 04, 2020 (II)]
E0
( xˆ - yˆ ) sin(kz - wt )
c
E
(d) 0 ( xˆ - yˆ ) cos(kz - wt )
c
The magnetic field of a plane electromagnetic wave is
ur
ˆ
B = 3 ´ 10-8 sin[200p ( y + ct )]iT
(c)
6.
E0
m0 e 0
cos(kx) ˆj
A plane electromagnetic wave, has frequency of 2.0 × 1010
Hz and its energy density is 1.02 × 10–8 J/m3 in vacuum.
The amplitude of the magnetic field of the wave is close to
Nm 2
1
= 9 ´109 2 and speed of light = 3 × 108 ms–1) :
4pe0
C
[Sep. 02, 2020 (I)]
(a) 150 nT
(b) 160 nT
(c) 180 nT
(d) 190 nT
(
P-378
7.
In a plane electromagnetic wave, the directions of electric
field and magnetic field are represented by k̂ and 2iˆ - 2 ˆj,
respectively. What is the unit vector along direction of
propagation of the wave.
[Sep. 02, 2020 (II)]
(a)
8.
Physics
1 ˆ ˆ
(i + j )
2
(b)
1
2
( ˆj + kˆ)
1 ˆ ˆ
1 ˆ
(2i + j )
(c)
(i + 2 ˆj )
(d)
5
5
The electric fields of two plane electromagnetic plane
waves in vacuum are given by
ur
ur
E1 = E0 ˆj cos(wt - kx ) and E2 = E 0 kˆ cos(wt - ky ) .
At t = 0, a particle of charge q is at origin with a velocity
r
v = 0.8 cjˆ (c is the speed of light in vacuum). The
instantaneous force experienced by the particle is:
[9 Jan 2020, I]
ˆ
(a) E0 q(0.8iˆ - ˆj + 0.4k ) (b) E0 q(0.4iˆ - 3 ˆj + 0.8kˆ)
9.
(c) E0 q(- 0.8iˆ + ˆj + kˆ) (d) E0 q( 0.8iˆ + ˆj + 0.2kˆ)
A plane electromagnetic wave is propagating along the
iˆ + ˆj
, with its polarization along the direction
direction
2
kˆ. The correct form of the magnetic field of the wave
would be (here B0 is an appropriate constant):
[9 Jan 2020, II]
æ
iˆ - ˆj
iˆ + ˆj ö
cos ç wt - k
(a) B0
è
2
2 ÷ø
(b) B0
ˆj - iˆ
æ
iˆ + ˆj ö
cos ç wt + k
è
2
2 ÷ø
æ
iˆ + ˆj ö
ˆ
(c) B0 k cos ç wt - k
è
2 ÷ø
æ
iˆ + ˆj ö
cos ç wt - k
è
2
2 ÷ø
A plane electromagnetic wave of frequency 25 GHz is
propagating in vacuum along the z-direction. At a
particular point in space and time, the magnetic field is
r
given by B = 5 ´ 10-8 ˆj T . The corresponding electric
r
field E is (speed of light c = 3 ´ l08 ms–l)
[8 Jan 2020, II]
(a) 1.66 ´ 10–16 iˆ V/m
(b) – 1.66 ´ 10–16 iˆ V/m
(d) B0
10.
11.
iˆ + ˆj
(c) –15 iˆ V/m
(d) 15 iˆ V/m
If the magnetic field in a plane electromagnetic wave is
ur
given by B = 3 ´ 10–8 sin (l.6 ´ 103x + 48 ´ 1010t) ĵ T, then
what will be expression for electric field?
[7 Jan 2020, I]
ur
(a) E = (60 sin (1.6 ´ l03x + 48 ´ l010t) k̂ v/m)
ur
(b) E = (9 sin (1.6 ´ l03x + 48 ´ l010t) k̂ v/m)
ur
(c) E = (3 ´ l0–8 sin (l.6 ´ l03x + 48 ´ l010t) k̂ v/m)
ur
(d) E = (3 ´ l0–8sin (l.6 ´ l03x + 48 ´ l010t) k̂ v/m)
12. The electric field of a plane electromagnetic wave is
given by
r E iˆ + ˆj cos(kz + wt )
E = 0 2
At t = 0, a positively charged particle is at the point
pö
æ
(x, y, z) = çè 0, 0, ÷ø . If its instantaneous velocity at (t = 0)
k
is v0 kˆ , the force acting on it due to the wave is:
[7 Jan 2020, II]
iˆ + ˆj
(b) zero
(a) parallel to
2
iˆ + ˆj
(c) antiparallel to
(d) parallel to k̂
2
13. An electromagnetic wave is represented by the electric
ur
field E = E n$ sin[w t + (6 y - 8z)] . Taking unit vectors in
0
x, y and z directions to be $i , $j , k$ , the direction of
propogation $s is :
3$i - 4 $j
(a) s$ =
5
[12 April 2019, I]
-4k$ + 3 $j
(b) s$ =
5
3 $j - 3k$
(d) s$ =
5
æ -3$j + 4k$ ö
(c) s$ = ç
÷÷
ç
5
è
ø
14. A plane electromagnetic wave having a frequency
v = 23.9 GHz propagates along the positive z-direction in
free space. The peak value of the Electric Field is 60 V/m.
Which among the following is the acceptable magnetic
field component in the electromagnetic wave ?
[12 April 2019, II]
ur
7
3
(a) B = 2 ´ 10 sin(0.5 ´ 10 z + 1.5 ´ 1011 t )$i
ur
(b) B = 2 ´ 10-7 sin(0.5 ´ 103 z - 1.5 ´ 1011 t )$i
ur
(c) B = 60sin(0.5 ´ 103 x + 1.5 ´ 1011 t )k$
ur
-7
2
11 $
(d) B = 2 ´ 10 sin(1.5 ´ 10 x + 0.5 ´ 10 t ) j
15. The electric field of a plane electromagnetic wave is given
ur
by E = E $i cos(kz) cos(w t)
0
The corresponding magnetic field is then given by :
[10 April 2019, I]
ur E 0
$j sin (kz) sin (w t)
(a) B =
C
ur E 0
$j sin(kz) cos ( w t)
(b) B =
C
ur E 0
$j cos (kz) sin (w t)
(c) B =
C
ur E 0
k$ sin (kz) cos( w t)
(d) B =
C
P-379
Electromagnetic Waves
16.
17.
Light is incident normally on a completely absorbing
surface with an energy flux of 25 Wcm–2. If the surface has
an area of 25 cm2, the momentum transferred to the surface
in 40 min time duration will be:
[10 April 2019, II]
(a) 6.3×10–4 Ns
(b) 1.4×10–6 Ns
(c) 5.0×10–3 Ns
(d) 3.5×10–6 Ns
The magnetic field of a plane electromagnetic wave is
given by:
r
B = B $i [ cos ( kz – w t ) ] + B $j cos ( kz + wt )
0
18.
1
Where B0 = 3 × 10–5 T and B1 = 2 × 10–6 T.
The rms value of the force experienced by a stationary
charge Q = 10–4 C at z = 0 is closest to: [9 April 2019 I]
(a) 0.6 N
(b) 0.1 N
(c) 0.9 N
(d) 3 × 10–2 N
A plane electromagnetic wave of frequency 50 MHz travels
in free space along the positive x-direction. At a particular
r
point in space and time, E = 6.3 ˆj V / m. The
r
corresponding magnetic field B , at that point will be:
[9 April 2019 I]
(a) 18.9 × 10–8 k̂T
(b) 2.1 × 10–8 k̂T
19.
20.
21.
(c) 6.3 × 10–8 k̂T
(d) 18.9 × 108 k̂T
50 W/m2 energy density of sunlight is normally incident
on the surface of a solar panel. Some part of incident
energy (25%) is reflected from the surface and the rest is
absorbed. The force exerted on 1m2 surface area will be
close to (c = 3 × 108 m/s):
[9 April 2019, II]
(a) 15 × 10–8 N
(b) 20 × 10–8 N
(c) 10 × 10–8 N
(d) 35 × 10–8 N
A plane electromagnetic wave travels in free space along
the x-direction. The electric field component of the wave
at a particular point of space and time is E = 6 Vm –1 along
y-direction. Its corresponding magnetic field component,
B would be:
[8 April 2019 I]
(a) 2 × 10–8 T along z-direction
(b) 6 × 10–8 T along x-direction
(c) 6 × 10–8 T along z-direction
(d) 2 × 10–8 T along y-direction
The magnetic field of an electromagnetic wave is given by:
ur
Wb
B = 1.6 ´ 10 –6 cos 2 ´ 10 7 z + 6 ´ 1015 t 2iˆ + ˆj 2
m
The associated electric field will be : [8 April 2019, II]
(
)(
)
V
ur
(a) E = 4.8 × 102 cos(2 × 107 z – 6 × 1015 t) 2iˆ + ˆj
m
ur
V
(b) E = 4.8 ´ 10 2 cos(2 ´ 10 7 z - 6 ´ 1015 t)( -2 $j + $i)
m
(
)
V
ur
(c) E = 4.8 × 102 cos(2 × 107 z + 6 × 1015 t) –iˆ + 2 ˆj
m
(
V
ur
(d) E = 4.8 × 102 cos(2 × 107 z + 6 × 1015 t) iˆ – 2 ˆj
m
The mean intensity of radiation on the surface of the Sun
is about 108 W/m2. The rms value of the corresponding
magnetic field is closest to :
[12 Jan 2019, II]
(a) 1 T
(b)102 T
(c) 10–2 T (d) 10–4 T
(
22.
)
)
23. An electromagnetic wave of intensity 50 Wm–2 enters in a
medium of refractive index ‘n’ without any loss. The ratio
of the magnitudes of electric fields, and the ratio of the
magnitudes of magnetic fields of the wave before and after
entering into the medium are respectively, given by :
[11 Jan 2019, I]
æ 1 1 ö
,
n, n
(a) ç
(b)
÷
è n nø
1 ö
æ
æ 1
ö
, n÷
(c) ç n ,
(d) ç
÷
nø
è
è n
ø
24. A 27 mW laser beam has a cross-sectional area of 10 mm 2.
The magnitude of the maximum electric field in this
electromagnetic wave is given by :
[Given permittivity of space Î0 = 9 × 10 –12 SI units, Speed
of light c = 3 × 108 m/s]
[11 Jan 2019, II]
(a) 2 kV/m
(c) 0.7 kV/m
(b) 1 kV/m
(d) 1.4 kV/m
25. If the magnetic field of a plane electromagnetic wave is
given by (The speed of light = 3 × 108 m/s)
(
)
é
xö ù
15 æ
B = 100 × 10–6 sin ê 2 p ´ 2 ´ 10 çè t - ÷ø ú
c û
ë
then the maximum electric field associated with it is:
[10 Jan. 2019 I]
4
4
(a) 6 × 10 N/C
(b) 3 × 10 N/C
(c) 4 × 104 N/C
(d) 4.5 104 N/C
26. The electric field of a plane polarized electromagnetic
wave in free space at time t = 0 is given by an expression
ur
E ( x, y ) = 10 ˆj cos [(6 x + 8 z)]
ur
The magnetic field B ( x, z, t ) is given by: (c is the
velocity of light)
[10 Jan 2019, II]
1 ˆ
ˆ cos [ (6 x - 8 z + 10ct ) ]
(a)
(6k + 8i)
c
1 ˆ
ˆ cos [ (6 x + 8 z - 10ct ) ]
(b)
(6k - 8i)
c
1 ˆ
ˆ cos [ (6 x + 8 z - 10ct ) ]
(c)
(6k + 8i)
c
1 ˆ
ˆ cos [(6 x + 8 z + 10ct ) ]
(d)
(6k - 8i)
c
27. An EM wave from air enters a medium. The electric fields
r
é
æ z öù
are E1 = E01 xˆ cos ê 2pv ç - t ÷ú in air and
è c øû
ë
r
E2 = E02 xˆ cos [ k (2 z - ct )] in medium, where the wave
number k and frequency v refer to their values in air. The
medium is nonmagnetic. If Îr1 and Îr refer to relative
2
permittivities of air and medium respectively, which of the
following options is correct?
[9 Jan 2019, I]
Î
Îr
r1
1 =4
=2
(a)
(b)
Îr
Îr
2
(c)
Îr
Îr
1
2
2
=
1
4
(d)
Îr
Îr
1
2
=
1
2
P-380
28.
Physics
The energy associated with electric field is (UE) and with
magnetic fields is (UB) for an electromagnetic wave in
free space. Then :
[9 Jan 2019, II]
UB
(b) UE > UB
2
(c) UE < UB
(d) UE = UB
29. A plane electromagnetic wave of wavelength l has an
intensity I. It is propagating along the positive Y–
direction. The allowed expressions for the electric and
magnetic fields are given by
[Online April 16, 2018]
ur
I
é 2p
ù r 1
cos ê (y - ct) ú ˆi; B = Ekˆ
(a) E =
e C
c
ël
û
(a) U E =
0
30.
ur
(b) E =
I
1
é 2p
ùˆ r
= - Eiˆ
cos ê (y - ct) ú k;B
e0 C
c
ël
û
ur
(c) E =
2I
1
é 2p
ùˆ r
cos ê (y - ct) ú k;B
= + Eiˆ
e0 C
c
ël
û
ur
(d) E =
2I
é 2p
ùˆ r 1 ˆ
cos ê (y + ct) ú k;
B = Ei
e0 C
c
ël
û
A monochromatic beam of light has a frequency
3
v=
´ 1012 Hz and is propagating along the direction
2p
iˆ + ˆj
. It is polarized along the k̂ direction. The acceptable
2
form for the magnetic field is: [Online April 15, 2018]
é 4 æ iˆ - ˆj ö r
E æ iˆ - ˆj ö
12 ù
(a) k 0 ç
÷ cos ê10 ç
÷ .r - (3 ´ 10 )t ú
C è 2 ø
è 2 ø
ëê
ûú
31.
32.
(b)
é 4 æ iˆ + ˆj ö r
E0 æ iˆ - ˆj ö
12 ù
ç
÷ cos ê10 ç
÷ .r - (3 ´10 )t ú
C è 2 ø
è 2 ø
ëê
ûú
(c)
é
æ iˆ + ˆj ö r
E0 ˆ
12 ù
k cos ê104 ç
÷ .r + (3 ´10 )t ú
C
è 2 ø
ëê
ûú
é
æ iˆ + ˆj ö r
E (iˆ + ˆj + kˆ)
12 ù
cos ê104 ç
(d) 0
÷ .r + (3 ´ 10 )t ú
C
3
è 2 ø
ëê
ûú
The electric field component of a monochromatic
radiation
ur is given by
E = 2 E0 $i cosur kz cos wt
Its magnetic field B is then given by :
[Online April 9, 2017]
2Eo $
2Eo $
j sin kz cos wt (b) (a)
j sin kz sin wt
c
c
2Eo $
2Eo $
j sin kz sin wt (d)
(c)
j cos kz cos wt
c
c
Magnetic field in a plane electromagnetic wave is given by
r
B = B sin(kx + wt)ˆjT
0
Expression for corresponding electric field will be :
Where c is speed of light.
[Online April 8, 2017]
33.
r
ˆ /m
(a) E = B0 csin(kx + wt)kV
r B0
ˆ /m
sin(kx + wt)kV
(b) E =
c
r
ˆ /m
(c) E = - B0 csin(kx + wt)kV
r
ˆ
(d) E = B0 csin(kx - wt)kV / m
Consider an electromagnetic wave propagating in vacuum.
Choose the correct statement : [Online April 10, 2016]
(a) For an electromagnetic wave propagating in +y
r
1
direction the electric field is E =
E yz (x, t)zˆ and
2
r
1
the magnetic field is B =
Bz (x, t)yˆ
2
(b) For an electromagnetic wave propagating in +y
r
1
E yz (x, t)yˆ and
direction the electric field is E =
2
r
1
the magnetic field is B =
B yz (x, t)zˆ
2
(c) For an electromagnetic wave propagating in
r
1
+x direction the electric field is E =
E yz (y, z, t)
2
( yˆ + zˆ )
and the magnetic field is
r
1
B=
Byz (y, z, t) ( yˆ + zˆ )
2
(d) For an electromagnetic wave propagating in +x
r
1
direction the electric field is E =
E yz (x, t) ( yˆ - zˆ )
2
r
and the magnetic field is B = 1 B yz (x, t) ( yˆ + zˆ )
2
34. For plane electromagnetic waves propagating in the
z-direction, which one of the following combination gives
ur
ur
the correct possible direction for E and B field
respectively?
[Online April 11, 2015]
(a) (2$i + 3$j) and ($i + 2$j) (b) (-2$i - 3$j) and (3$i - 2 $j)
(c) (3$i + 4 $j) and (4$i - 3$j) (d) ($i + 2$j) and (2$i - $j)
35. An electromagnetic wave travelling in the x-direction has
frequency of 2 × 1014 Hz and electric field amplitude of 27
Vm–1. From the options given below, which one describes
the magnetic field for this wave ? [Online April 10, 2015]
r
-8
(a) B ( x, t ) = 3 ´ 10 T ˆj
(
)
sin é 2p(1.5 × 10 x – 2 × 1014 t) ù
û
ur ë
-8
(b) B ( x,t ) = 9×10 T iˆ
sin é 2p(1.5 × 10 –8 x – 2 × 1014 t) ù
û
ur ë
-8
(c) B ( x, t ) = 9 ´10 T ˆj
–8
(
)
(
)
sin é1.5 × 10 –6 x – 2 × 1014 t) ù
û
ur ë
-8
ˆ
(d) B ( x, t ) = 9 ´ 10 T k
(
sin é 2p(1.5 × 10
ë
)
–6
x – 2 × 1014 t) ù
û
P-381
Electromagnetic Waves
36.
37.
38.
39.
40.
41.
42.
43.
During the propagation of electromagnetic waves in a
medium:
[2014]
(a) Electric energy density is double of the magnetic
energy density.
(b) Electric energy density is half of the magnetic energy
density.
(c) Electric energy density is equal to the magnetic energy
density.
(d) Both electric and magnetic energy densities are zero.
A lamp emits monochromatic green light uniformly in all
directions. The lamp is 3% efficient in converting electrical
power to electromagnetic waves and consumes 100 W of
power. The amplitude of the electric field associated with
the electromagnetic radiation at a distance of 5 m from the
lamp will be nearly:
[Online April 12, 2014]
(a) 1.34 V/m
(b) 2.68 V/m
(c) 4.02 V/m
(d) 5.36 V/m
An electromagnetic wave of frequency 1 × 1014 hertz is
propagating along z-axis. The amplitude of electric field is
4 V/m. If e0 = 8.8 × 10–12 C2/N-m2, then average energy
density of electric field will be: [Online April 11, 2014]
(a) 35.2 × 10–10 J/m3
(b) 35.2 × 10–11 J/m3
–12
3
(c) 35.2 × 10 J/m
(d) 35.2 × 10–13 J/m3
The magnetic field in a travelling electromagnetic wave
has a peak value of 20 nT. The peak value of electric field
strength is :
[2013]
(a) 3 V/m
(b)6 V/m
(c) 9 V/m
(d) 12 V/m
A plane electromagnetic wave in a non-magnetic dielectric
ur ur
medium is given by E = E 0 (4 ´ 10 -7 x - 50t ) with
distance being in meter and time in seconds. The dielectric
constant of the medium is :
[Online April 22, 2013]
(a) 2.4
(b)5.8
(c) 8.2
(d) 4.8
Select the correct statement from the following :
[Online April 9, 2013]
(a) Electromagnetic waves cannot travel in vacuum.
(b) Electromagnetic waves are longitudinal waves.
(c) Electromagnetic waves are produced by charges
moving with uniform velocity.
(d) Electromagnetic waves carry both energy and
momentum as they propagate through space.
An electromagnetic wave in vacuum has the electric and
r
r
magnetic field E and B , which are always perpendicular
r
to each other. The direction of polarization is given by X
r
and that of wave propagation by k . Then
[2012]
r
r r r
r
(a) X || B and k || B ´ E
r r r
r r
(b) X || E and k || E ´ B
r r r
r r
(c) X || B and k || E ´ B
r
r r
r r
(d) X || E and k || B ´ E
An electromagnetic wave with frequency w and
wavelength l travels in the + y direction. Its magnetic field
is along + x-axis. The vector equation for the associated
electric field (of amplitude E0) is [Online May 19, 2012]
®
2p
æ
(a) E = - E0 cos ç wt +
è
l
ö
y ÷ xˆ
ø
®
2p
æ
(b) E = E0 cos çè wt l
ö
y ÷ xˆ
ø
®
2p
æ
(c) E = E0 cos ç wt è
l
ö
y÷ zˆ
ø
®
2p
æ
(d) E = - E0 cos ç wt +
è
l
ö
y÷ zˆ
ø
44. An electromagnetic wave of frequency v = 3.0 MHz
passes from vacuum into a dielectric medium with
permittivity Î = 4.0. Then
[2004]
(a) wave length is halved and frequency remains
unchanged
(b) wave length is doubled and frequency becomes half
(c) wave length is doubled and the frequency remains
unchanged
(d) wave length and frequency both remain unchanged.
45. Electromagnetic waves are transverse in nature is evident
by
[2002]
(a) polarization
(b) interference
(c) reflection
(d) diffraction
TOPIC 2 Electromagnetic Spectrum
46. The correct match between the entries in column I and
column II are :
[Sep. 05, 2020 (II)]
I
II
Radiation
Wavelength
(A) Microwave
(i) 100 m
(B) Gamma rays
(ii) 10–15 m
(C) A.M. radio waves
(iii) 10–10 m
(D) X-rays
(iv) 10–3 m
(a) (A)-(ii), (B)-(i), (C)-(iv), (D)-(iii)
(b) (A)-(i), (B)-(iii), (C)-(iv), (D)-(ii)
(c) (A)-(iii), (B)-(ii), (C)-(i), (D)-(iv)
(d) (A)-(iv), (B)-(ii), (C)-(i), (D)-(iii)
47. Chosse the correct option relating wavelengths of different
parts of electromagnetic wave spectrum :
[Sep. 04, 2020 (I)]
(a) l visible < l micro waves < l radio waves < l X - rays
(b) l radio waves > l micro waves > l visible > l x-rays
(c) l x- rays < l micro waves < l radio waves < l visible
(d) l visible > l x-rays > l radio waves > l micro waves
48. Given below in the left column are different modes of
communication using the kinds of waves given in the
right column.
[10 April 2019, I]
A. Optical Fibre
P. Ultrasound
Communication
B. Radar
Q. Infrared Light
C. Sonar
R. Microwaves
D. Mobile Phones
S. Radio Waves
P-382
49.
50.
51.
From the options given below, find the most appropriate
match between entries in the left and the right column.
(a) A – Q, B – S, C – R, D – P
(b) A – S, B – Q, C – R, D – P
(c) A – Q, B – S, C – P, D – R
(d) A – R, B – P, C – S, D – Q
Arrange the following electromagnetic radiations per
quantum in the order of increasing energy :
[2016]
A : Blue light B : Yellow light
C : X-ray
D : Radiowave.
(a) C, A, B, D
(b) B, A, D, C
(c) D, B, A, C
(d) A, B, D, C
Microwave oven acts on the principle of :
[Online April 9, 2016]
(a) giving rotational energy to water molecules
(b) giving translational energy to water molecules
(c) giving vibrational energy to water molecules
(d) transferring electrons from lower to higher energy
levels in water molecule
Match List - I (Electromagnetic wave type) with List - II
(Its association/application) and select the correct option
from the choices given below the lists:
[2014]
List 2
(i) To treat muscular
strain
2. Radio waves
(ii) For broadcasting
3. X-rays
(iii) To detect fracture of
bones
4. Ultraviolet rays
(iv) Absorbed by the
ozone layer of the
atmosphere
1
2
3
4
(a) (iv)
(iii)
(ii)
(i)
(b) (i)
(ii)
(iv)
(iii)
(c) (iii)
(ii)
(i)
(iv)
(d) (i)
(ii)
(iii)
(iv)
If microwaves, X rays, infrared, gamma rays, ultra-violet,
radio waves and visible parts of the electromagnetic
spectrum are denoted by M, X, I, G, U, R and V then which
of the following is the arrangement in ascending order of
wavelength ?
[Online April 19, 2014]
(a) R, M, I, V, U, X and G
(b) M, R, V, X, U, G and I
(c) G, X, U, V, I, M and R
(d) I, M, R, U, V, X and G
Match the List-I (Phenomenon associated with
electromagnetic radiation) with List-II (Part of
electromagnetic spectrum) and select the correct code from
the choices given below this lists:[Online April 11, 2014]
1.
52.
53.
Physics
List 1
Infrared waves
I
List I
Doublet of sodium
List II
(A) Visible radiation
II Wavelength
corresponding to
temperature associated
with the isotropic
radiation filling all space
(B) Microwave
III Wavelength emitted by
atomic hydrogen in
interstellar space
(C) Short radio wave
IV Wavelength of radiation (D) X-rays
arising from two close
energy levels in hydrogen
(a) (I)-(A), (II)-(B), (III)-(B), (IV)-(C)
(b) (I)-(A), (II)-(B), (III)-(C), (IV)-(C)
(c) (I)-(D), (II)-(C), (III)-(A), (IV)-(B)
(d) (I)-(B), (II)-(A), (III)-(D), (IV)-(A)
54. Match List I (Wavelength range of electromagnetic
spectrum) with List II (Method of production of these
waves) and select the correct option from the options given
below the lists.
[Online April 9, 2014]
List I
Lis t II
(1) 700 nm to
1 mm
(i) Vibration of atoms
and molecules .
(2) 1 nm to
400 nm
(ii) Inner s hell electrons
in atoms moving
from one energy
level to a lower level.
(3) < 10–3 nm (iii) Radioactive decay of
the nucleus .
(4) 1 mm to
0.1 m
(iv) Magnetron valve.
(a) (1)-(iv), (2)-(iii), (3)-(ii), (4)-(i)
(b) (1)-(iii), (2)-(iv), (3)-(i), (4)-(ii)
(c) (1)-(ii), (2)-(iii), (3)-(iv), (4)-(i)
(d) (1)-(i), (2)-(ii), (3)-(iii), (4)-(iv)
55. Photons of an electromagnetic radiation has an energy
11 keV each. To which region of electromagnetic spectrum
does it belong ?
[Online April 9, 2013]
(a) X-ray region
(b) Ultra violet region
(c) Infrared region
(d) Visible region
56. The frequency of X-rays; g-rays and ultraviolet rays are
respectively a, b and c then
[Online May 26, 2012]
(a) a < b; b > c
(b) a > b ; b > c
(c) a < b < c
(d) a = b = c
P-383
Electromagnetic Waves
1.
(a) Relation between electric field E0 and magnetic field
B0 of an electromagnetic wave is given by
Directiono f wave propagation
E
c= 0
B0
\ Ê = - kˆ
\ E = E0 sin[200p( y + ct )](- kˆ) V/m
or, E = -9sin[200p( y + ct )]kˆ V/m
(c) Relation between electric field and magnetic field for
( E ´ B ) || C
Bˆ = iˆ and Cˆ = - ˆj
(Here, c = Speed of light)
Þ E0 = B0 ´ c = 1.2 ´ 10-7 ´ 3 ´ 108 = 36
As the wave is propagating along x-direction, magnetic
field is along z-direction
5.
an electromagnetic wave in vacuum is B0 =
and ( Eˆ ´ Bˆ ) || Cˆ
r
\ E should be along y-direction.
r
r
So, electric field E = E0 sin E × ( x, t )
= [ -36sin (0.5 ´ 103 x + 1.5 ´ 1011t ) ˆj ]
2.
(c) In electromagnetic wave,
m0 e0
Here, m0 = absolute permeability, e0 = absolute permittivity
E
E0
\ B0 = 0 =
= E0 m 0 e 0
c 1/ m 0 e 0
V
m
As the electromagnetic wave is propagating along x
direction and electric field is along y direction.
\ Eˆ ´ Bˆ || Cˆ (Here, Ĉ = direction of propagation of wave)
r
\ B should be in k̂ direction.
E0
=C
B0
E0
C
qV0 E0
C
(Given V0 = 0.1 C and E0 = 30)
\ B = E 0 m0 e0 cos (wt – kx) k̂
At t = 0
Fmax = qVBmax sin 90° =
=
3.
1.6 ´ 10
-19
8
´ 0.1 ´ 3 ´ 10 ´ 30
3 ´ 108
B = E 0 m0 e0 cos (kx) k̂
= 4.8 ´ 10-19 N
6.
iˆ + ˆj
Unit vector in the direction of electric field, Eˆ =
2
The direction of electromagnetic wave is perpendicular to
both electric and magnetic field.
\ kˆ = Eˆ ´ Bˆ
æ iˆ + ˆj ö ˆ
-iˆ + ˆj
Þ kˆ = ç
´ B Þ Bˆ =
÷
2
è 2 ø
r E
\ B = 0 (- xˆ + yˆ )sin(kz - wt )
c
ˆ
(d) Given : B = 3 ´ 10-8 sin[200p( y + ct )]iT
\ B0 = 3 ´ 10
-8
E0 = CB0 Þ E0 = 3 ´ 108 ´ 3 ´ 10 -8 = 9 V/m
(b) Energy density =
ÞB=
r
(a) E = E0 ( xˆ + yˆ ) sin(kz - wt )
m0 =
Direction of propagation of em wave = + k̂
4.
1
In free space, its speed c =
\ Maximum value of magnetic field, B0 =
E0
.
c
7.
1 B2
2 m0
2 ´ m 0 ´ Energy density
1
2
C e0
= 4p ´ 10 -7
\ B = 2 ´ 4p ´ 10-7 ´ 1.02 ´ 10 -8 = 160 ´ 10 -9
= 160 nT
(a) Electromagnetic wave will propagate perpendicular to
the direction of Electric and Magnetic fields
Cˆ = Eˆ ´ Bˆ
Here unit vector Ĉ is perpendicular to both Ê and B̂
ur
ur
Given, E = k$ , B = 2i$ - 2 $j
\ Cˆ = Eˆ ´ Bˆ =
iˆ + ˆj
Þ Cˆ =
2
1
2
iˆ
ˆj
kˆ
0
0
1 =
1 -1 0
iˆ + ˆj
2
P-384
8.
Physics
r
(d) Given: E1 = E0 ˆj cos ( wt - kx )
r r
i.e., Travelling in +ve x-direction E ´ B should be in xdirection
r
\ B is in K̂
r
E
\ B1 = 0 cos ( wt - kx ) kˆ
C
E0 ö
æ
çèQ B0 = C ÷ø
11.
r
(b) Given, B = 3 ´ 10 -8 sin(1.6 ´ 103 x + 48 ´ 1010 t)
Using, E0 = B0 ´ C = 3 ´ 10 –8 ´ 3 ´ 108 = 9 V/m
\ Electric field,
r
E = 9sin(1.6 ´ 103 x + 48 ´ 1010 t ) kˆ V /m
12. (c) At t = 0, z =
p
k
r E
E
E = 0 (iˆ + ˆj ) cos[p ] = – 0 (iˆ + ˆj )
2
2
r
r
FE = qE
\
Force due to electric field will be in the direction
Force due to magnetic field is in direction
r
r r
r r
q(v ´ B) and v || k . Therefore, it is parallel to E .
r
E2 = E0 kˆ cos ( wt - ky )
r
E
B2 = 0 iˆ cos ( wt - ky )
C
\ Travelling in +ve y-axis
r r
E ´ B should be in y-axis
r
r
r r
\ Net force F = qE + q v ´ B
r
r
r
r
q E1 + E2 + q 0.8cjˆ ´ B1 + B2
(
(
) (
Þ
)
(
E0
60
=
C 3 ´ 108
= 20 × 10–8 T = 2 × 10–7 T
)
14. (b) B0 =
r
r
E
E
B1 = 0 kˆ
B2 = 0 iˆ
c
c
r
E
\ Fnet = qE0 ˆj + kˆ + q ´ 0.8c ´ 0 ˆj ´ kˆ + iˆ
C
ˆ
ˆ
= qE ˆj + k + 0.8 qE iˆ - k
)
( )
0(
= qE0 ( 0.8iˆ + ˆj + 0.2 kˆ )
0
9.
10.
(
w 2pf 2p ´ 23.9 ´ 109
=
K= =
= 500
v
v
3 ´ 108
)
)
(a) Direction of polarisation = Ê = kˆ
$ $
µ´B
µ = i+ j
Direction of propagation = E
2
r r
But E.B = 0 \ Bˆ = iˆ - j
2
(d) Amplitude of electric field (E) and Magnetic field (B)
of an electromagnetic wave are related by the relation
E
=c
B
Þ E = Bc
Þ E = 5 × 10–8 × 3 × 108 = 15 N/C
r
Þ E = 15iˆ V / m
iˆ + ˆj
r
r
r
Fnet = FE + FB is antiparallel to
2
6 ˆj + 8kˆ
-3 ˆj + 4kˆ
ˆ
=
13. (c) S = 2
5
6 + 82
If t = 0 and x = y = 0
r
r
E1 = E0 ˆj
E2 = E0 kˆ
(
-(iˆ + ˆj )
®
Therefore, B = B sin(kz - wt )
0
= 2 × 10–7 sin(0.5 ´ 103 z - 1.5 ´ 1011 t )i
15. (a)
E0
=C
B0
E0
C
r
Given that E = E0 cos(kz) cos(w t) iˆ
Þ B0 =
r E
E = 0 éëcos ( kz – wt ) ˆi – cos ( kz + wt ) iˆùû
2
Correspondingly
r B
B = 0 éëcos ( kz – wt ) ˆj – cos ( kz + wt ) ˆjùû
2
r B
B = 0 ´ 2sin kz sin wt
2
r æE
ö
B = ç 0 sin kz sin wt ÷ ˆj
è C
ø
2
P-385
Electromagnetic Waves
16.
(c) Pressure, P =
I
C
22. (d) I =
F I
=
A C
IA Dp
ÞF=
=
C
Dt
I
Þ Dp = ADt
C
Þ
=
17.
Þ Brms =
3 ´ 108
N-s
(a) B0 = B02 + B12 = 302 + 2 2 ´ 10 -6
= 30 × 10–6T
;
Which is closest to 10–4.
23. (c) The speed of electromagnetic wave in free space is
given by
C=
=
IA
C
E 20 kV n 2 E 0
=
= Þ = n
E2 C n
E
similarly
Þ
24.
(1 + 0.25) ´ 50 ´ 1
B20 C B 2 v B0 1
=
Þ =
2m0 2m0
B
n
(d) EM wave intensity
Power 1
= e 0 E 02 c
Area 2
[where E0= maximum electric field]
Þ I=
8
3 ´ 10
(a) The relation between amplitudes of electric and
magnetic field in free space is given by
E
6
B0 = 0 =
= 2 ´ 10 -8 T
c
3 ´ 108
Propagation direction = Eˆ ´ Bˆ
iˆ = ˆj ´ Bˆ
21.
...(i)
1
...(ii)
k Î0 m 0
Dividing equation (i) by (ii), we get
C
\
= k =n
V
1
1
Î0 E 02C = intensity = Î0 kE 2 v
2
2
2
2
\ E 0C = kE v
; 20 ´ 10-8 N
20.
1
m0 Î0
In medium, v =
E0
9
=
´ 103V / m
V2
2
Force on the charge,
9
F = EQ =
´ 103 ´ 10 -4 ; 0.64 N
2
(b) As we know,
r
r |E|
6.3
=
= 2.1 ´ 10-8 T
| B |=
C
3 ´ 108
ˆ =C
ˆ
and Eˆ ´ B
(b) F = (1 + r )
3 ´ 108
6 × 10–4 T
ˆ = iˆ [Q EM wave travels along +(ve) x-direction.]
Jˆ ´ B
r
ˆ
\ B̂ = kˆ or B = 2.1 ´ 10 –8 kT
19.
Iµ0
C
108 ´ 4 p ´ 10 -7
=
\ E0 = CB = 3 ´ 108 ´ 30 ´ 10 -6
= 9 × 103 V/m
18.
B20 Iµ0
=
2
C
Þ
(25 ´ 25) ´ 104 ´ 10–4 ´ 40 ´ 60
= 5 × 10–3 N-s
B20
·C
2µ0
Þ B̂ = kˆ
\ The magnetic field component will be along z direction.
(c) E0 = cB0 = 3 × 108 × 1.6 × 10–6 = 4.8 × 102 V/m
uur uur uur
Also S Þ E ´ B
uur uur
or - K Þ E ´ (2iˆ + ˆj )
uur
Therefore direction of E ® - iˆ + 2 ˆj
(
)
Þ
27 ´ 10 –3
10 ´10
–6
1
= ´ 9 ´ 10 –12 ´ E 20 ´ 3 ´ 108
2
Þ E 0 = 2 ´103 kV / m =1.4kV / m
25. (b) Using, formula E0 = B0 × C
= 100 × 10–6 × 3 × 108
= 3 × 104 N/C
Here we assumed that
B0 = 100 × 10–6 is in tesla (T) units
r
ˆ
é 6iˆ + 8kˆ . xiˆ + zkˆ ù
26. (b) E =10jcos
ë
û
r r
ˆ
=10 jcos éë K . r ùû
r
ˆ direction of waves travel
\ K =6iˆ + 8K;
(
i. e. direction of ‘c’.
)(
)
P-386
Physics
29. (c) If E0 is magnitude of electric field then
E(10 ˆj)
1
2I
e0 E2 ´ C = 1 Þ E0 =
2
Ce 0
E0 =
r r
Direction of E ´ B will be along + ĵ .
3iˆ + 4kˆ
= ĉ
5
B
ˆ should give the direction of wave propagation
30. (c) Eˆ ´ B
ˆ ˆ
ˆ
ˆ ˆ ˆ ˆ
æ ˆ ˆö ˆ
)
ˆ P i´ jÞK
ˆ ´ i + j = j - ( - i) = i + j P i + j
Þ K´B
ç
÷
2
2
2
2
è 2ø
Option (a), option (b) and option (d) does not satisfy.
ˆi + ˆj
Wave propagation vector K̂ should along
.
2
31. (c) Given, Electric field component of monochromatic
r
radiation, (E) = 2E ˆi coskzcos wt
–4iˆ + 3kˆ
Cˆ ´ Eˆ =
5
r E 10
B= =
C C
r 10 æ –4iˆ + 3kˆ ö æ –8iˆ + 6kˆ ö
\ B çç
÷= ç
÷
Cè
5 ÷ø çè C ÷ø
r
1
or, magnetic field B ( x, z, t ) =
C
6kˆ – 8iˆ cos ( 6x + 8z –10ct )
(
27.
0
)
(c) Velocity of EM wave is given by v =
1
mÎ
w
=C
k
C
Velocity in medium =
2
Here, m1 = m2 = 1 as medium is non-magnetic
1
Îr1 1
Îr1
C
=
\
=
=2 Þ
1
Îr2 4
æCö
Îr2 çè 2 ÷ø
Velocity in air =
28.
(d) Average energy density of magnetic field,
uB =
B02
4m 0
.
Average energy density of electric field,
uE =
e0 E 20
4
Now, E0 = CB0 and C2 =
1
m0 Î0
2
e0
´ C 2 B 20 = e0 ´ 1 ´ B02 = B0 = u B
4
4 m 0 e0
4m 0
\ uE = uB
Since energy density of electric and magnetic field is same,
so energy associated with equal volume will be equal i.e.,
uE = uB
uE =
E0
C
dE
dB
We know that,
=dz
dt
dE
dB
= -2E 0 k sin kz cos w t = dz
dt
dB = + 2E0k sin kz cos wt dt
..... (i)
Integrating eq.n (i), we have
B = +2E 0 k sin kz ò cos wt dt
Magnetic field is given by,
k
sin kz sin wt
w
We also know that,
E0 w
= =c
B0 k
Magnetic field vector,
r 2E
B = 0 ˆj sin kz sin wt
c
= +2E 0
32. (a) Speed of EM wave in force space (c) =
E0
B0
r
or E = cB0 sin (kx + wt)kˆ
33. (d) Wave in X-direction means E and B should be function
of x and t.
) ) ) )
y,z ^ y∗z
uur ur
uur ur
34. (b) As we know, E . B = 0 Q [ E ^ B ]
uur ur
and E ´ B should be along Z direction
As (–2$i – 3 $j ) ´ (3$i – 2 $j ) = 5k$
Hence option (b) is the correct answer.
35. (d) As we know,
E
27
B0 = 0 =
= 9 ´10 –8 tesla
8
C 3 ´ 10
P-387
Electromagnetic Waves
Oscillation of B can be only along ĵ or k̂ direction.
w = 2pf = 2p × 2 × 1014 Hz
ur
\ B ( x, t ) = (9 ´ 10 –8 T )kˆ sin[2p(1.5 ´ 10 –6 ´ –2 ´104 t )]
36.
(c) E0 = CB0 and C =
Electric energy density =
1
m 0e 0
1
e0 E02 = m E
2
Magnetic energy density =
37.
1 Bo 2
= mB
2 m0
Thus, mE = mB
Energy is equally divided between electric and magnetic
field.
(b) Wavelength of monochromatic green light
= 5.5 × 10–5 cm
Power
Area
Intensity I =
=
100 ´ ( 3 /100 )
=
3
Wm -2
100p
4p ( 5 )
Now, half of this intensity (I) belongs to electric field and
half of that to magnetic field, therefore,
I 1
= e0 E 02 C
2 4
or E 0 =
=
=
38.
39.
2
2I
e0 C
æ 3 ö
2´ç
p÷
è 100 ø
1
æ
ö
´ 3 ´108
ç
9÷
è 4p ´ 9 ´10 ø
(
6
´ 30 =
25
)
7.2
\ E 0 = 2.68 V / m
(c) Given: Amplitude of electric field,
E0 = 4 v/m
Absolute permitivity,
e0 = 8.8 × 10–12 c2/N-m2
Average energy density uE = ?
Applying formula,
1
2
Average energy density uE = e0 E
4
1
-12
2
Þ uE = ´ 8.8 ´ 10 ´ (4)
4
= 35.2 × 10–12 J/m3
(b) From question,
B0 = 20 nT = 20 × 10–9T
(Q velocity of light in vacuum C = 3 × 108 ms–1)
r
r
r
E0 = B0 ´ C
r
r r
| E0 |=| B | ×| C |= 20 ´ 10 -9 ´ 3 ´ 108
= 6 V/m.
40. (b)
41. (d) Electromagnetic waves do not required any medium
to propagate. They can travel in vacuum. They are
transverse in nature like light. They carry both energy and
momentum.
A changing electric field produces a changing magnetic
field and vice-versa. Which gives rise to a transverse wave
known as electromagnetic wave.
42. (b) Q The E.M. wave are transverse in nature i.e.,
r r
k ´E r
=H
=
…(i)
m
r
r B
where H =
m
r r
r
k ´H
and
…(ii)
= -E
we
r
r
r
r
k is ^ H and k is also ^ to E
r r
The direction of wave propagation is parallel to E ´ B.
The direction of polarization is parallel to electric field.
43. (c) In an electromagnetic wave electric field and
magnetic field are perpendicular to the direction of
propagation of wave. The vector equation for the electric
field is
r
E = E0 cos æç wt - 2p y ö÷ zˆ
è
l ø
44. (a) Frequency remains unchanged during refraction
Velocity of EM wave in vacuum
1
=C
Vvacuum =
m0 Î0
vmed =
l med
l vacuum
1
µ0 Î0 ´4
=
vmed
v vacuum
=
c
2
=
c/2 1
=
2
c
\ Wavelength is halved and frequency remains
unchanged
45. (a) The phenomenon of polarisation is shown only by
transverse waves. The vibration of electromagnetic wave
are restricted through polarization in a direction
perpendicular to wave propagation.
46. (d) Energy sequence of radiations is
Eg -Rays > EX-Rays > Emicrowave > EAM Radiowaves
\ l g -Rays < l X-Rays < l microwave < l AM Radiowaves
From the above sequence, we have
(a) Microwave ® 10 -3 m (iv)
(b) Gamma Rays ® 10-15 m (ii)
(c) AM Radio wave ® 100 m (i)
(d) X-Rays ® 10-10 m (iii)
P-388
47.
Physics
(b) The orderly arrangement of different parts of EM wave
in decreasing order of wavelength is as follows:
l radiowaves > l microwaves > l visible > l X-rays
48.
49.
(c) Optical Fibre Communication – Infrared Light
Radar – Radio Waves
Sonar – Ultrasound
Mobile Phones – Microwaves
(c)
E, Decreases
g-rays X-rays uv-rays Visible rays IR rays
Radio
VIBGYOR Microwaves waves
Radio wave < yellow light < blue light < X-rays
(Increasing order of energy)
50.
51.
52.
(c) Microwave oven acts on the principle of giving
vibrational energy to water molecules.
(d)
(1) Infrared rays are used to treat muscular strain because
these are heat rays.
(2) Radio waves are used for broadcasting because these
waves have very long wavelength ranging from few
centimeters to few hundred kilometers.
(3) X-rays are used to detect fracture of bones because
they have high penetrating power but they can't penetrate
through denser medium like dones.
(4) Ultraviolet rays are absorbed by ozone of the
atmosphere.
(c) Gamma rays < X-rays < Ultra violet < Visible rays
< Infrared rays < Microwaves < Radio waves.
53. (d) Wavelength emitted by atomic hydrogen in interstellar
space - Part of short radio wave of electromagnetic
spectrum.
Doublet of sodium - visible radiation.
54. (d) Vibration of atoms and molecules 700 nm to 1 mm
Radioactive decay of the nucleus < 10–3 nm
Magnetron valve 1 mm to 0.1 m
55. (a) E =
Þ l=
hc
hc
Þ l=
l
E
6.6 ´ 10-34 ´ 3 ´108
11´1000 ´ 1.6 ´10 -19
= 12.4 Å
Increasing order of frequency
x-rays u-v rays visible Infrared
wavelength range of visible region is 4000Å to 7800Å.
56. (a) Frequency range of g-ray,
b = 1018 – 1023 Hz
Frequency range of X-ray,
a = 1016 – 1020 Hz
Frequency range of ultraviolet ray,
c = 1015 – 1017 Hz
\ a < b; b > c
23
Ray Optics and
Optical Instruments
below. The distance over which the man can see the image
of the light source in the mirror is:
[12 Jan. 2019 I]
Plane Mirror, Spherical Mirror
TOPIC 1
and Reflection of Light
1.
When an object is kept at a distance of 30 cm from a concave
mirror, the image is formed at a distance of 10 cm from the
mirror. If the object is moved with a speed of 9
cms–1, the speed (in cms–1) with which image moves at
that instant is ________.
[NA Sep. 03, 2020 (II)]
d
L
2L
2.
(a) d
Object
20
(cm)
16
12
8
4
A spherical mirror is obtained as shown in the figure from
a hollow glass sphere. If an object is positioned in front of
the mirror, what will be the nature and magnification of the
image of the object? (Figure drawn as schematic and not
to scale)
[Sep. 02, 2020 (I)]
(a) Inverted, real and magnified
(b) Erect, virtual and magnified
(c) Erect, virtual and unmagnified
(d) Inverted, real and unmagnified
4.
A concave mirror for face viewing has focal length of 0.4 m.
The distance at which you hold the mirror from your face
in order to see your image upright with a magnification of
5 is:
[9 April 2019 I]
(a) 0.24 m
(b) 1.60 m (c) 0.32 m
(d) 0.16 m
A point source of light, S is placed at a distance L in front
of the centre of plane mirror of width d which is hanging
vertically on a wall. A man walks in front of the mirror
along a line parallel to the mirror, at a distance 2L as shown
5.
6.
(b) 2d
d
2
Two plane mirrors are inclined to each other such that a
ray of light incident on the first mirror (M1) and parallel to
the second mirror (M2) is finally reflected from the second
mirror (M2) parallel to the first mirror (M1). The angle
between the two mirrors will be:
[9 Jan. 2019 II]
(a) 45°
(b) 60°
(c) 75°
(d) 90°
(c) 3d
3.
S
(d)
An object is gradually moving away from the focal point
of a concave mirror along the axis of the mirror. The
graphical representation of the magnitude of linear
magnification (m) versus distance of the object from the
mirror (x) is correctly given by
(Graphs are drawn schematically and are not to scale)
[8 Jan. 2020 II]
(a)
P-390
Physics
(b)
10. To get three images of a single object, one should have
two plane mirrors at an angle of
[2003]
(a) 60º
(b) 90º
(c) 120º
(d) 30º
11. If two plane mirrors are kept at 60° to each other, then the
number of images formed by them is
[2002]
(a) 5
(b) 6
(c) 7
(d) 8
Refraction of Light at Plane
TOPIC 2 Surface and Total Internal
Reflection
(c)
12. An observer can see through a small hole on the side of a
jar (radius 15 cm) at a point at height of 15 cm from the
bottom (see figure). The hole is at a height of 45 cm. When
the jar is filled with a liquid up to a height of 30 cm the same
observer can see the edge at the bottom of the jar. If the
refractive index of the liquid is N/100, where N is an integer,
the value of N is ___________. [NA Sep. 03, 2020 (I)]
(d)
45 cm
7.
A particle is oscillating on the X-axis with an amplitude
2 cm about the point x0 = 10 cm with a frequency w. A concave
mirror of focal length 5 cm is placed at the origin (see figure)
Identify the correct statements: [Online April 15, 2018]
(A) The image executes periodic motion
(B) The image executes non-periodic motion
(C) The turning points of the image are asymmetric w.r.t
the image of the point at x = 10 cm
(D) The distance between the turning points of the
oscillation of the image is
8.
9.
100
21
x0 = 10 cm
x=0
(a) (B), (D)
(b) (B), (C)
(c) (A), (C), (D)
(d) (A), (D)
60 cm
(b) –24 cm
(c)
– 60 cm
(d) 24 cm
A car is fitted with a convex side-view mirror of focal length
20 cm. A second car 2.8 m behind the first car is overtaking
the first car at a relative speed of 15 m/s. The speed of the
image of the second car as seen in the mirror of the first
one is :
[2011]
(a)
1
m/s
15
(b) 10 m/s
(c) 15 m/s
15 cm
13. A light ray enters a solid glass sphere of refractive index
m = 3 at an angle of incidence 60°. The ray is both
reflected and refracted at the farther surface of the sphere.
The angle (in degrees) between the reflected and refracted
rays at this surface is ___________.
[NA Sep. 02, 2020 (II)]
14. A vessel of depth 2h is half filled with a liquid of refractive
index 2 2 and the upper half with another liquid of
You are asked to design a shaving mirror assuming that a
person keeps it 10 cm from his face and views the magnified
image of the face at the closest comfortable distance of 25
cm. The radius of curvature of the mirror would then be :
[Online April 10, 2015]
(a)
15 cm
(d)
1
m/s
10
refractive index 2. The liquids are immiscible. The
apparent depth of the inner surface of the bottom of vessel
will be:
[9 Jan. 2020 I]
h
h
(a)
(b)
2( 2 + 1)
2
(c)
h
3 2
(d)
3
h 2
4
15. There is a small source of light at some depth below the
4
surface of water (refractive index = ) in a tank of large
3
cross sectional surface area. Neglecting any reflection from
the bottom and absorption by water, percentage of light
that emerges out of surface is (nearly):
[Use the fact that surface area of a spherical cap of height
h and radius of curvature r is 2prh]
[9 Jan. 2020 II]
(a) 21%
(b) 34%
(c) 17%
(d) 50%
P-391
Ray Optics and Optical Instruments
16. The critical angle of a medium for a specific wavelength, if
the medium has relative permittivity 3 and relative
(a)
2 3
+ 2b
a
(b) 2a +
2b
3
4
for this wavelength, will be: [8 Jan. 2020 I]
3
(a) 15°
(b) 30°
(c) 45°
(d) 60°
17. A concave mirror has radius of curvature of 40 cm. It is at
the bottom of a glass that has water filled up to 5 cm (see
figure). If a small partricle is floating on the surface of
water, its image as seen, from directly above the glass, is at
a distance d from the surface of water. The value of d is
close to :
[12 Apr. 2019 I]
(Refractive index of water = 1.33)
20. In figure, the optical fiber is l = 2 m long and has a diameter
of d = 20 mm. If a ray of light is incident on one end of the
fiber at angle q1 = 40°, the number of reflections it makes
before emerging from the other end is close to :
(a) 6.7 cm
(b) 13.4 cm (c) 8.8 cm
(d) 11.7 cm
18. A transparent cube of side d, made of a material of refractive
index m2, is immersed in a liquid of refractive index
m1(m1< m2). A ray is incident on the face AB at an angle q
(shown in the figure). Total internal reflection takes place
at point E on the face BC.
(a) 55000
(b) 66000 (c) 45000
(d) 57000
21. A light wave is incident normally on a glass slab of
refractive index 1.5. If 4% of light gets reflected and the
amplitude of the electric field of the incident light is 30 V/
m, then the amplitude of the electric field for the wave
propogating in the glass medium will be:[12 Jan. 2019 I]
permeability
Then q must satisfy :
-1
(a) q < sin
(c) q < sin -1
m1
m2
m 22
m12
[12 Apr. 2019 II]
(b) q > sin -1
-1
(d) q > sin -1
m 22
m12
(c) 2a +
2b
(refractive index of fiber is 1.31 and sin 40° = 0.64)
[8 April 2019 I]
(a) 30 V/m
(b) 10 V/m
(c) 24 V/ m
(d) 6 V/m
22. Let the refractive index of a denser medium with respect to
a rarer medium be n 12 and its critical angle be qC. At an
angle of incidence A when light is travelling from denser
medium to rarer medium, a part of the light is reflected and
the rest is refracted and the angle between reflected and
refracted rays is 90°. Angle A is given by :
[Online April 8, 2017]
-1
m1
m2
19. A ray of light AO in vacuum is incident on a glass slab at
angle 60o and refracted at angle 30o along OB as shown in
the figure. The optical path length of light ray from A to B
is :
[10 Apr. 2019 I]
(d) 2a + 2b
3
(a)
1
-1
cos (sin qC )
(c) cos–1 (sin qC)
(b)
1
-1
tan (sin qC )
(d) tan–1 (sin qC)
23. A diver looking up through the water sees the outside
world contained in a circular horizon. The refractive index
4
of water is , and the diver’s eyes are 15 cm below the
3
surface of water. Then the radius of the circle is:
[Online April 9, 2014]
(a)
15 ´ 3 ´ 5 cm
(b) 15 ´ 3 7 cm
(c)
15 ´ 7
cm
3
(d)
15 ´ 3
7
cm
P-392
24.
25.
Physics
A printed page is pressed by a glass of water. The refractive
index of the glass and water is 1.5 and 1.33, respectively. If
the thickness of the bottom of glass is 1 cm and depth of
water is 5 cm, how much the page will appear to be shifted
if viewed from the top ?
[Online April 25, 2013]
(a) 1.033 cm
(b) 3.581 cm
(c) 1.3533 cm
(d) 1.90 cm
A light ray falls on a square glass slab as shown in the
diagram. The index of refraction of the glass, if total internal
reflection is to occur at the vertical face, is equal to :
[Online April 23, 2013]
(c)
æ
æ
1ö
1 ö
ç1 + ÷ h2 - ç 1 + ÷ h1
è m1 ø
è m2 ø
æ
æ
1 ö
1 ö
ç 1 - ÷ h2 + ç 1 - ÷ h1
è m1 ø
è m2 ø
29. A transparent solid cylindrical rod has a refractive index of
2
. It is surrounded by air. A light ray is incident at the
3
mid-point of one end of the rod as shown in the figure.
(d)
q
45° Incident ray
The incident angle q for which the light ray grazes along
the wall of the rod is :
[2009]
æ
ö
2
1
-1
3/2
÷
(b) sin ç
(a) sin
è 3ø
(
)
æ 1 ö
sin -1 ç
÷
(d) sin -1 (1/ 2 )
è 3ø
A fish looking up through the water sees the outside world
(c)
30.
5
2
26.
27.
28.
3
3
(a) ( 2 + 1) (b)
(c)
(d)
2
2
2
Light is incident from a medium into air at two possible
angles of incidence (A) 20° and (B) 40°. In the medium
light travels 3.0 cm in 0.2 ns. The ray will :
[Online April 9, 2013]
(a) suffer total internal reflection in both cases (A) and
(B)
(b) suffer total internal reflection in case (B) only
(c) have partial reflection and partial transmission in case
(B)
(d) have 100% transmission in case (A)
Let the x-z plane be the boundary between two transparent
media. Medium 1 in z ³ 0 has a refractive index of 2 and
medium 2 with z < 0 has a refractive index of 3 . A ray of
r
light in medium 1 given by the vector A = 6 3iˆ + 8 3 ˆj - 10kˆ
is incident on the plane of separation. The angle of
refraction in medium 2 is:
[2011]
(a) 45°
(b) 60°
(c) 75°
(d) 30°
A beaker contains water up to a height h1 and kerosene
of height h2 above water so that the total height of
(water + kerosene) is (h1 + h2). Refractive index of water is
m1and that of kerosene is m2. The apparent shift in the
position of the bottom of the beaker when viewed from
above is
[2011 RS]
æ
æ
1ö
1 ö
(a) ç1 + m ÷ h1 - ç1 + m ÷ h2
è
1ø
è
2ø
æ
æ
1 ö
1 ö
(b) ç 1 - m ÷ h1 + ç 1 - m ÷ h2
è
1ø
è
2ø
contained in a circular horizon. If the refractive index of
4
water is and the fish is 12 cm below the surface, the
3
radius of this circle in cm is
[2005]
(a)
36
7
(b)
36 7
(c)
4 5
(d) 36 5
31. Consider telecommunication through optical fibres. Which
of the following statements is not true?
[2003]
(a) Optical fibres can be of graded refractive index
(b) Optical fibres are subject to electromagnetic
interference from outside
(c) Optical fibres have extremely low transmission loss
(d) Optical fibres may have homogeneous core with a
suitable cladding.
32. Which of the following is used in optical fibres? [2002]
(a) total internal reflection (b) scattering
(c) diffraction
(d) refraction.
TOPIC 3
Refraction at Curved Surface
Lenses and Power of Lens
33. A point like object is placed at a distance of 1 m in front of
a convex lens of focal length 0.5 m. A plane mirror is placed
at a distance of 2 m behind the lens. The position and
nature of the final image formed by the system is :
[Sep. 06, 2020 (I)]
(a) 2.6 m from the mirror, real
(b) 1 m from the mirror, virtual
(c) 1 m from the mirror, real
(d) 2.6 m from the mirror, virtual
P-393
Ray Optics and Optical Instruments
34.
A double convex lens has power P and same radii of curvature R of both the surfaces. The radius of curvature of a
surface of a plano-convex lens made of the same material
with power 1.5 P is :
[Sep. 06, 2020 (II)]
R
3R
R
(c)
(d)
2
2
3
35. For a concave lens of focal length f, the relation between
object and image distances u and v, respectively, from its pole
can best be represented by (u = v is the reference line) :
(a) 2R
(b)
R
m1 - m2
(b)
2R
m1 - m2
(c)
2R
2(m1 - m2 )
(d)
R
2 - (m1 - m2 )
=
u
(a)
40. The graph shows how the magnification m produced by a
thin lens varies with image distance v. What is the focal
length of the lens used ?
[10 Apr. 2019 II]
u
=
v
f
u
v
f
(a)
v
[Sep. 05, 2020 (I)]
v
f
39. One plano-convex and one plano-concave lens of same
radius of curvature ‘R’ but of different materials are joined
side by side as shown in the figure. If the refractive index
of the material of 1 is m1 and that of 2 is m2, then the focal
length of the combination is :
[10 Apr. 2019 I]
(b)
u
=
v
f
u
v
f
(c)
u
(b)
b2c
a
(c)
a
c
(d)
b
c
u
=
41. A convex lens of focal length 20 cm produces images of
the same magnification 2 when an object is kept at two
distances x1 and x2 (x1 > x2) from the lens. The ratio of x1
and x2 is:
[9 Apr. 2019 II]
(d)
36.
b2
ac
v
f
v
f
(a)
u
f
The distance between an object and a screen is 100 cm. A
lens can produce real image of the object on the screen for
two different positions between the screen and the object.
The distance between these two positions is 40 cm. If the
æ N ö
power of the lens is close to ç
D where N is an
è 100 ÷ø
integer, the value of N is ___________.
[NA Sep. 04, 2020 (II)]
(a) 2 : 1
(b) 3 : 1
(c) 5 : 3
(d) 4 : 3
42. A thin convex lens L (refractive index = 1.5) is placed on a
plane mirror M. When a pin is placed at A, such that OA =
18 cm, its real inverted image is formed at A itself, as shown
in figure. When a liquid of refractive index µi is put between
the lens and the mirror, the pin has to be moved to A’, such
that OA’ = 27 cm, to get its inverted real image at A’ itself.
The value of µi will be:
[9 Apr. 2019 II]
37. A point object in air is in front of the curved surface of a
plano-convex lens. The radius of curvature of the curved
surface is 30 cm and the refractive index of the lens material
is 1.5, then the focal length of the lens (in cm)
is__________.
[NA 8 Jan. 2020 I]
38. A thin lens made of glass (refractive index = 1.5) of focal
length f = 16 cm is immersed in a liquid of refractive index
1.42. If its focal length in liquid is fl ,then the ratio fl /f is
closest to the integer:
[7 Jan. 2020 II]
(a) 1
(b) 9
(c) 5
(d) 17
(a)
4
3
(b)
3
2
(c)
3
(d)
2
P-394
Physics
43. An upright object is placed at a distance of 40 cm in front
of a convergent lens of focal length 20 cm. A convergent
mirror of focal length 10 cm is placed at a distance of 60 cm
on the other side of the lens. The position and size of the
final image will be :
[8 April 2019 I]
(a) 20 cm from the convergent mirror, same size as the object
(b) 40 cm from the convergent mirror, same size as the object
(c) 40 cm from the convergent lens, twice the size of the
object
(d) 20 cm from the convergent mirror, twice the size of the
object
44. A convex lens (of focal length 20 cm) and a concave mirror,
having their principal axes along the same lines, are kept
80 cm apart from each other. The concave mirror is to the
right of the convex lens. When an object is kept at a
distance of 30 cm to the left of the convex lens, its image
remains at the same position even if the concave mirror is
removed. The maximum distance of the object for which
this concave mirror, by itself would produce a virtual image
would be :
[8 Apr. 2019 II]
(a) 30 cm
(b) 25 cm (c) 10 cm
(d) 20 cm
45. What is the position and nature of image formed by lens
combination shown in figure? (f1, f2 are focal lengths)
[12 Jan. 2019 I]
2 cm
A
20 cm
B
f1 = + 5 cm
f2 = –5 cm
(a) 70 cm from point B at left; virtual
(b) 40 cm from point B at right; real
20
cm from point B at right, real
3
(d) 70 cm from point B at right; real
Formation of real image using a biconvex lens is shown
below :
[12 Jan. 2019 II]
(c)
46.
2f
2f
47.
f
screen
f
If the whole set up is immersed in water without disturbing
the object and the screen positions, what will one observe
on the screen ?
(a) Image disappears
(b) Magnified image
(c) Erect real image
(d) No change
A plano-convex lens (focal length f2, refractive index µ2,
radius of curvature R) fits exactly into a plano-concave
lens (focal length f 1 , refractive index µ1 , radius of
curvature R). Their plane surfaces are parallel to each other.
Then, the focal length of the combination will be :
[12 Jan. 2019 II]
(a) f1 – f2
(b)
R
µ2 - µ1
2 f1 f 2
(d) f1 + f2
f1 + f 2
48. An object is at a distance of 20 m from a convex lens of
focal length 0.3 m. The lens forms an image of the object. If
the object moves away from the lens at a speed of 5m/s,
the speed and direction of the image will be :
[11 Jan. 2019 I]
(a) 2.26 × 10–3 m/s away from the lens
(b) 0.92 × 10–3 m/s away from the lens
(c) 3.22 × 10–3 m/s towards the lens
(d) 1.16 × 10–3 m/s towards the lens
49. A plano convex lens of refractive index m1 and focal
length f1 is kept in contact with another plano concave
lens of refractive index m2 and focal length f2 If the
radius of curvature of their spherical faces is R each
and f1 = 2f2, then m1 and m2 are related as:
[10 Jan. 2019 I]
(a) m1 + m2 = 3
(b) 2m1 – m2 = 1
(c)
(c) 3m2 – 2m1 = 1
(d) 2m2 – m1 = 1
50. The eye can be regarded as a single refracting surface.
The radius of curvature of this surface is equal to that
of cornea (7.8 mm). This surface separates two media of
refractive indices 1 and 1.34. Calculate the distance from
the refracting surface at which a parallel beam of light
will come to focus.
[10 Jan. 2019 II]
(a) 1 cm
(b) 2 cm
(c) 4.0 cm
(d) 3.1 cm
51. A convex lens is put 10 cm from a light source and it
makes a sharp image on a screen, kept 10 cm from the lens.
Now a glass block (refractive index 1.5) of 1.5 cm thickness
is placed in contact with the light source. To get the sharp
image again, the screem is shifted by a distance d. Then d
is:
[9 Jan. 2019 I]
(a) 1.1 cm away from the lens
(b) 0
(c) 0.55 cm towards the lens
(d) 0.55 cm away from the lens
52. A planoconvex lens becomes an optical system of 28 cm
focal length when its plane surface is silvered and
illuminated from left to right as shown in Fig-A. If the same
lens is instead silvered on the curved surface and
illuminated from other side as in Fig. B, it acts like an optical
system of focal length 10 cm. The refractive index of the
material of lens if:
[Online April 15, 2018]
Fig. A
(a) 1.50
(b) 1.55
Fig. B
(c) 1.75
(d) 1.51
P-395
Ray Optics and Optical Instruments
53.
54.
55.
A convergent doublet of separated lenses, corrected for
spherical aberration, has resultant focal length of 10cm.
The separation between the two lenses is 2cm. The focal
lengths of the component lenses [Online April 15, 2018]
(a) 18cm, 20cm
(b) 10cm, 12cm
(c) 12cm, 14cm
(d) 16cm, 18cm
In an experiment a convex lens of focal length 15 cm is
placed coaxially on an optical bench in front of a convex
mirror at a distance of 5 cm from it. It is found that an
object and its image coincide, if the object is placed at a
distance of 20 cm from the lens. The focal length of the
convex mirror is :
[Online April 9, 2017]
(a) 27.5 cm (b) 20.0 cm (c) 25.0 cm (d) 30.5 cm
A hemispherical glass body of radius 10 cm and refractive
index 1.5 is silvered on its curved surface. A small air bubble
is 6 cm below the flat surface inside it along the axis. The
position of the iamge of the air bubble made by the mirror
is seen :
[Online April 10, 2016]
10 cm
f1 and f2 are close to :
(a) f1 = 7.8 cm
(b) f1 = 12.7 cm
(c) f1 = 15.6 cm
(d) f1 = 7.8 cm
f2 = 12.7 cm
f2 = 7.8 cm
f2 = 25.4 cm
f2 = 25.4 cm
58. A thin convex lens of focal length ‘f’ is put on a plane
mirror as shown in the figure. When an object is kept at
a distance ‘a’ from the lens - mirror combination, its
a
image is formed at a distance
in front of the
3
combination. The value of ‘a’ is :
[Online April 11, 2015]
6cm
O
Silvered
56.
(a) 14 cm below flat surface
(b) 20 cm below flat surface
(c) 16 cm below flat surface
(d) 30 cm below flat surface
A convex lens, of focal length 30 cm, a concave lens of
focal length 120 cm, and a plane mirror are arranged as
shown. For an object kept at a distance of 60 cm from the
convex lens, the final image, formed by the combination, is
a real image, at a distance of :
[Online April 9, 2016]
(a) 3f
(b)
3
f
2
(c) f
(d) 2f
3ö
æ
59. A thin convex lens made from crown glass ç m = ÷ has
2ø
è
focal length f. When it is measured in two different
4
5
and , it has the focal
3
3
lengths f1 and f2 respectively. The correct relation between
the focal lengths is:
[2014]
liquids having refractive indices
(a) f1 = f2 < f
(b) f1 > f and f2 becomes negative
|Focal length| |Focal length|
= 30 cm
= 120 cm
60cm
20cm
57.
70 cm
(a) 60 cm from the convex lens
(b) 60 cm from the concave lens
(c) 70 cm from the convex lens
(d) 70 cm from the concave lens
To find the focal length of a convex mirror, a student records
the following data :
[Online April 9, 2016]
Object Pin Convex Lens Convex Mirror Image Pin
22.2cm
32.2cm
45.8cm
71.2cm
The focal length of the convex lens is f1 and that of mirror
is f2. Then taking index correction to be negligibly small,
(c) f2 > f and f1 becomes negative
(d) f1 and f2 both become negative
60. The refractive index of the material of a concave lens is m.
It is immersed in a medium of refractive index m1. A parallel
beam of light is incident on the lens. The path of the
emergent rays when m1 > m is:
[Online April 12, 2014]
m
m1
(a)
m1
m
P-396
m1
Physics
m1
(b)
m
m1
m1
(c)
m
m1
(c) remain same
(d) does not depend on colour of light
66. In an optics experiment, with the position of the object
fixed, a student varies the position of a convex lens and
for each position, the screen is adjusted to get a clear
image of the object. A graph between the object distance u
and the image distance v, from the lens, is plotted using
the same scale for the two axes. A straight line passing
through the origin and making an angle of 45° with the xaxis meets the experimental curve at P. The coordinates of
P will be
[2009]
æ f fö
(a) ç , ÷
(b) ( f, f )
è 2 2ø
(c) ( 4 f, 4 f )
(d) ( 2 f, 2 f )
67. A student measures the focal length of a convex lens by
putting an object pin at a distance ‘u’ from the lens and
measuring the distance ‘v’ of the image pin. The graph
between ‘u’ and ‘v’ plotted by the student should look
like
[2008]
m1
(d)
v(cm)
(a)
(b)
O
61.
62.
63.
64.
65.
An object is located in a fixed position in front of a screen.
Sharp image is obtained on the screen for two positions of
a thin lens separated by 10 cm. The size of the images in
two situations are in the ratio 3 : 3. What is the distance
between the screen and the object?
[Online April 11, 2014]
(a) 124.5 cm
(b) 144.5 cm
(c) 65.0 cm
(d) 99.0 cm
Diameter of a plano-convex lens is 6 cm and thickness at
the centre is 3 mm. If speed of light in material of lens is
2× 108 m/s, the focal length of the lens is
[2013]
(a) 15 cm (b) 20 cm (c) 30 cm (d) 10 cm
The image of an illuminated square is obtained on a screen
with the help of a converging lens. The distance of the
square from the lens is 40 cm. The area of the image is 9
times that of the square. The focal length of the lens is :
[Online April 22, 2013]
(a) 36 cm (b) 27 cm (c) 60 cm (d) 30 cm
An object at 2.4 m in front of a lens forms a sharp image on
a film 12 cm behind the lens. A glass plate 1 cm thick, of
refractive index 1.50 is interposed between lens and film
with its plane faces parallel to film. At what distance
(from lens) should object shifted to be in sharp focus of
film?
[2012]
(a) 7.2 m
(b) 2.4 m
(c) 3.2 m
(d) 5.6 m
When monochromatic red light is used instead of blue
light in a convex lens, its focal length will
[2011 RS]
(a) increase
(b) decrease
v(cm)
O
u(cm)
u(cm)
v(cm)
v(cm)
(c)
O
u(cm)
(d)
O
u(cm)
68. Two lenses of power –15 D and +5 D are in contact with
each other. The focal length of the combination is [2007]
(a) + 10 cm
(b) – 20 cm
(c) – 10 cm
(d) + 20 cm
69. A thin glass (refractive index 1.5) lens has optical power of
– 5 D in air. Its optical power in a liquid medium with
refractive index 1.6 will be
[2005]
(a) – 1D
(b) 1 D
(c) – 25 D (d) 25 D
70. A plano convex lens of refractive index 1.5 and radius of
curvature 30 cm. Is silvered at the curved surface. Now this
lens has been used to form the image of an object. At what
distance from this lens an object be placed in order to have
a real image of size of the object
[2004]
(a) 60 cm (b) 30 cm (c) 20 cm (d) 80 cm
TOPIC 4 Prism and Dispersion of Light
71. The surface of a metal is illuminated alternately with
photons of energies E1 = 4 eV and E2 = 2.5 eV respectively.
The ratio of maximum speeds of the photoelectrons emitted
in the two cases is 2. The work function of the metal in (eV)
is _______________.
[NA Sep. 05, 2020 (II)]
P-397
Ray Optics and Optical Instruments
72.
The variation of refractive index of a crown glass thin prism
with wavelength of the incident light is shown. Which of
the following graphs is the correct one, if Dm is the angle
of minimum deviation ?
[11 Jan. 2019, I]
1.535
1.530
n2
1.525
1.520
1.515
1.510
400
500
600
700
l (nm)
Dm
(a)
Dm
400 500 600 700
l (nm)
73. A monochromatic light is incident at a certain angle on an
equilateral triangular prism and suffers minimum deviation.
3,
[11 Jan. 2019 II]
If the refractive index of the material of the prism is
then the angle of incidence is :
(a) 90°
(b) 30°
(c) 60°
(d) 45°
74. A ray of light is incident at an angle of 60° on one face of
a prism of angle 30°. The emergent ray of light makes an
angle of 30° with incident ray. The angle made by the
emergent ray with second face of prism will be:
[Online April 16, 2018]
(a) 30°
(b) 90°
(c) 0°
(d) 45°
75. In an experiment for determination of refractive index of
glass of a prism by i – d, plot it was found thata ray incident
at angle 35°, suffers a deviation of 40° and that it emerges
at angle 79°. In that case which of the following is closest
to the maximum possible value of the refractive index?
[2016]
(a) 1.7
(b) 1.8
(c) 1.5
(d) 1.6
76. Monochromatic light is incident on a glass prism of angle
A. If the refractive index of the material of the prism is µ,
a ray, incident at an angle q, on the face AB would get
transmitted through the face AC of the prism provided :
[2015]
A
q
(b)
B
400 500 600 700
l (nm)
Dm
C
é
æ
æ 1 öù
(a) q > cos -1 êµsin ç A + sin -1 ç ÷ ú
è µ ø ûú
è
ëê
é
æ
æ 1 öù
(b) q < cos-1 êµsin ç A + sin -1 ç ÷ ú
è µ ø ûú
è
ëê
æ
-1 é
-1 æ 1 ö ù
(c) q > sin êµsin ç A - sin ç ÷ ú
è µ ø ûú
è
ëê
(c)
400 500 600 700
l (nm)
Dm
æ
-1 é
-1 æ 1 ö ù
(d) q < sin êµsin ç A - sin ç ÷ ú
è µ ø ûú
è
ëê
77. The graph between angle of deviation (d) and angle of
incidence (i) for a triangular prism is represented by[2013]
d
(d)
(a)
400 500 600 700
l (nm)
o
i
P-398
Physics
d
(b)
o
i
d
(c)
o
i
d
(d)
o
78.
i
A beam of light consisting of red, green and blue colours
is incident on a right-angled prism on face AB. The
refractive indices of the material for the above red, green
and blue colours are 1.39, 1.44 and 1.47 respectively. A
person looking on surface AC of the prism will see
[Online May 26, 2012]
A
80. Which of the following processes play a part in the
formation of a rainbow?
[Online May 7, 2012]
(i) Refraction
(ii) Total internal reflection
(iii) Dispersion
(iv) Interference
(a) (i), (ii) and (iii)
(b) (i) and (ii)
(c) (i), (ii) and (iv)
(d) (iii) and (iv)
81. The refractive index of a glass is 1.520 for red light and
1.525 for blue light. Let D1 and D2 be angles of minimum
deviation for red and blue light respectively in a prism of
this glass. Then,
[2006]
(a) D1 < D2
(b) D1 = D2
(c) D1 can be less than or greater than D2 depending
upon the angle of prism
(d) D1 > D2
82. A light ray is incident perpendicularly to one face of a 90°
prism and is totally internally reflected at the glass-air
interface. If the angle of reflection is 45°, we conclude that
the refractive index n
[2004]
45°
45°
45°
45°
B
79.
C
(a) no light
(b) green and blue colours
(c) red and green colours
(d) red colour only
A glass prism of refractive index 1.5 is immersed in water
4
(refractive index ) as shown in figure. A light beam
3
incident normally on the face AB is totally reflected to
reach the face BC, if
[Online May 19, 2012]
B
A
q
C
(a)
(c)
5
9
8
sin q >
9
sin q >
(b)
(d)
2
3
1
sin q >
3
sin q >
(a)
n>
(c)
n<
1
2
1
2
(b)
n> 2
(d)
n< 2
TOPIC 5 Optical Instruments
83. A compound microscope consists of an objective lens of
focal length 1 cm and an eye piece of focal length 5 cm with
a separation of 10 cm.
The distance between an object and the objective lens, at
n
which the strain on the eye is minimum is
cm.
40
The value of n is ______.
[NA Sep. 05, 2020 (I)]
84. In a compound microscope, the magnified virtual image is
formed at a distance of 25 cm from the eye-piece. The focal
length of its objective lens is 1 cm. If the magnification is
100 and the tube length of the microscope is 20 cm, then the
focal length of the eye-piece lens (in cm) is __________.
[NA Sep. 04, 2020 (I)]
85. The magnifying power of a telescope with tube length 60
cm is 5. What is the focal length of its eye piece?
[8 Jan. 2020 I]
(a) 20 cm
(b) 40 cm
(c) 30 cm
(d) 10 cm
P-399
Ray Optics and Optical Instruments
86. If we need a magnification of 375 from a compound
microscope of tube length 150 mm and an objective of
focal length 5 mm, the focal length of the eye-piece, should
be close to:
[7 Jan. 2020 I]
(a) 22mm
(b) 12mm
(c) 2 mm
(d) 33mm
87. An observer looks at a distant tree of height 10 m with a
telescope of magnifying power of 20. To the observer the
tree appears :
[2016]
(a) 20 times taller
(b) 20 times nearer
(c) 10 times taller
(d) 10 times nearer
88. To determine refractive index of glass slab using a
travelling microscope, minimum number of readings
required are :
[Online April 10, 2016]
(a) Two
(b) Four
(c) Three (d) Five
89. A telescope has an objective lens of focal length 150 cm
and an eyepiece of focal length 5 cm. If a 50 m tall tower at
a distance of 1 km is observed through this telescope in
normal setting, the angle formed by the image of the tower
is q, then q is close to :
[Online April 10, 2015]
(a) 30°
(b) 15°
(c) 60°
(d) 1°
90. In a compound microscope, the focal length of objective
lens is 1.2 cm and focal length of eye piece is 3.0 cm. When
object is kept at 1.25 cm in front of objective, final image is
formed at infinity. Magnifying power of the compound
microscope should be:
[Online April 11, 2014]
(a) 200
(b) 100
(c) 400
(d) 150
91. The focal lengths of objective lens and eye lens of a
Galilean telescope are respectively 30 cm and 3.0 cm.
telescope produces virtual, erect image of an object
situated far away from it at least distance of distinct vision
from the eye lens. In this condition, the magnifying power
of the Galilean telescope should be:
[Online April 9, 2014]
(a) + 11.2 (b) – 11.2 (c) – 8.8
(d) + 8.8
92. This question has Statement-1 and Statement-2. Of the
four choices given after the Statements, choose the one
that best describes the two Statements.
Statement 1: Very large size telescopes are reflecting
telescopes instead of refracting telescopes.
Statement 2: It is easier to provide mechanical support to
large size mirrors than large size lenses.
[Online April 23, 2013]
(a) Statement-1 is true and Statement-2 is false.
(b) Statement-1 is false and Statement-2 is true.
93.
94.
95.
96.
97.
98.
(c) Statement-1 and statement-2 are true and Statement2 is correct explanation for statement-1.
(d) Statements-1 and statement-2 are true and Statement2 is not the correct explanation for statement-1.
The focal length of the objective and the eyepiece of a
telescope are 50 cm and 5 cm respectively. If the telescope
is focussed for distinct vision on a scale distant 2 m from
its objective, then its magnifying power will be:
[Online April 22, 2013]
(a) – 4
(b) – 8
(c) + 8
(d) – 2
–2
A telescope of aperture 3 × 10 m diameter is focused on
a window at 80 m distance fitted with a wire mesh of spacing
2 × 10–3 m. Given: l = 5.5 × 10–7 m, which of the following
is true for observing the mesh through the telescope?
[Online May 26, 2012]
(a) Yes, it is possible with the same aperture size.
(b) Possible also with an aperture half the present
diameter.
(c) No, it is not possible.
(d) Given data is not sufficient.
We wish to make a microscope with the help of two positive
lenses both with a focal length of 20 mm each and the
object is positioned 25 mm from the objective lens. How
far apart the lenses should be so that the final image is
formed at infinity?
[Online May 12, 2012]
(a) 20mm
(b) 100 mm
(c) 120 mm
(d) 80mm
An experment is performed to find the refractive index of
glass using a travelling microscope. In this experiment
distances are measured by
[2008]
(a) a vernier scale provided on the microscope
(b) a standard laboratory scale
(c) a meter scale provided on the microscope
(d) a screw gauge provided on the microscope
The image formed by an objective of a compound
microscope is
[2003]
(a) virtual and diminished
(b) real and diminished
(c) real and enlarged
(d) virtual and enlarged
An astronomical telescope has a large aperture to [2002]
(a) reduce spherical aberration
(b) have high resolution
(c) increase span of observation
(d) have low dispersion
P-400
1.
Physics
Magnification, m =
2.
Let angle between the two mirrors be q.
Ray PQ P mirror M1 and Rs P mirror M2
\ M1Rs = ÐORQ = Ð M1OM2= q
Similarly, ÐM2QP = ÐOQR = Ð M2OM1= q
(1)
Distance of object, u = – 30 cm
Distance of image, v = 10 cm
-v (-10) 1
=
=
u
-30
3
\
1
Speed of image = m2 × speed of object = ´ 9 = 1 cm s–1
9
(d) Object is placed beyond radius of curvature (R) of
concave mirror hence image formed is real, inverted and
diminished or unmagnified.
6.
C hi
3.
F
Image
(c) Using mirror formula, magnification is given by
f
–1
=
u – f 1– u
f
At focus magnification is ¥
And at u = 2f, magnification is 1.
Hence graph (d) correctly depicts ‘m’ versus distance of
object ‘x’ graph.
P
hi < h0
v
Þ v = -5u
u
1 1 1
Using + =
v u f
(c)
+5 = -
7.
(c) When object is at 8 cm
f ´ u 5´8
40
cm
=
=u -f 8-5
3
When object is at 12 cm
1
1
1
+ =
-5u u 0.4
\ u = 0.32 m
Image V1 =
or
Image V2 =
3d
2
4.
180°
= 60°
3
m=
h0
O
In DORQ, 3q = 180° Þ q =
S
L
(c)
L
f ´ u 5 ´ 12
60
=
=cm
u - f 12 - 5
7
40 60 100
cm
=
3
7
21
So A, C and D are correct statements.
(c) Convex morror is used as a shaving mirror.
Separation = V1 - V2 =
8.
3d
2
L
O
Total distance =
5.
(b)
10 cm
3d 3d
+
= 3d
2
2
15 cm
M1
q
R
From question : v = 15 cm, u = – 10 cm
Radius of curvature, R = 2f = ?
P
q
O
q
Using mirror formula,
q
q
Q
M2
1 1 1
+ =
v u f
1
1
1
+
=
Þ f = – 30 cm
15 (-10) f
Therefore radius of curvature,
R = 2f = – 60 cm
P-401
Ray Optics and Optical Instruments
9.
(a) From mirror formula
1 1 1
+ =
v u f
Differentiating the above equation, we get
13. (90.00)
In the figure, QR is the reflected ray and QS is refracted
ray. CQ is normal.
dv
v 2 æ du ö
=- 2ç ÷
dt
u è dt ø
Q
Also,
v
f
=
u u– f
r'
P
2
æ f ö du
dv
= -ç
Þ
è u - f ø÷ dt
dt
r
60°
2
10.
11.
12.
m= 3
dv 1
=
m/s
dt 15
(b) The number of images formed is given by
360
n=
-1
q
360
-1 = 3
Þ
q
360°
Þ q=
= 90°
4
(a) When two plane mirrors are inclined at each other at
an angle q then the number of the images (n) of a point
object kept between the plane mirrors is
360°
- 1,
n=
q
360°
(if
is even integer)
q
360°
-1 = 5
\Number of images formed =
60º
(158)
15
From figure, sin i =
and sin r = sin 45°
2
15 + 30 2
From Snell's law, m ´ sin i = 1 ´ sin r
Þm´
15
2
15 + 30
2
= 1 ´ sin 45° =
r
30 cm
1
Apply Snell's law at P
1sin 60° = 3 sin r
1
Þ r = 30°
2
From geometry, CP = CQ
Þ sin r =
\ r ' = 30°
Again apply snell's law at Q,
3 sin r ' = 1sin e
3
= sin e Þ e = 60°
2
From geometry
(As angles lies on a straight line)
r '+ q + e = 180°
Þ
Þ 30° + q + 60° = 180° Þ q = 90°.
14. (d) Apparent depth,
h
m1 = 2
m2 = 2 2
t1 t2
h
h
3h
3h 2
+
=
+
=
=
m1 m2
4
2 2 2 2 2
15. (c) Given,
Refractive index, m =
45° 15 cm
P
15 cm
45 cm
4
3
4
sin q = 1sin 90°
3
3
Þ sinq =
4
30 cm
1
2 = 158 ´ 10 -2 = N
\m =
15
100
1125
Hence, value of N ; 158 .
h
D=
2
15 cm
i
q
S
r' R
reflected ray
C
dv æ 0.2 ö
=ç
Þ
÷ ´ 15
dt è 2.8 - 0.2 ø
Þ
e
cos q =
7
4
Solid angle, W = 2p(1– cosq) = 2p(1– 7 / 4)
Fraction of energy transmitted
P-402
Physics
2p(1– cos q) 1 – 7 / 4
=
= 0.17
4p
2
Percentage of light emerges out of surface
= 0.17 × 100 = 17%
(b) Here, from question, relative permittivity
=
16.
er =
19. (d) From the given figure
As sin 60o = m sin 30o
e
= 3 Þ e = 3 e0
e0
4
m 4
Relative permeability m r = m = 3 Þ m = 3 m0
0
\ me = 4m0 e0
m 0 e0 v 1 æ
1
= = çQ c =
c 2 çè
me
m 0 e0
ö
÷
÷
ø
c
17.
Critical angle, qc = 30°
Þ
x = 20 µm × cot q
v=
\ Number of reflections =
20
35
+ 5 = cm
=
3
3
Using,
18.
0.64
= 0.49 » 0.5
1.31
q = 30°
1 1
1
+
=
v -5 -20
20
cm
3
Distance of this image from water surface
\
2b
Þ sin q =
1 1 1
+ =
v u f
\
= 3
= 2a + 2b
3
20. (d) Using Snell’s law of refraction,
1 × sin 40° = 1.31 sin q
(c) If v is the distance of image formed by mirror, then
or
sin 30o
= 2a + ( 3)
1 1
Þ sin qc = =
n 2
\
sin 60o
a
= cos 60o Þ AO = 2a
AO
b
2b
= cos30o Þ BO =
BO
3
Optical path length = AO + mBO
4
´3 = 2
3
n = mr er =
1
And n = sin q
Þm=
=
RD
=µ
AD
AD = d =
21.
RD (35 / 3)
=
= 8.8 cm
m
1.33
(c) Using, sin qmax = µ1 µ22 - µ12 =
æ 2
ö
-1 ç µ2
÷
sin
1
or qmax =
ç µ12
÷
è
ø
æ 2
ö
-1 ç µ2
÷
sin
1
For T1 R, q <
ç µ12
÷
è
ø
2 ´ 10 6
µ12
-1
20 ´10 ´ cot q
= 57735 » 57000
20 ´ 3
(c) As 4% of light gets reflected, so only (100 – 4 = 96%)
of light comes after refraction so,
Prefracted =
µ22
2
-6
96
PI
100
Þ K 2 A t2 =
Þ r2 A t2 =
Þ A 2t =
At
96
K1Ai2
100
96
r1Ai2
100
96 1
´ ´ (30) 2
100 3
2
64
´ (30)2 = 24
100
P-403
Ray Optics and Optical Instruments
22.
(d)
Incident ray
denser
A A
(µD)
90°
rarer
(µR)
r
m R sin i
..... (i)
=
m D sin r
Q Ði = A and Ðr = (90° - A)
m
We also know that, sin qC = R
mD
sin A
n
From eq (i), sin qC =
sin(90° - A)
sin A
sin q C =
cos A
sin qC = tan A
or
A = tan–1 (sin qC)
4
23. (d) Given, µ =
3
h = 15 cm
R= ?
From Snell's law,
Þ
Þ
or,
or,
24.
(c)
1
sin C = =
m
h
3
=
2
2
4
R +h
2
2
2
O
16 R = 9 R + 9 h
7R2 = 9 h2
3
3
R=
h=
´ 15cm
7
7
Real depth = 5 cm + 1cm = 6 cm
Water
m = 1.33
R
\ (sin 90° - r) =
Þ cos r =
5cm
1cm
5
1
=
+
1.33 1.5
; 3.8 + 0.7 ; 4.5 cm
\ Shift = 6 cm – 4.5 cm @ 1.5 cm
(d) At point A by Snell’s law
sin 45°
1
m=
Þ sin r =
… (i)
sin r
m 2
At point B, for total internal reflection,
sin i1 =
1
m
1
m
1
m
… (ii)
Now cos r = 1 - sin 2 r = 1 R
=
1
2m 2
2m 2 - 1
… (iii)
2m2
From eqs (ii) and (iii)
1
2m 2 - 1
=
m
2m 2
Squaring both sides and then solving, we get
3
2
26. (b) Velocity of light in medium
m=
3cm
3 ´ 10-2 m
=
= 1.5 m/s
0.2ns 0.2 ´ 10-9 s
Refractive index of the medium
Vmed =
d1 d2
+
+ ....
Apparent depth =
m1 m2
25.
A
i1
90° m
From figure, i1 = 90° - r
m=
m = 1.5
Glass
i2
B
C
sin 90°
=m
sin C
45°
Air
Vair
3 ´108
=
= 2 m/s
Vmed
1.5
As µ =
1
sin C
\ sin C =
1 1
= = 30°
m 2
Condition of TIR is angle of incidence i must be greater
than critical angle. Hence ray will suffer TIR in case of (B)
(i = 40° > 30°) only.
27. (a) As refractive index for z > 0 and z £ 0 is different xy
plane should be the boundry between two media.
Angle of incidence is given by
cos (p–i) =
(6
)
3iˆ + 8 3 ˆj - 10 kˆ .kˆ
20
P-404
Physics
Applying Snell’s Law for air and medium inside the cylinder
at P we get
sin q
n=
sin a
1
2
Þ Ð i = 60 °
From Snell's law,
sin i u 2
=
sin r u1
– cos i = –
Þ
Þ sin q = n ´ sin a = n2 - 1 ; [from (i)]
2
æ 2 ö
4
1
-1 =
\ sin q = ç ÷ - 1 =
è 3ø
3
3
sin i
3
=
sin r
2
æ 1 ö
q = sin -1 ç
÷
è 3ø
30. (a) From the figure it is clear that
or
Þ 2 sin i = 3sin r
Þ
2 sin 60° =
3
AB
OA
Þ R = OA tan qc
tan q c =
3
Þ 2´
= 3 sin r
2
ÞÐ r = 45°
28.
Þ R=
(b)
m2
Kerosene
h2
m1
Water
h1
Þ R=
OA sin qc
cos qc
OA sin qc
1 - sin 2 qc
Þ tan qc =
Apparent shift of the bottom due to water,
é
1ù
Dh1 = h1 ê1 - ú
ë m1 û
Apparent shift of the bottom due to kerosene, Dh2
Q sin qc =
sin qc
R
=
12
1 - sin 2 q c
1 3
=
m 4
Þ tan qc =
é
1 ù
= h2 ê1 - ú
ë m2 û
Thus, total apparent shift :
= Dh1 + Dh2
3
16 - 9
=
3
7
=
R
A
qc qc
æ
æ
1ö
1 ö
= h1 ç1 - ÷ + h2 ç1 - ÷
m
m
è
è
1ø
2ø
R
12
B
12 cm
O
Q
29.
(c)
q
a
n
90 – a
Þ R=
P
Applying Snell’s law
for medium inside the cylinder and air at Q we get
n=
sin 90°
1
=
sin(90º -a ) cos a
\ cos a =
1
n
\ sin a = 1 - cos 2 a = 1 -
1
n2
=
n2 - 1
...(i)
n
36
cm
7
31. (b) Optical fibres form a dielectric wave guide and are free
from electromagnetic interference or radio frequency
interference. There is extremely low transmission loss in
optical fibre.
32. (a) In an optical fibre, light is sent through the fibre
without any loss by the phenomenon of total internal
reflection. Total internal reflection of light waves confine
the light rays inside the optical fiber.
33. (d) Focal length of the convex lens, f = 0.5 m
Object is at 2f so, image (I1) will also be at 2f.
Image of I1 i.e., I2 will be 1 m behind mirror.
Now I2 will be object for lens.
P-405
Ray Optics and Optical Instruments
1m
1m
1m
1m
I1
Object
f = 0.5m
I2
1 1 1
- =
v u f
1 1 1
1
1
3
= + =
+
or v = = 0.6 m
v f u +0.5 -3
5
Using lens formula,
Hence, distance of image from mirror
= 2 + 0.6 = 2.6 m and real.
(d) Given, using lens maker's formula
1
1 ö
æ1
= (m - 1) ç è R - R ÷ø
f
...(i)
For plano convex lens,
R1 = R ', R2 = ¥
Using lens maker's formula again, we have
æ 1 1ö
1.5 P = ( m - 1) ç - ÷
è R ' ¥ø
...(ii)
3
m -1
Þ P=
2
R'
From (i) and (ii),
35.
3 R'
R
=
Þ R' =
2 2R
3
(d) From lens formula,
1 1 1
uf
- = Þv=
v u f
u+ f
Case-I : If v = u Þ f + u = f Þ u = 0
36.
Case-II : If u = ¥ then v = f.
Hence, correct u versus v graph, that satisfies this
condition is (a).
(476.19)
Given,
Distance between an object and screen, D = 100 cm
Distance between the two position of lens, d = 40 cm
Focal length of lens,
f =
æ 1
1
1 ö
= (m –1) ç –
÷
f
è R1 R2 ø
For plano-convex lens
R1 ® ¥ then R2 = – R
R
30
=
= 60 cm
m –1 1.5 –1
38. (b) Using lens maker’s formula
Here, R1 = R2 = R (For double convex lens)
1
2
Þ P = = (m - 1)
f
R
37. (60) Given : m = 1.5; Rcurved = 30 cm
Using, Lens-maker formula
\f =
æ 1
1
1ö
= (k - 1) ç - ÷
f
è R1 R2 ø
\
1 100
N
=
=
f
21 100
\ N = 476.19.
Mirror
\ u = (-1) + ( -1) + ( -1) = -3 m
34.
Power, P =
D 2 - d 2 100 2 - 402 (100 + 40)(100 - 40)
=
=
= 21 cm
4D
4(100)
4(100)
öé 1
1 æ mg
1 ù
= çç
–1÷÷ ê –
ú
f è ma
R
R
2û
øë 1
Here, mg and ma are the refractive index of glass and air
respectively
æ 1
1
1 ö
= (1.5 –1) ç –
÷
f
è R1 R2 ø
When immersed in liquid
Þ
...(i)
öæ 1
1 æ mg
1 ö
= çç
–1÷÷ ç –
÷
fl è m l
ø è R1 R2 ø
[Here, ml = refractive index of liquid]
1 æ 1.5
1 ö
öæ 1
=ç
–1÷ ç –
÷
fl è 1.42 ø è R1 R2 ø
Dividing (i) by (ii)
Þ
Þ
...(ii)
fl (1.5 –1)1.42 1.42 142
=
=
=
»9
f
0.08
0.16 16
39. (a) Focal length of plano-convex lens1
1 ö m1 –1
æ1
= ( m1 –1) ç –
÷=
f1
R
è ¥ –R ø
Þ f1 =
R
( m1 – 1)
Focal length of plano-concave lens 1
1 ö m –1
æ 1
= ( m 2 –1) ç
– ÷= 2
f2
–R
è –R ¥ ø
Þ f2 =
–R
( m2 – 1)
For the combination of two lens1
1 1 m –1 m2 –1
= + = 1
–
f eq f1 f 2
R
R
P-406
=
Physics
when liquid is put between, then
m1 – m2
R
1
2 2
= +
f2 fl f
R
m1 – m2
(d) From the equation of line
m = k1v + k2 (Q y = mx + c)
Þ f eq =
40.
1
2 2
or (27 / 2) = 18 + f
or f = – 54 cm
Þ
v
= k1v + k2
u
vö
æ
çQ m = ÷
uø
è
Now -
Þ
k
1
= k1 + 2
u
v
(Dividing both sides by v)
æ 1 ö
= (m1 - 1) ´ ç
÷
è -18 ø
k
1
Þ 2 - – k1
v u
\m1 =
1 1 1
Comparing with lens formula v - u = f , we get
k1 =
41.
(b) Using, M <
or
,2 <
v
u
1
4
+1 =
3
3
40 ´ 20
43. (Bouns) v1 = ( 40 – 20 ) = 40 cm
u2 = 60 – 40 = 20 cm
1
and k2 = 1
–f
1
b
\ f=
= –
slope of m - v graph
c
1
1
= (m1 - 1) ´
54
R
\v2 =
20 ´10
( 20 –10 )
= 20 cm
\ Image traces back to object itself as image formed by
lens is a centre of curvature of mirror.
44. (c) For lens
v1
Þ v1 < ,2 x1
x1
1 1 1
We have v , u < f
1
1
1
, <
,2 x1 x1 20
x1 = 30 cm
or
1 1 1
- =
v u f
1
1
1
And 2 x , x < 20
2
2
or x2 = – 10 cm
x1 30
<
<3
x2 10
1
2
(a) f = f
1
l
So,
42.
Here 2f1 = 18 cm or f1 = 9 cm
So,
1 2
=
or fl = 18 cm
9 fl
1
æ2ö
= (m - 1) ç ÷
Using,
fl
èRø
1
æ2ö
= (1.5 - 1) ç ÷
18
èRø
\ R = 18 cm
or
1
1
1
=
v -30 20
\ v = + 60 cm
According to the condition, image formed by lens
should be the centre of curvature of the mirror, and so
2f’ = 20 or f’ = 10 cm
1 1 1
(d) By lens’s formula, - =
V u f
For first lens, [u1 = –20]
or
45.
1
1
1
20
= Þ VI =
V1 -20 5
3
Image formed by first lense will behave as an object for
second lens
so, u 2 =
20
14
-2=
3
3
1
1
1
=
Þ V2 = 70 cm
V2 14 -5
3
P-407
Ray Optics and Optical Instruments
46.
47.
(a) According to lens maker's formula,
æ 1
1 ö
1
= (mrel – 1) ç R – R ÷
f
è 1
2ø
æ 1
1
1 ö
= (µ - 1) ç
÷
f
R
R
è 1
2ø
Focal length of lens will change due to change in refrective
index mrel. So, image will be formed at new position. Hence
image disappears
1
1 ö
1
æ1
= (µ1 - 1) ç ÷=
f1
è ¥ -R ø 2f2
(b)
Similarly, for plano-concave lens
1
æ +1ö
= (µ2 - 1) ç ÷
è Rø
f2
1
1ö
æ 1
= ( µ2 - 1) ç
- ÷
f2
è -R ¥ ø
( -1)
1
= (µ1 - 1)
f1
R
Now when combined the focal length is given by
= (µ1 - 1)
R
( -1) +
R
+1
(µ2 - 1) R
=
1
éµ 2 - 1 - µ1 + 1 ù
û
Rë
=
µ 2 - µ1
R
Þ f=
1
1
Dividing f by f we get,
1
2
(µ1 - 1) (µ2 - 1)
1 1 1
= +
f f1 f2
48.
49. (b) From lens maker’s formula,
=
2R
or, 2m1 – µ2 = 1
50. (d) using,
m2 m1 m 2 – m1
– =
v u
R
R = 7.8 mm
R
µ2 - µ1
m1 = 1 m2 = 1.34
1.34 1 1.34 –1
– =
[Qu = ¥]
V ¥
7.8
\ V = 30.7 mm = 3.07 cm ; 3.1 cm
(d) By lens formula
Þ
1 1 1
- =
v u f
51. (d)
1
1
10
=
v (-20) 3
1 10 1
= v 3 20
1 197
60
=
; v=
v 60
197
Magnification of lens (m) is given by
Using lens formula
æ 60 ö
ç
÷
æ v ö è 197 ø
m=ç ÷ =
è uø
20
1 1 1
1
1
1
- = Þ = Þ f = 5cm
v u f
10 -10 f
velocity of image wrt. to lens is given by
vI/L = m2vO/L
direction of velocity of image is same as that of object
vO/L = 5 m/s
2
v I/L
æ 60 ´ 1 ö
=ç
(5)
è 197 ´ 20 ÷ø
= 1.16 × 10–3 m/s towards the lens
æ 1ö
Shift due to slab, = t çè1 - ÷ø in the direction of incident
m
ray
æ 2ö
or, d = 1.5 ç1 - ÷ = 0.5
è 3ø
Now, u = – 9.5
Again using lens formulas
1
1
1
=
v -9.5 5
P-408
Þ
Physics
1 1 2
9
= - =
v 5 19 95
95
= 10.55cm
9
Thus, screen is shifted by a distance
d = 10.55 – 10 = 0.55 cm away from the lens.
or, v =
52.
(b) Case-1
+
1 æ m –1ö
=ç
÷
f1 è R ø
P = 2P1 + P2 Þ
f = –28
1
æ m –1 ö
= 2ç
÷
28
è R ø
(Q Power, P =
1
& fplane mirror = ¥)
f
1 1 1
1
1
1
+ = Þ +
=
Þ v = 20 cm.
v u f
v -4 -5
+
f2 = –
R
2
f = –10 cm
1
æ m –1ö 2
= 2ç
÷+
10
è 2 ø R
1
1 2
=
+ Þ 2 = 1 - 2 = 18
10 28 R
R 10 28 280
280
R=
cm
9
P = 2P1 + P2 Þ
or,
or,
or,
5
9
1 1
1
d
= +
and solving, we get f1, f2 18
F f1 f 2 f1 f 2
cm and 20 cm respectively.
(a) Given, focal length of lens (f) = 15 cm
object is placed at a distance (u) = – 20 cm
By lens formula,
f = 15 cm
1 1 1
= f v u
1 1
1 1 1
= + = v f u 15 20
1 4 -3
=
v
60
v = 60 cm
Apparent height, ha = hr
m1
1
= 30 ´
= 20 cm below
m2
1.5
flat surface.
56. (a) Len's formula is given by
1 1 1
= f v u
For convex lens,
1
1
1
1
1
= +
Þ
=
30 v 60
60 v
Similarly for concave lens
m = 1+
Using
54.
Þ m–1=
5 14
=
= 1.55
9 9
(a) For minimum spherical aberration separation,
d = f1 – f2 = 2 cm
Resultant focal length = F = 10 cm
\
53.
1
æ m –1 ö
= 2ç
÷9
28
è 280 ø
10
= –5 cm
2
u = (10 – 6) = –4 cm.
By using mirror formula,
\ Focal length f =
Case-2
1 æ m –1 ö
=ç
÷
f1 è R ø
The image I gets formed at 60 cm to the right of the lens
and it will be inverted.
The rays from the image (I) formed further falls on the
convex mirror forms another image. This image should
formed in such a way that it coincide with object at the
same point due to reflection takes place by convex mirror.
Distance between lens and mirror will be
d = image distance (v) – radius of curvature of convex
mirror.
5 = 60 – 2f
2f = 60 – 5
55
f=
= 27.5 cm (convex mirror)
2
55. (b) Given, radius of hemispherical glass R = 10 cm
d = 5 cm
1
1 1
1
1
= Þ =
- 120 v 40
v 60
Virtual object 10 cm behind plane mirror.
Hence real image 10 cm infront of mirror or, 60 cm from
convex lens.
57. (a) Taking f2 = 12.07
Using Mirror's formula
1 1 1
= +
f
v u
1
1
1
1
1
1
Þ
=
+ Þ
=
12.7 25.4 u
12.7 25.4 u
u = 25.4 = v'
Now using Len's formula
1 1 1
1
1
1
= - Þ
=
+
f v u
f1 25.4 + 13.6 10
1
1
1
390
=
+
Þ f1 =
= 7.96
f1 39 10
49
The closest answers is (a) as option (c) and (d) are not
possible.
Þ
P-409
Ray Optics and Optical Instruments
58.
(d) When object is kept at a distance ‘a’ from thin
covex lens
æ 1
1
1 ö
= ( a m l - 1) ç
÷
fa
è R1 R 2 ø
O
Þ
I1 v
If m1 > m, then fm and fa have opposite signs and the
nature of lens changes i.e. a convex lens diverges the light
rays and concave lens converges the light rays. Thus given
option (a) is correct.
61. (d) Given: Separation of lens for two of its position,
d = 10 cm
Ratio of size of the images in two positions
a
1 1 1
By lens formula : v – u = f
1
1
1
–
=
V (– a) f
I1 3
=
I2 2
1 1 1
or, = –
...(i)
v f a
Mirror forms image at equal distance from mirror
v
I1
v
Distance of object from the screen, D = ?
Applying formula,
I1 ( D + d )2
=
I 2 ( D - d )2
I2
Þ
Now, again from lens formula
I3
3
D 2 + 100 + 20 D
= 2
2
D + 100 - 20 D
Þ 3D2 + 300 – 60D = 2D2 + 200 + 40D
Þ D2 – 100D + 100 = 0
On solving, we get D = 99 cm
Hence the distance between the screen and the object is 99 cm.
v
3 1 1
– =
a V f
59.
62. (c) \ n =
[From eqn. (i)]
3
2
32 + (R – 3mm)2 = R2
Þ 32 + R2 – 2R(3mm) + (3mm)2 = R2
Þ R » 15 cm
öæ 2 ö
1 æ m
=ç
- 1÷ ç ÷
f è mL
øè R ø
60.
4
, f1 = 4 R
3
5
for m L 2 = , f 2 = -5 R
3
Þ f2 = (–) ve
(a) If a lens of refractive index m is immersed in a medium
of refractive index m1, then its focal length in medium is
given by
æ 1
1 ö
- 1) ç
÷
è R1 R 2 ø
If fa is the focal length of lens in air, then
1
=
fm
(
m ml
Velocity of light in vacuum
Velocity of light in medium
\ n=
Hence, a = 2f
(b) By Lens maker's formula for convex lens
for, m L1 =
3 ( D + 10) 2
=
2 ( D - 10) 2
Þ
I2
a/3
3 1 1 1
– + =
a f a f
f m ( a ml - 1)
=
fa ( m ml - 1)
63.
1 æ 3 öæ 1 ö
= ç –1 ÷ç ÷ Þ f = 30 cm
f è 2 øè 15 ø
(d) If side of object square = l
and side of image square = l¢
From question,
l '2
=9
l
l'
=3
l
i.e., magnification m = 3
u = – 40 cm
v = 3 × 40 = 120 cm
f= ?
or
R = 3cm
3mm
P-410
Physics
From formula,
1 1 1
- =
v u f
Experimental
curve
66. (d) |v|
1
1
1
=
120 -40 f
or,
64.
Straight
line
1
1
1 1+ 3
=
+
=
\ f = 30 cm
f 120 40 120
(d) The focal length of the lens
1 1 1
= f v u
1
1
+
12 240
20 + 1 21
=
=
240
240
1 1 1
- =
v u f
240
cm
21
When glass plate is interposed between lens and film, so
shift produced will be
æ
1ö
Shift = t ç 1 - ÷
è mø
1 ö
1
æ
1ç 1 = 1´
è 3 / 2 ÷ø
3
Now image should be form at
v' = 12 - 1 = 35 cm
3 3
Now the object distance u.
Using lens formula again
1
1 1
< ,
f
v' u
1 1 1
Þ < ,
u v' f
1 3
21 1 é 3 21 ù
=
Þ =
u 35 240 5 êë 7 48 úû
1 1 é 48 - 49 ù
Þ = ê
u 5 ë 7 ´ 16 úû
Þ u = –7 ×16 × 5 = – 560 cm = – 5.6 m
(a) From the Cauchy
B C
Formula, µ = A + 2 + 1
l
l
1
\µµ
l
As, lblue < lred
\ lblue > µred
From lens maker's formula
Þ
æ1
1
1ö
= (m - 1) ç - ÷
f
è R1 R2 ø
1
1
>
fB
fR
45°
From lens formula
f =
and
(2f, 2f)
|u|
For the graph to intersect y = x line. The value of | v | and |
u | must be equal.
=
65.
P
Þ fR > fB.
When u = -2 f , v = 2 f
f
Also v =
f
1+
u
As |u| increases, v decreases for |u| > f. The graph between
|v| and |u| is shown in the figure. A straight line passing
through the origin and making an angle of 45°with the
x-axis meets the experimental curve at P (2f, 2f ).
1 1 1
= f v u
This graph suggest that when
67. (c) From the lens formula
u = – f, v = + µ
When u is at – µ , v = f.
v (cm)
f
–f
u (cm)
When the object is moved further away from the lens, v
decreases but remains positive.
68. (c) When two thin lenses are in contact coaxially, power
of combination is given by
P = P1 + P2
= (– 15 + 5) D
= – 10 D.
Also, P =
Þ f =
1
f
1
1
metre
=
P -10
æ1
ö
\ f = - ç ´ 100÷ cm = -10 cm.
è 10
ø
P-411
Ray Optics and Optical Instruments
69.
(b) According to lens maker's formula in air
æ 1
1
1 ö
= a µg -1 ç - ÷
fa
è R1 R2 ø
(
Þ
)
1 æ 1.5 ö
=ç
- 1÷
fa è 1
ø
æ 1
1 ö
ç ÷ .... (i)
è R1 R2 ø
Using lens maker's formula in liquid medium,
öæ 1
1 æ mg
1 ö
= çç
- 1÷÷ ç ÷
fm è mm
ø è R1 R2 ø
Þ 1 = æ 1.5 - 1ö
ç
÷
f m è 1.6 ø
æ 1
1 ö .... (ii)
ç ÷
R
R
è 1
2ø
Dividing (i) by (ii),
æ
ö
f m ç 1.5 - 1 ÷
=ç
÷ =–8
f a ç 1.5 ÷
1
ç
÷
è 1.6 ø
Pa = - 5 =
1
fa
Þ fa = -
1
5
1 8
Þ f m = -8 ´ f a = - 8 ´ - =
5 5
Pm =
70.
m 1.6
=
´ 5 = 1D
fm
8
(c) The focal length(F) of the final mirror is
1
2
1
=
+
F fl f m
Using lens maker's formula
æ 1
1
1 ö
= ( µ - 1) ç Here
÷
fl
è R1 R2 ø
Here, R1 = ¥
R2 = 30 cm
1 ù 1
é1
= (1.5 - 1) ê =
ë ¥ -30 úû 60
1
1
1
1
\
= 2´ +
=
F
60 30 / 2 10
\ F = 10 cm
Real image will be equal to the size of the object if the
object distance
u = 2F = 20 cm
71. 2
From the Einstein's photoelectric equation
Energy of photon
= Kinetic energy of photoelectrons + Work function
Þ Kinetic energy = Energy of Photon – Work Function
Let f0 be the work function of metal and v1 and v2 be the
velocity of photoelectrons. Using Einstein's photoelectric
equation we have
1 2
mv1 = 4 - f 0
2
...(i)
1 2
mv2 = 2.5 - f0
2
...(ii)
1 2
mv1
4 - f0
Þ 2
=
1 2 2.5 - f0
mv2
2
Þ (2)2 =
4 - f0
Þ 10 - 4f 0 = 4 - f0
2.5 - f 0
f0 = 2eV
72. (a) When angle of prism is small, then angle of deviation
is given by Dm = (µ – 1)A
So, if wavelength of incident light is increased, µ decreases
and hence Dm decreases.
73. (c) For minimum deviation:
A
r1 = r2 = = 30o
2
by Snell’s law m1sin i = m2sin r
1
3
Þ i = 60
1 × sin i = 3 ´ =
2 2
74. (c) Angle of prism, A = 30°, i = 60°,
angle of deviation, d = 30°
Using formula, d = i + e – A
30°
30°
60°
Þ e = d +A – i
= 30° + 30° – 60° = 0°
\ Emergent ray will be perpendicular to the face
So it will make angle 90° with the force through which it
emerges.
75. (c) We know that i + e – A = d
35° + 79° – A = 40°
\ A = 74°
+
d
æA m ö
æ 74 + d m ö
sin ç
sin ç
÷
÷
2
è 2 ø=
è
ø
But m =
74
sinA / 2
sin
2
dm ö
5 æ
= sin ç 37° +
3 è
2 ÷ø
P-412
Physics
5
5
. That is m max is less than = 1.67
3
3
But dm will be less than 40° so
m max can be
m<
76.
5
5
sin 57° < sin 60° Þ m = 1.5
3
3
(c) When r2 = C, ÐN2RC = 90°
Where C = critical angle
As sin C =
or
m<
Þ
m < 2 Þ m < 1.414
79. (c) For total internal reflection on face AC
q > critical angle (C)
and sinq ³ sinC
sin q ³
1
= sin r2
m
N1
N2
Q r1
r2
P
1
wm g
4
mw
sin q ³
Þ sin q ³ 3
3
mg
2
8
\ sin q ³ .
9
80. (a) Rainbow is formed due to the dispersion of light
suffering refraction and total internal reflection (TIR) in
the droplets present in the atmosphere.
A
q
1
1
Þm<
sin q
sin 45°
R
81. (a) When angle of prism is small,
B
Angle of deviation, D = (m – 1) A
C
Applying snell's law at ‘R’
µ sin r2 = 1 sin90°
...(i)
Applying snell's law at ‘Q’
1 × sin q = µ sin r 1
...(ii)
But r1 = A – r2
So, sin q = µ sin (A – r 2)
sin q = µ sin A cos r 2 – cos A
From (1)
2
cos r2 = 1 – sin r2 = 1–
Since lb < lr
Þ mr < mb
82. (b) For total internal reflection
Incident angle (i) > critical angle (ic),
...(iii)
1
µ2
[using (i)]
1
µ2
...(iv)
- cos A
é
–1 æ 1 ö ù
q = sin–1 êm sin(A – sin ç µ ÷ ú
è øû
ë
So, for transmission through face AC
78.
é
–1 æ 1 ö ù
q > sin–1 êm sin(A – sin ç µ ÷ ú
è øû
ë
(c) For the prism as the angle of incidence (i) increases,
the angle of deviation (d) first decreases goes to minimum
value and then increases.
(d) For light to come out through face 'AC', total
internal reflection must not take place.
i.e., q < c Þ sin q < sin c
1
Þ sin q <
m
\ sin 45° >
Þ
on further solving we can show for ray not to transmitted
through face AC
77.
\ sin i > sin ic
Þ sin 45° > sin ic Þ sin ic =
By eq. (iii) and (iv)
sin q = µsin A 1 -
Þ D1 < D2
1
2
>
1
n
1
n
1
n
Þn> 2
83. (50)
Given : Length of compound microscope, L = 10 cm
Focal length of objective f0 = 1 cm and of eye-piece,
fe = 5 cm
u0 = fe = 5 cm
Final image formed at infinity ( ¥ ), ve = ¥
v0 = 10 – 5 = 5
Using lens formula,
1 1 1
- =
v u f
1 1
1
1 1 1
5
=
Þ = Þ u0 = - cm
v0 u0
f0
5 u0 1
4
or,
5 N
=
4 40
\N =
200
= 50 cm.
4
P-413
Ray Optics and Optical Instruments
84.
(4.48)
According to question, final image i.e., v2 = 25 cm,
f0 = 1 cm, magnification, m = m1m2 = 100
fe
f0=1cm
O
25cm
I1
x
I2
O
I1
I2
86. (a) According question, M = 375
L = 150 mm, f0 = 5 mm and fe = ?
L æ
Dö
Using, magnification, M ; f ç 1 + f ÷
0è
eø
object
image formed by 1st lens
image formed by 2nd lens
Þ 375 =
v1
Objective
Eye-piece
20 cm
150 æ 250 ö
ç1 +
÷
5 è
f e ø (Q D = 25 cm = 250 mm)
Þ 12.5 = 1 +
Using lens formula,
For first lens or objective =
Also magnification | m1 |=
1 1 1
x
= Þ v1 =
v1 - x 1
x -1
v1
1
=
u1
x -1
Angular magnification | m A |=
87.
88.
89.
25
D
=
u2 | u 2 |
MP =
Total magnification m = m1m A = 100
25
= 100 Þ 1 = 80( x - 1) - 4 x
20( x - 1) - x
Þ 76 x = 81 Þ x =
æ
ç
Þ u2 = - ç 20 çç
è
81
76
81 ö
76 ÷ = -19
÷
81 ÷
5
-1÷
76 ø
85.
(d) For telescope
Tube length (L) = fo + fe = 60
f
and magnification (m) = o = 5 Þ f 0 = 5 f e
fe
\ fo = 50 cm and fe = 10 cm
Hence focal length of eye-piece, fe = 10 cm
f o 150
=
= 30
fe
5
a=
50
1
rad
=
1000 20
\
b = q = MP × a = 30 ´
90.
1
3
= = 1.5 rad
20 2
180°
; 84°
p
(a) Given : f0 = 1.2 cm; fe = 3.0 cm
u0 = 1.25 cm; M¥ = ?
or, b = 1.5 ´
From
1
1 1
= f 0 v0 u0
Þ
1
1
1
=
1.2 v0 ( -1.25)
Þ
1
1
1
=
v0 1.2 1.25
Again using lens formula for eye-piece
1
1
1
25 ´ 19
=
Þ fe =
» 4.48cm
-25 - 19 f e
106
5
b (angle subtended by image at eye piece)
a (angle subtended by object on objective)
Also, MP =
æ
ö
÷
25
æ 1 öç
÷ = 100
ç
֍
è x - 1 ø ç 20 - x ÷
ç
÷
x -1 ø
è
Þ
250
= 21.7 » 22 mm
11.5
(b) A telescope magnifies by making the object appearing
closer.
(c) Reading one Þ without slab
Reading two Þ with slab
Reading three Þ with saw dust
Minimum three readings are required to determine refractive
index of glass slab using a travelling microscope.
(c) Magnifying power of telescope,
Þ fe =
For 2nd lens or eye-piece, this is acting as object
x ö
æ
\ u2 = -(20 - v1 ) = - ç 20 ÷ and v2 = – 25 cm
x -1 ø
è
250
fe
Þ v0 = 30 cm
Magnification at infinity,
M¥ = =
v0 D
´
u0 f e
30 25
´
1.25 3
P-414
91.
Physics
(Q D = 25 cm least distance of distinct vision) = 200
Hence the magnifying power of the compound microscope
is 200
(d) Given, Focal length of objective, f0 = 30 cm
focal length of eye lens, fe = 3.0 cm
Magnifying power, M = ?
Magnifying power of the Galilean telescope,
f æ f ö
M D = 0 ç1 - e ÷
fe è
Dø
Limit of resolution, Dq =
1.22 ´ 5.5 ´ 10-7
= 2.23 × 10–5 rad.
3 ´10-2
At a distance of 80 m , the telescope is able to resolve
between two points which are separated by 2.23 × 10–5
× 80 m
= 1.78 × 10–3 m
=
30 æ
3 ö
ç1 - ÷ [Q D = 25 cm]
3 è 25 ø
=
22
= 8.8 cm
25
(c) One side of mirror is opaque and another side is
reflecting this is not in case of lens hence, it is easier to
provide mechanical support to large size mirrors than large
size lenses. Reflecting telescopes are based on the same
principle except that the formation of images takes place
by reflection instead of refraction.
1.22l
d
Eye-Piece
Objective
= 10 ´
92.
93.
(d) Given : f0 = 50 cm, f e = 5cm
d = 25 cm, u0 = –200 cm
Magnification M = ?
As
1
1
1
=
v0 u 0 f 0
Þ
1
1 1
1
1
4 -1
3
= +
=
=
=
v0 f 0 u 0 50 200 200 200
v0 =
or
200
cm
3
Now ve = d = –25cm
From,
1
1
1
=
ve u e fe
–
1
1 1
= u e f e ve
=
1 1
6
+
=
5 25 25
-25
cm
6
Magnification M = M0 × Me
or,
ve =
v v
-200 / 3 -25
= 0´ e =
´
u0 ue
200
-25 / 6
94.
1
= - ´ 6 = -2
3
(a) Given : d = 3 × 10–2 m
l = 5.5 × 10–7 m
95. (c)
O
F0
20 mm
25 mm
I
V0
fe
To obtain final image at infinity, object which is the image
formed by objective should be at focal distance of eyepiece.
By lens formula (for objective)
1
1
1
=
v0 u0
f0
or,
1
1
1
=
v0 -25 20
1
1
1
5- 4
1
=
=
mm
v0 20 25 = 100 100
\ v0 = 100 mm
Therefore the distance between the lenses
= v0 + fe = 100 mm + 20 mm = 120 mm
96. (a) To find the refractive index of glass using a travelling
microscope, a vernier s