QUESTION PAPERS L-2 - I
Difficulty level –L2
Class XII
Physics
SAMPLE PAPER XII - PHYSICS
Time : Three Hours
Max. Marks : 70
General Instructions
(a) All questions are compulsory.
(b) There are 30 questions in total. Questions 1 to 8 carry one mark each, questions 9 to 18 carry
two marks each,questions 19 to 27 carry three marks each and questions 28 to 30 carry five
marks each.
(c) There is no overall choice. However, an internal choice has been provided in one question of
two marks, one
(d) question of three marks and all three questions of five marks each. You have to attempt only
one of the given choices in such questions.
(e) Use of calculators is not permitted.
(f) You may use the following physical constants wherever necessary :
c = 3 x 108ms-1
h = 6.6 x 10-34Js
e = 1.6 x 10-19 C
Boltzmann constant k = 1.38 x 1023 JK-1
Avogadro’s number NA = 6.023 x 1023/mole
Mass of neutron mn = 1.6 x 10-27 kg
1. Write the expression for torque on a magnetic dipole placed in a uniform magnetic field.
Ans. Torque = MBSinθ
2. Show that JA-1 is SI unit of magnetic flux.
Ans. We have JA-1 = work/current =F.L/current=BIL.L/I = BL2 =B.S = Magnetic flux.
in an AC circuit A the current leads the voltage by 30 degree and circuit B the current lags
behind the voltage by 30 degree. What is the ratio of their power factors.
Ans 1 : 1
3.
What is the range of wavelength of E.M. waves used for TV broadcast?
Ans: Between 1.36m to 3m.
4.
State any two factors on which the refractive index of a medium depend.
Ans.
Factors May be any two :1. Nature of the medium
2. Wavelength of light used
3. Temperature and
4. Nature of surrounding medium
The kinetic energy of -particles incident on gold foil is doubled. How does the distance of
closest approach change?
Ans. The distance of closest approach is inversely proportional to the kinetic energy of the
incident -particles, so the distance of closest approach is halved when the kinetic energy of particles is doubled
5.
7. Why is germanium preferred over silicon for making semiconductor devices?
Ans: Because the forbidden energy gap for Ge (0.7 eV) is smaller than the energy gap for Si (1.1 eV). So
the semiconductors devices made by germanium can be operated at low voltage
8. In the following diagram, is the diode forward or reverse biased?
Ans: Reversed biased.
9. A conductor A with a cavity is given a charge Q. Another conducting body B having
charge q is inserted into the cavity keeping it insulated from A. What will be the total
charge on the outside surface of A?
Ans: q will induce –q on the inner surface of A and +q on the outer surface of A. so the total
charge on
outer surface of a is Q + q
10. A wire shown in the figure having negative tolerance is stretched by one-tenth of its original
length.Then its resistance will be
Ans- 470 +- 5%
.
11. Two primary cells of emf E1 and E2 (E1>E2) are connected to the potentiometer AB as
shown in the fig. If the balancing for two combinations of the cells are 250cm and
400cm, find the ratio of E1 and E2.
400
250
B
A
E1
E2
G
E1
E2
G
Ans- I case- gives result is
E1-E2 =250K ( K is potential gradient)
II case –
E1 + E2 = 400K
Adding case I and case 2
E1 = 325 K
Subtracting case 2 – case 1- E2 = 75 K
E1/E2 =325 K/75 K = 13/3
12. Write Ampere’s circuital law and prove it ?
13. Name any four constituent radiation of electromagnetic waves spectrum which
(a) is used in satellite communication,
(b) is used for studying crystal structure,
(c) is similar to the radiation emitted during decay of a radioactive nucleus, (d) is absorbed
from sunlight by ozone layer,
(e) produces intens heating effect and
(f) has its wavelength range between 390 nm to 770 nm.
Ans:
(a) Microwaves,
(b) X-rays,
(c) γ rays,
(d) UV rays,
(e) infra-red-rays and
(f) visible light.
14. An object PQ is placed in front of an isosceles right -angled prism of crown glass with critical
angle equal to 410. Trace the path of two rays from P and Q normal to the hypotenuse.
Q
45o
90o
P
Ans
45o
15. The graphs between the stopping potential ‘V’ and frequency ‘’ of the incident radiation
of two different metal plates P and Q are shown in the fig.
Q
P
V
O
V
Which metal has greater value of work function? What does the slope of the line depict?
Ans.
(a) Metal Q has the greater value of work function because its threshold frequency is
higher than that of metal P.
(b) Slope of the line = V /  = h/e
constant and e=electronic charge.
[since eV= h]
here
h
=
Planck’s
16. For photoelectric effect in a metal, the fig. shows the plot of cut off voltage versus frequency of
incident radiation.
3
Cut off voltage
2
1
1 2 3 4 5 6 7 8 9 10 × ( 10 14 Hz)
Frequency in Hz
Calculate (i) the threshold frequency and (ii) the work function for the given metal
Ans. The threshold frequency (ν0) is equal to the intercept made by the graph on frequency axis.
ν 0 = 4.5 x 1014 Hz.

work function of metal
W= h ν 0 = 6.62 x 10-34 x 4.5 × 10 14
= 2.98 x 10-19 J
= 2.98 x 10-19 = 1.86 eV
1.6 x 10-19
17. Deduce the relation between decay constant and half life period.
Ans.
Expression: Consider that a radioactive sample contains No, atoms at time t = 0 , then
the number of atoms left behind after time ‘ t ‘ is given by
N = Noe-λ
t
----------------
(i)
From definition of half life,
When t = T , N = N0/2
Setting the above condition in eq. (i) , we have
No/2 = Noe -λ
e-
T
T
= ½ or e
λT
=2
λ T = log e 2 = 2.303 log102= 2.3030 × .3010 = .693
T = .693/λ
Half life of a radioactive substance is inversely proportional to its decay constant.
18. A T.V tower has a height of 150m. How much population will it cover if the population density
of area is 104 m2. Radius of earth is 6400km?
[Hint: Pupulation = π d2 x population density and d 
2 Rh Ans; 6.0 x 1013]
19. State and prove Gauss’s theorem
Ans
20. Write a short note on Van de Graaff generator giving its principle and and draw a labeled
diagram.
Ans:- PRINCIPLE: If a charged conductor is brought into internal contact with a second hollow
conductor, all of its charge is transferred to the hollow conductor; all of its charge is
transferred to the hollow conductor, no matter how high the potential of the latter may rise.
Consequently, the charge and hence the potential of the hollow conductor can be raised to a
very high value (limited only by the breaking of the insulation of air around) by successively
adding charges to it by internal contact.
CONSTRUCTION: -
21. Calculate resistance between A & B of the given network.
Answer- (i) It is a wheat stone bridge as
2 4
 . Hence
1 2
2 Ohm
(ii) It is a parallel combination of three 1 resistances hence 1/R = 1/1+1/1+1/1 = 3 or
R = 1/3
Ohm
22. State Lenz’s law. Show that Lenz’s law is consequence of law of conservation of energy.
Answer. Lenz’s Law It states that direction of induced current in the closed circuit is such that it will oppose
the cause (change in magnetic flux) which is producing it. So Faraday’s 2nd law is written as e = - d/dt For
numericals e = - (2 - 1)/t
Lenz’s law is consequence of law of conservation of energy
In Faraday’s experiment when N- pole of magnet is moved towards the coil, the face of the coil towards the
approaching magnet acquires N – polarity. Work has to be done against force of repulsion in bringing the
magnet closer to the coil. Similarly, when N- pole of magnet is moved away the coil, the face of the coil towards
the receding magnet acquires
South polarity.  Work has to be done against force of attraction in taking the magnet away from the coil. It is
this mechanical work done on the magnet against the force of repulsion or attraction in moving the magnet
w.r.t. the coil that changes into electrical energy producing induced e.m.f. and induced current. When the
magnet is stationary, no work done and so no induced current. Hence Lenz’s Law obeys the principle of
conservation of energy.
23. A 200V a. c. source of variable frequency is connected to a series combination of L= 5H, C =80F and
R =40. Calculate (1) angular frequency of the source to get maximum current in the circuit, (2) the
current amplitude at resonance and (3) the power dissipation in the circuit.
Answer. (i) We know  =1/LC; putting the values we get
 = 1/ 5x80x10-6 =1000/20 = 50 rads-1 .
(ii) Irms = Vrms / Z = Vrms / R ( At resonance Z =R ) putting the values we get
Irms = 2000/40 =5A.
Current amplitude, I0 = 2 Irms = 52A.
(iii) Power dissipation P = Vrms Irms cos ; putting the values we get P =
200x5x1 or P = 1000W.
24. Derive Lens maker formula. State the assumptions used.
Ans. Assumptions:
(i)
(ii)
(iii)
The lens used is thin.
The aperture of the lens is small.
For appoint object lies on the principal axis of the lens
Let a thin lens ‘XY’ of negligible thickness and made up of material of refractive index µ2 is placed in a
medium of refractive index µ1. Let a point object ‘O’ is kept at a distance ‘u’ from the first surface of the
lens xp1y. Assuming that there is no second surface then first surface will form image I’ at a distance v’
from the P1. For the refraction at the face XP1Y
µ2/v’ - µ1/u = (µ2- µ1)/R1 -------------------------(i)
where R1 is the radius of curvature of the first surface.
Image I’ serves as the object for the second surface and it forms its real image at a distance v from it. So
µ1/v - µ2/v’ = (µ1 - µ2)/R2 --------------------------(ii)
Where R2 is the radius of the curvature of the second surface.
on adding equation (i) and (ii)
µ1/v - µ1/u = (µ2 - µ1)/R1 + (µ1 - µ2)/R2
µ1 (1/v – 1/u) = (µ2 - µ1)(1/R1 – 1/R2)
(1/v – 1/u) = (µ2/µ1 – 1)(1/R1 – 1/R2)
If u →∞ then v = f
So 1/f = (µ2/µ1 – 1)(1/R1 – 1/R2)
25. Explain the laws of reflection on the basis of Huygens wave theory.
Laws of Reflection are:
1.
2.
Angle of incidence = Angle of reflection
The incident ray, the normal and the reflected ray lie in the same plane.
A’B’ represents reflected wave front and lies normal to it i.e. A’T and B’V are reflected
rays further
Ð RAP = i and Ð QB’V = r represents angle of incidence and angle of reflection.
Since angle between two lines = angle between their perpendicular
\ Ð BAB’ = Ð RAP = i and Ð A’B’A = Ð QB’V = r
In Δ ABB’ and Δ A’AB’
Ð ABB’ = Ð AA’B’ = each equal to 90°.
AB’ = AB’ (Common)
AA’ = BB’ (distance covered is air in same time = ct)
\ Δ ABB’ and Δ A’AB’ are congruent.
\ Ð BAB’ = Ð A’B’A
Þ Ði = Ðr
So the first law is proved.
Further from the figure it is clear that incident rays reflected ray and the normal are
drawn in the plane of the paper hence they all are in a plane. Hence second law of
reflection is also verified.
26 Draw graph between mass number and average binding energy per nucleon
Answer-
27. What is modulation? List three reasons why a signal needs to be modulated using high frequency
carrier waves?
Ans. The process of varying one of the characteristics of the carrier wave according to the instantaneous
value of the modulating signal is called modulation It is essential for efficient operation of
communication due the following reasons:
(A)to have a manageable antenna size
(B)to increase the power of the signal as the power is inversely proportional to the fourth power of
the wavelength.
(C)to avoid inter-mixing of different signals.
28.
(a)Draw diagrams to depict the behavior of magnetic field lines near a boar of
(i)Copper (ii) Aluminum (iii) Mercury cooled to a very low temperature 94.2 K)
(b) What is the radius of the path of an electron (mass 9 × 10-31 kg and charge 1.6
× 10–19 C) moving at a speed of 3 ×107 m/s in a magnetic field of 6 × 10–4 T
perpendicular to it? What is its frequency? Calculate its energy in keV.
( 1 eV = 1.6 × 10–19 J).
Answer(a)- For Copper (Diamagnetic)
For Aluminum
Mercury cooled to a very low temperature (4.2 K0 at this temperature Hg
becomes purely diamagnetic substance.
Answer(b)-
29. Explain compound microscope with ray diagram. Derive an expression for its
magnification.
Answer. Construction of a Compound Microscope
A compound microscope consists of the following parts:

Objective lens -The objective lens of a compound microscope is a convex lens of very short focal
length (fo) that is fo < 1cm. The object to be seen is kept very close to the objective lens.


Eye piece-The eye piece of a compound microscope is also a convex lens of short focal length fe.
But fe > fo.
Microscope tube-The objective lens and the eyepiece are mounted coaxially (having a common
axis) at the ends of two brass tubes which can be made to slide into each other so that the
distance between the two lenses can be adjusted.
Working
The ray diagram given below gives the principle of a compound microscope. The object is
mounted on the stand below the microscope tube. The objective lens forms a real, inverted and
magnified image (I1) of the object. The image I1 acts as an object for the eye piece. The position of
the eyepiece is so adjusted that the image lies within the focus of the eyepiece (Fe). The eyepiece
acts like a magnifying glass and forms a virtual erect and magnified image of the object.
Image Formation in a Compound Microscope




The object (O) is placed just outside Fo, the principal focus of the objective lens.
Fe is the principal focus of the eye lens.
A real, inverted magnified image I1 is formed. The magnified image I1 acts as an object for the
eye lens.
The final image I2 is virtual and is magnified still further. It is inverted compared with the object.
I2 may appear 1000 times larger than the object.
Magnifying Power of a Compound Microscope
The magnifying power of a compound microscope is defined as the ratio of the size of the final
image (I2) as seen through the microscope to the size of the object as seen with a naked eye.
Magnifying power of compound microscope is defined as the ratio of the angle subtended on the eye
by the final image (A”B” here) to the angle subtended by the object on the eye when seen directly,
when both are placed as least distance of distinct vision. Imagine the object to the shifted to A1B” so
that it is also located by least distance of distinct vision from the eye. If Ð A”C2B” = β and Ð A1C2B” = α,
then by definition.
Magnifying power, m =
m=
β
. For small angles β = tan β and α2 = tanα, hence
α
tanβ A"B"/C2 B" A"B" A"B" A'B' A"B" A'B'
=
=
=
×
=
×
[Q A1B"= AB]
tanα A1B"/C2 B" A1B" A'B' A1B" A'B' AB
\ m = mi  mo =
ö v
v æ
çç1+ D ÷
÷=ç
ø u
- u è fe ÷
æ Dö
çç1+ ÷
÷
çè fe ÷
ø
OR
Find an expression for the fringe width in Young’s double slit experiment.
Ans. Expression for Fringe Width in Young's Double Slit Experiment
Let A and B be two fine slits, a small distance 'd' apart. Let them be illuminated by a
monochromatic light of wavelength l.
MN in the screen is at a distance D from the slits AB. The waves from A and B superimpose upon
each other and an interference pattern is obtained on the screen. The point C is equidistant from
A and B and therefore the path difference between the waves will be zero and so the point C is of
maximum intensity. It is called the central maximum.
For another point P at a distance 'x' from C, the path difference at P = BP - AP.
Now AB = EF = d, AE = BF = D
In triangle BPF
[Pythagoras theorem]
Similarly in D APE
Hence Path difference.
(on expanding Binomially)
For bright fringes (constructive wavelength) the path difference is integral multiple of
wavelength i.e., path difference is nil.
(x therefore represents distance of nth bright fringe from C)Now
And so on.
Therefore separation between the centers of two consecutive bright fringes is the width of a
dark fringe.
Similarly for dark fringes,
The separation between the centers of two consecutive dark interference fringes is the
width of a bright fringe.
The separation between the centers of two consecutive dark interference fringes is the
width of a bright fringe.
All bright and dark fringes are of equal width as 1 = 2.
Note:
The intensity of all bright bands is the same. All dark bands also have same (zero)
intensity. The intensity distribution Vs distance is shown as:
30. Draw the circuit diagram showing use of p-n junction
(a) as a – (i) half wave rectifier (ii) full wave rectifier
(b) on what principal does it work?
(c) Describe the working of full wave rectifier.
Draw the input and output wave forms of full wave rectifier.
Ans.(a) (i) p-n junction as half wave rectifier
(ii) p-n junction as full wave rectifier
(b) The p-n junction allows the current to flow only when it is forward biased..
(c) Working of full wave rectifier:
(i) Let for Ist half cycle (+ve) of the input, end A be (+ve) and end B (-ve), Diode D1 will conduct and
Diode D2 will not conduct. The potential drop across RL gives output.
(ii) For 2nd half of input cycle, end A will be (-ve) and end B will be (+ve), Diode D1 will not conduct
and Diode D2 will conduct. Again potential drop across RL gives output.
OR
(a)Describe the working of a transistor as a switch:
(b)Given below is the circuit symbol of a NAND gate with inputs shorted. The signal shown below is
given as input to the above gate. Draw the output wave form . and draw the truth table
Ans.
When the input voltage is less than certain value such that the transistor does not conduct, then the
collector current is zero and the output voltage is maximum. In this mode the transistor is in switched
off position.
When the input voltage is beyond that voltage ;the transistor conducts .The transistor is active. The
output decreases linearly with the input voltage. When the input voltage is such that the transistor is
active and the collector current has reached its optimum value, then the output voltage reduces to the
minimum or zero.There is no output voltage.At this mode the transistor is in switched on position.
Hence the transistor acts as switch if it is used at cutoff region and saturation region.
Ans (b) The signal shown below is given as input to the above gate. Draw the output wave form . and
hence draw the truth table
The gate is NOT gate The output waveform for the given input is as shown:
Truth table
--------------------------------------------------------------------------------------------------------------
QUESTION PAPERS L-2 - 2
.
Class XII- Physics.
Time-3 Hrs.
M.Marks-70
General Instructions.
i)
ii)
iii)
iv)
v)
vi)
vii)
All questions are compulsory.
There is no overall choice. However an internal choice has been provided in one question of
2 marks, one question of 3 marks, and in all the questions of 3 marks.
Q.no.1 to 8 are very short answer questions and carry 1 mark each.
Q.no.9 to 18 are short answer questions carry 2 marks each
Q.no.19 to 27 carry 3marks each
Q.no.28 to 30 carry 5 marks each
If necessary you may use the following value of physical constants.
1) What happens to the neutron to proton ratio after the emission of α-particle.?
2) At a place, the horizontal and vertical components of earth’s magnetic field are equal. What is
the angle of dip at that place?
3) Write truth table for NAND gate
4) How does the angular separation of interference fringes change in Young’s experiment if the
distance between the slit is increased?
5) Name the part of electromagnetic spectrum which is used for taking photographs of earth under
foggy conditions from great height.
6) Four nuclei of an element fuse together to form a heavier nucleus, if the process is accompanied
by release of energy, which of the two, the parents or the daughter nucleus would have a higher
binding energy/nucleons.?
7) If a coil of metallic wire is placed at rest in a non-uniform magnetic field. Does any e.m.f is
induced in the coil?
8) In a semi-conductor the concentration of electron is 8x10 13 cm -3 and that of hole is 5x1012 cm -3
Is it p-type or n-type semi-conductor.?
9) What are equipotential surfaces? Show that electric field is always directed perpendicular to the
equipotential surface.
10) Draw V-I graph for ohmic and non-ohmic material. Give one example of each.
11) Write any two difference between Diamagnetism and Ferromagnetism
12) Define electrical conductivity of conductor and give its S.I. units.
13) Which constituent radiation of electromagnetic spectrum is used?
i)
In radar.
ii)
For taking photograph of the sky during night and foggy conditions.
14) State the reasons for the following observations recorded from the surface of moon.
i)
Sky appears dark.
ii)
Rainbow is never seen.
15) An electron and α-particle have same K.E. How is the de-Broglie wavelength associated with
them related?
16) In a photoemission surface has threshold freq. is 4.6x1014 Hz. Calculate the minimum energy of
photon (in e.v.) Which will emit photoelectrons?
Or
Why should gases be insulators at ordinary pressure and start conducting at very low pressure?
17) What is meant by detection of modulated carrier wave? Describe briefly the essential steps for
detection.
18) Prove that the instantaneous rate of change of the activity of a radio active substance is
inversely proportional to the square of its half-life.
19) State Hygen’s principle of secondary wavelets. Using Hygen’s principle obtains law of reflection.
Or
A ray of light when moves from denser to rare medium undergo total internal reflection. Derive
the expression for the critical angle in terms of speed of light in respective medium.
20) Define half-life and decay constant. Derive the relation N= N0e –λT.
21) Show a labelled ray-diagram showing the formation of image in a compound microscope. Write
the expression for its magnifying power.
22) The instantaneous current in an A.C.circuit is i=0.5sin314t. What is r.m.s. value and frequency of
current?
23) State and prove Gauss theorem in electrostatics.
24) Derive a mathematical expression for resistivity of a conductor in terms of number of density of
a charge carrier in the conductor and relaxation time.
25) Explain briefly the principle of transmitting signal using a satellite. State two main advantage of
using satellite for transmitting signal.
26) Show that energy stored in parallel plate capacitor is ½ CV2.Hence, derive an expression for
energy density of a capacitor.
27) Describe briefly the principle, construction and working of transformer.
28) State Biot- Savart law. Using this law, find an expression for the magnetic field at the centre of a
circular coil of N- turns , radius ’r’,carrying current ‘I’.Sketch the magnetic field for a circular
current loop cleary, indicating the direction of the field.
Or
a) With the help of neat and labelled diagram explain the underlying principle
and working of the moving coil galvanometer.
b) What is the function of i) uniform radial field ii) soft iron core in such a device.
29) Draw the circuit diagram of a common emitter amplifier using an npn-transistor. What is the
phase difference between its input signal and output voltage? Draw the input and output
waveforms of the signals. Write the expression of its voltage gain. State two reasons why a
common-emitter amplifier is preferred to a common-base amplifier?
Or
With the help of a suitable circuit diagram explain working of transistor as an oscillator ?
Find the voltage gain and power gain in a CE transistor amplifier when β =66 and the input and output
resistances are 0.5 K  and
50 K 
?
(5)
30) a) Derive a Lens-makers formula for a double convex lens.
b) Double convex lenses are to be manufactured from a glass of refractive
index 1.55 with
both faces of the same radii of curvature. What is the radius of curvature required if the focal
length to be 20 cm?
Or
Describe an astronomical telescope and find expression for its magnifying power. A small
telescope has an objective lens of focal length 1.44 cm. and eye piece of focal length 6 cm. What
is the magnifying power of the telescope? What is the separation between the objective and eye
piece?
MARKING SCHEME
1. Increases
1 mark
2. At 45
1 mark
3. correct truth table
1 mark
4 d increased, β is decreased
1 mark
5. Infrared rays
1 mark
6 .Daughter nucleus will have higher BE/ nucleon
7 .No
1 mark
1 mark
8 .n-type semi-conductor
1 mark
9 .Surface having same potential
dw = - E. dr = 0
1mark
E is perpendicular to the surface at each point
1 mark
10. for graph of ohmic and non ohmic
Ohmic—Tunguston
Non-ohmic—Jn. diode
1\2×4 = 2
11.For each difference 1 mark
12.For def.
1 mark
For unit
1 mark
13.In Radar
--- microwave
1 mark
In taking photograph---- infrared
1 mark
14.i. Moon has no atmosphere so no scattering light reach on ones eye. 1 mark
ii. There is no water vapour on moon
1 mark
15. λe = h \ 2meEk
1\2mark
λα = h\ 2mαEk
1\2 mark
λe = 86.5 × λα
1 mark
16.W = hυ0
= 6.6 × 10 × 4.6 × 10
1\2 mark
joule
1\2 mark
=1.5 eV.
1 mark
OR
.At low press, ions are far away. So, they have possibility to reach their respective electrodes and
constitute the current, while at ordinary pressure, ions collide freq. and recombine with the opposite
charge ions.
2 marks
17) definition
1 mark
Explanation
1 mark
18) Formula for rate of change of activity
1 mark
Result
1 mark
19) For correct statement. -------1 mark
Fig and proof -------2 marks
Or
When light travels from denser to rare medium from Snell’s law
sin i nr
=
sin r nd
By definition nr =
(1)
-------1 mark
c
c
and nd =
vr
vd
c
sin c
from equation (1) gives
= vr
sin 90 c
vd
c=
------1 marks
vd
vr
c= sin -1 (
vd
)
vr
-------1 mark
20) Rutherford and Soddy statement---------1 mark
According to Rutherford and Soddy
dN
dN
N i.e.
= - λN -------1 mark
dt
dt
log N = - λt + c
For correct observation and graph.
-------1 mark
21) For correct fig. -------1 mark
 v0
u0
Magnifying power M =
=

D
1  
fe 

L
D
1   ------1 mark.
fo 
fe 
For final image at distance of distinct vision
M=
 vo D  L D
=
uo fe
fo fe
For final image at infinity
-----1 mark
22) Standard equation of current I = Iosinωt
i) Irms =
ii) f =
io
2
=
0 .5
2
-----1 mark
= .35A -----1 mark

3.14
=
= 50 Hz -----1 mark
2
2 x3.14
23) For correct statement -----1 mark
Proof
-----2 marks
24) The relation between vd (drift velocity and current) is
I = -neAvd
-----1 mark
Where l, A, v and I have their usual meanings
Correct derivation -----2 marks
25) Principle of transmitting signal using satellite -----1 mark
Main advantage of satellite communication -----2 marks
26 ) for proof of expression of energy -----1 1
For energy density -----1 1
27) Principle of transformer ----- 1
Fig
----- 1
2
2
2
2
marks
marks
marks
marks
Construction -----1 mark
Working -----1 mark
28) statement -----1 mark
Correct fig. -----1/2 mark
Finding mag. Field -----------2 1
2
marks
Magnetic field lines due to circular current loop ---------- 1 mark
Or
For correct diagram --------------- 1 mark
Principle --------------- 1 mark
Working --------------- 1 1
2
marks
Radial mag.field and soft iron core --------------- 1 1
2
marks
29) correct fig. --------------- 1 mark
Input and output signal are out of phase--------------- 1 mark
Fig. of input and output amplified signal --------------- 1 mark
Formula for voltage gain --------------- 1 mark
Reasons --------------- 1 mark
Or
Circuit Diagram
Working
(21/2)
(21/2)
30) a) Fig. --------------- 1 mark
Correct derivation --------------- 2 marks
b) Putting values in the formula --------------- 1 mark
Correct ans. --------------- 1 mark
Or
a) for correct fig. --------------- 1 mark
expression for magnifying power. --------------- 2 marks
b) formula for magnification and putting values ½ + ½ marks
reason for –ve sign ½ marks
correct ans for length of tube L ½ marks
QUESTION PAPERS L-2 - 3
QUESTION PAPER - L2
FOR CLASS XII - PHYSICS
BLUE PRINT
SL.
No
VSA
(1 Mark)
UNIT
SA I
SA II
(2 Marks) (3 Mark
1
Electrostatics
2(1)
6(2)
2
Current Electricity
4(2)
3(1)
3
Magnetic Effect of Current and Magnetism
1(1)
2(1)
4
E$lectromagnetic Induction and Alternating Currents
1(1)
4(2)
5
Electromagnetic Waves
1(1)
2(1)
6
Optics
1(1)
2(1)
7
Dual Nature of Matter
1(1)
8
Atoms and Nuclei.
1(1)
9
Electronic Devices
2(2)
10
Communication Systems
TOTAL
PHYSICS FOR CLASS XII
L-2
8(8)
3(1)
6(2)
3(1)
2(1)
3(1)
2(1)
3(1)
20(10)
27(9)
General Instructions:
(a)
All questions are compulsory
(b)
There are 30 questions in total. Qns. 1 to 8 carry 1 mark each. Qns. 9 to 18 carry 2 marks each,
Qns. 19 to 27 carry 3 mark each and Qns. 28 to 30 carry 5 marks each.
(c)
There is no overall choice. However, an internal choice has been provided in one question of 2
marks, one question of 3 marks and all three questions of 5 marks each. You have to attempt only
one of the given choices in such questions.
(d)
Use of calculators is not permitted.
(e)
You may use the following physical constants wherever necessary.
c = 3 x 108 ms-1
h = 6.6 x 10-34 Js
e = 1.6 x 10-19C
µo = 4π x 10-7 TmA-1
1/4πεo = 9 x 109 N m2C-2
Avogadro number
NA = 6.023 x 1023 mol-1
Mass of the neutron
= 1.675 x 10-27 kg
Boltzmann constant, k = 1.38 x 1023 J K-1
1
In each of the diodes shown, forward resistance is 50 Ω. Then find current
through the 100 Ω resistor in the given circuit.
D1
150 Ω
D2
50 Ω
6V
100 Ω
2
A convex lens and a concave lens of same focal length are combined in
contact. What would be the resulting focal length?
3
What orientation will a freely suspended magnetic needle have at (i)
geographic south and (ii) magnetic south pole of the earth?
4
What is the power factor of a pure inductor?
5
Compare the velocities of X – rays of wavelength 1 Ả and red light of wave
length 6000 Ả in vacuum.
6
In photo electric emission from the surface of a metal, when the wave length
of incident light is λ, the maximum kinetic energy of photo electron is E and
the number of emitted electrons is N. If now the intensity is doubled, keeping
λ fixed, what will be the effect on kinetic energy and number of photo
electrons emitted?
7
Calculate the ratio of energy of electron in the fourth and fifth orbits of
hydrogen atom.
8
Why is the base region of transistor thin and lightly doped ?
9
A 80 µF capacitor is charged by a 50 V battery. The capacitor is disconnected
from the battery and then connected across another uncharged 320 µF
capacitor. Calculate the charge on the second capacitor.
10 A potential difference V is applied to a copper wire of diameter D and length
l. What is the effect on the drift velocity of electrons when:
(i)
doubling V (ii)
doubling l
(iii)
doubling D?
11 For what value of unknown resistance X, the current flowing through the
galvanometer would be zero in the circuit shown below.
1Ω
6Ω
6 ΩB
G
1Ω
1Ω
X
6Ω
A
C
D
1Ω
12 State Ampere’s circuital law in magnetism and use it to find the magnetic
field inside a current carrying toroidal solenoid.
13 The reactance of an inductor is 20 Ω. What does it mean? What would be its
reactance, if the frequency of the a c is doubled? What opposition does the
inductor offer to the flow of a d c?
14 State Lenz’s law in electromagnetic induction and explain how it supports law
of conservation of energy.
15 Which part of the electromagnetic radiation:
(i)
is used for satellite communication
(ii)
has is frequency range 4 x 1014 Hz to 7 x 1014 Hz
(iii)
produce intense heating effect
(iv)
used for purification of water.
16 An equiconvex lens has a power of 5 D. If it is made of glass of refractive
index 1.5, calculate the radius of curvature of each surface.
OR
Draw a labelled ray diagram of an astronomical telescope used in the normal
adjustment position.
Two astronomical telescopes T1 and T2 have the same magnifying power. The
ratio of aperture of their objectives is 3:2. Which one of these two produces
image of greater intensity?
17 When a deuteron of mass 2.0141 u and negligible kinetic energy is absorbed
by a lithium (3Li6) nucleus of mass 6.0155 u, the compound nucleus
disintegrates spontaneously into two alpha particles, each of mass 4.0026 u.
Calculate the energy in joules carried by each alpha particle. ( 1 u = 1.66 x 1027
kg)
18 Give reasons for the following:
(i)
Long distance radio broadcasts use short-wave bands.
(ii)
Satellites are used for long distance TV transmission.
19 A metallic square loop ABCD of size 15cm and negligible resistance is moved
at a uniform velocity of v m/s in a uniform magnetic field of 2T. The field lines
being normal to the plane of the paper. The loop is connected to an electrical
network of resistances each of resistance 2 Ω. Calculate the speed of the
loop for which 2mA current flows in the loop.
20 An electric kettle has two windings. When one of them is switched on , the
water begins to boil in 15 min. and other is switched on in 30 min.
In
how much time will the water in the kettle begin to boil if the two windings
are switched on (i) in series (ii) in parallel.
21 A large vessel is filled with a liquid of dielectric constant k. Two vertical plates
touch the liquid. The dimensions of plates are a and b ,the distance between
them is d. The plates have been charged by applying the voltage V0 and then
disconnected from the source. To what height will the liquid rise.
22 Three charges -2µC , +22µ
-2µ
as shown in the fig. Calculate the total electric field intensity due to all these
three charges at point P.
P
1m
A
B
C
-2µC
-2µC
1m
22µC
1m
23 The lens combination of three lenses L1 and L2 and L3 shown in the fig. forms
an image at 30 cm to the right of the lens L3 . The focal lengths of lenses L1
and L2 are 10 cm and -10 cm respectively. What is the focal length of lens L3.
24 What changes will occur in the focal length of a (i) concave mirror and (ii)
convex lens when the incident violet light is replaced with red light. Give
reason.
25 What is the de-Broglie wavelength of an electron in an atom at absolute
temperature T K.
The wavelength λ of a photon and de-Broglie wavelength of an electron
have the same value. Show that energy of photon is
2mc
times the kinetic
h
energy of electron.
26 state radioactive decay law and establish the relation between decay
constant and half life period of a radio active substance.
OR
Write the nuclear decay equations for:
(i)
α – decay of 88 Ra 226
(ii)
α – decay of 94 Pu 242
(iii)
β- – decay of 15 P 32
(iv)
β- – decay of 83 Bi 210
(v)
β+ – decay of 6 C 11
(vi)
β+ – decay of 43 Tc 97
27 Define the term Modulation . Name three different type of modulations used
for a message signal, using sinusoidal continuous carrier wave. Explain the
meaning of any one of these.
28 Explain the method of obtaining a n – type semiconductor from a sample of
intrinsic semi conductor. Draw the energy band diagram of the n – type semi
conductor.
A semiconductor has equal electron and hole concentration of 6 x 108/m3. On
doping with certain impurity, electron concentration increases to 9 x 1012/m3.
(i)
Identify the new semiconductor obtained after doping.
(ii)
Calculate the new hole concentration.
OR
What is an AND gate? Write its truth table and Boolean expression for its
output. Draw the logic symbol
With the help of a neat circuit diagram, explain the realization of an AND
gate using semiconductor diodes.
Write the truth table for the following combination of four NOR gates
arranged as shown in figure.
A
B
Y
29 With the help of a suitable diagram, describe Young’s double slit experiment
for obtaining interference pattern. Write an expression for the fringe width
of the pattern. Represent graphically the variation of intensity in the pattern.
What will be the effect on interference fringes obtained in Young’s double
slit experiment, if
(i)
(ii)
(iii)
(iv)
source of light of longer wavelength is used
distance between two slits be increased to larger extent
two slits are illuminated by different sources of light of same λ
if the distance of the source and the double slit is increased?
OR
Differentiate between plane polarized and un polarized light. Describe a
method to get a plane polarized light by reflection.
Discuss the intensity of transmitted light when a polaroid sheet is rotated
between two crossed polaroids?
What is the Brewster angle for air to glass transition? (Refractive index of
glass=√3.)
30 Two circular coils X and Y having radii R and R/2 respectively are placed in
horizontal plane with their centres coinciding with each other. Coil X has a
current I flowing through it in the clockwise sense. What must be the current
in coil Y to make the total magnetic field at the common centre of the two
coils, zero?
With the above currents flowing in the two coils, if the coil Y is now lifted
vertically upwards through a distance R, what would be the net magnetic
field at the centre of coil Y?
OR
Define current sensitivity and voltage sensitivity of a moving coil
galvanometer.
Two moving coil meters, M1 and M2 have the following particulars:
R1 = 10 Ω, N1 = 30, A1 = 3.6 × 10–3 m2, B1 = 0.25 T
R2 = 14 Ω, N2 = 42, A2 = 1.8 × 10–3 m2, B2 = 0.50 T
(The spring constants are identical for the two meters). Determine the ratio
of (a) current sensitivity and (b) voltage sensitivity of M2 and M1.
How can a galvanometer be converted into (i) an ammeter and (ii) a
voltmeter of required range?
*******************************************
Paper L 2
PHYSICS FOR CLASS XII
Qn.
No.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
Value points
Current I = V/Reff
I = 24 mA
Resulting focal length is infinity
(i)
Pointing in the resultant magnetic field depending on angle of dip
(ii)
Vertical orientation
Zero
Ratio of velocities : 1
(i)
Kinetic energy remains unaltered
(ii)
Number of photo electrons gets doubled
Equation for energy in the orbit En = - 13.6/n2
E4/E5 = 52/42 = 25/16
Reason for the base region to be thin and lightly doped
Calculation of common potential V = C1 V1/( C1 + C2)
Result v = 10 V
Calculation of charge q2 = C2 V
Result q2 = 3.2 μC
Equation for drift velocity vD = I / n A e = V/ρle
(i)
Drift velocity increases
(ii)
Reduces to half
(iii)
remains constant
Balancing condition of the bridge
Calculation of unknown resistance X
Result X = 6Ω
Statement of Ampere’s law
Derivation
Result B = μo n I
Opposition offered to the flow a c
Reactance gets doubled
Zero
Statement of Lenz’s law
Explanation
(i)
Microwaves
(ii)
Visible
(iii)
I R rays
(iv)
U V rays
Marks
½
½
1
½
½
1
1
½
½
½
½
1
½
½
½
½
½
½
½
½
½
1
½
½
1
½
½
1
½
1
1
½
½
½
½
Total
Marks
1
1
1
1
1
1
1
1
2
2
2
2
2
2
2
16
Lens maker’s formula
Calculation for radius of curvature
Result R = 20 cm
½
1
½
2
OR
17
18
19
20
21
22
23
24
25
26
27
28
Diagram
T1 produces image of higher intensity
Total mass of reactants and Total mass of products
Calculation of mass defect
Calculation of enery released
Result Energy of each α – particle = 1.822 x 10-22J
(i)
Reason for use of short wave
(ii)
Reason for use of satellites
Equation Blv = I Reff
Calculation of effective resistance Reff = 2Ω
Calculation of velocity
Result v = 1/75 ms-1
Equation Q = V2 t1/R1 = V2 t2/R2
In series t = [R1 + R2]Q/V2
Result T = 45 min
In parallel t =RP Q/V2
Calculation
Result t = 10 min
Energy stored in the capacitor = ½ CV2 = mgh
Capacity C = Kεoab/d
Calculation for height
Result h = [K a εo Vo2/2dρg]1/2
Diagram showing orientation of fields
Calculation of E1 and E3
Result E1 = E3 = K x 10-6/4√2
Resultant of E1 and E2 = K x 10-6 = 9 x 103
Calculation of E2
Result E2 = 2K√2 x 10-6 = 25.45 x 103
Net field = 16.45 x 103 N/C
Calculation of image distance for the first lens using lens equation as 15 cm
Image distance for the second lens as infinity
Image distance for the third lens as 30 cm
(Page No. 330 of NCERT Text book)
(i)
No change
(ii)
Focal length increases
Correct reasons
λ = h/√(3mkT)
Energy of photon E = hc/λ
K E of electron Ek = ½ mv2 = p2/2m =h2/λ2.2m
Finding E/Ek and statement of result
Statement of decay law
Mathematical form
Derivation of relationship
Result T1/2 = 0.693/λ
OR
Six correct nuclear reactions showing emission of neutrino or anti neutrino for β-emission
Definition
Types of modulations
Explanation with diagram of any one type
Explanation for making n – type semiconductor by doping with penta valent impurity
Energy band diagram showing donor level near conduction band
1
1
½
½
½
½
1
1
½
1
1
½
½
½
½
½
½
½
½
½
1½
½
½
½
½
½
½
½
1
1
1
1
1
½+½
½
½
1½
½
½
½
1½
½
6x½
½
1½
1
1½
1
2
2
2
3
3
3
3
3
3
3
3
3
3
Identification as n – type semiconductor
Calculation using equation ni2 = ne nh
Result nh = 4 x 104 m-3
1
1
½
5
OR
29
30
Definition of AND gate
Truth table
Logic symbol
Boolean expression
Circuit diagram
Expalnation
Truth table for the combination given
Diagram of Young’s double slit arrangement
Derivation of equation for fringe width
Intensity distribution graph
(i)
Fringe width increases
(ii)
No interference pattern is observed
(iii)
No interference pattern is produced
(iv)
No interference pattern is produced
OR
Differences between polarised and unpolarised light
Description of polarisation by reflection stating the idea of Brewster angle
Variation of intensity from 0 to maximum to 0 and so on
Brewster equation
Calculation
Result θ = 60o
Equation for magnetic field B = μoI/2r
Calculation of current in coil Y
Result I’ = I/2
Equations B1 = μoIR2/2[R2 + R2]3/2 anb B2 = - μoI/[4R]
Calculation
Result B = μo I(1 - √2)/4R√2
OR
Current sensitivity
Voltage sensitivity
(a) ratio of current sensitivities = 5:7
(b) ratio of voltage sensitivities = 1:1
Conversion of GM to ammeter
Conversion of G M to voltmeter
QUESTION PAPERS L-2 - 4
½
½
½
½
1
1
1
1
1
1
½
½
½
½
1
1
1
½
1
1/2
½
1
½
1
1
1
½
½
1
1
1
1
5
5
5
5
5
SAMPLE QUESTION PAPER (Level II)
FOR CLASS XII - PHYSICS
BLUE PRINT
SL.
No
UNIT
VSA
(1
Mark)
SA I
(2
Marks)
1(1)
2(1)
1
Electrostatics
2
Current Electricity
3
Magnetic Effect of Current and Magnetism
1(1)
2(1)
4
E$lectromagnetic Induction and Alternating Currents
1(1)
4(2)
5
Electromagnetic Waves
1(1)
2(1)
6
Optics
1(1)
2(1)
7
Dual Nature of Matter
1(1)
8
Atoms and Nuclei.
9
Electronic Devices
10
Communication Systems
4(2)
SA II
(3
Marks
3(1)
3(1)
6(2)
3(1)
6(2)
2(2)
TOTAL
8(8)
2(1)
3(1)
2(1)
3(1)
20(10)
27(9)
SAMPLE PAPER (Level II)
PHYSICS FOR CLASS XII
General Instructions:
(a)
All questions are compulsory
(b)
There are 30 questions in total. Qns. 1 to 8 carry 1 mark each. Qns. 9 to 18 carry 2 marks each,
Qns. 19 to 27 carry 3 mark each and Qns. 28 to 30 carry 5 marks each.
(c)
There is no overall choice. However, an internal choice has been provided in one question of 2
marks, one question of 3 marks and all three questions of 5 marks each. You have to attempt only
one of the given choices in such questions.
(d)
Use of calculators is not permitted.
(e)
You may use the following physical constants wherever necessary.
c = 3 x 108 ms-1
h = 6.6 x 10-34 Js
e = 1.6 x 10-19C
µo = 4π x 10-7 TmA-1
1/4πεo = 9 x 109 N m2C-2
Avogadro number
NA = 6.023 x 1023 mol-1
Mass of the neutron
= 1.675 x 10-27 kg
Boltzmann constant, k = 1.38 x 1023 J K-1
1
Two point charges q1 and q2 are placed close to each other in air. What is the
nature of the force between them when (i) q1 q2 > 0 and (ii) q1 q2 < 0
2
An electron and a proton moving parallel to each other in the same direction
with equal momenta, enter into a uniform magnetic field which is at right
angles to their velocities. Trace their trajectories in the magnetic field.
3
In a series L C R circuit the voltage across an inductor, capacitor and the
resistor are 30 V, 30 V and 60 V respectively. What is the phase difference
between the applied voltage and current in the circuit?
4
What physical quantity is the same for X – rays of wavelength 10-10 m, red
light of wavelength 6800Ả and radio waves of wavelength 500 m?
5
The refractive index of a medium is √3. What is the angle of refraction, if the
unpolarised light is incident on it at the polarizing of the medium?
6
The following graph shows the variation of stopping potential Vo with the
frequency ν of the incident radiation for two photosensitive metals X and Y.
Vo
O
X
0.5
Y
1.0
ν (x 1015 s-1)
Which of the metals has larger threshold wavelength? Give reason.
7
How is the band gap, Eg, of a photo diode related to the maximum
wavelength, λm , that can be detected by it?
8
An unknown input (A) and the input (B) shown below, are used as the two
inputs in a NAND gate. The output Y, has the form shown below. Identify the
intervals over which the input ‘A’ must be ‘low’.
1
B
0
1
Y
0
0
9
t1
t2
t3
t4
t5
Two point electric charges of values q and 2q are kept at a distance ‘d’ apart
from each other in air. A third charge Q is kept along the same line such a
way that the net force acting on q and 2q is zero. Calculate the position of
the charge ‘Q’ in terms of ‘q’ and ‘d’.
OR
Two point charges 4µC and – 2 µC are separated by a distance of 1 m in air.
At what point on the line joining the charges is the electric potential zero?
10 Two cells of e m f E1 and E2 are connected together in two ways as shown.
E1
E2
E1
E2
The balance points in a given potentiometer experiment for these two
combinations of cells are found to be at 351 cm and 70.2 cm respectively.
Calculate the ratio of the e m fs of the two cells.
11 A set of ‘n’ identical resistors, each of resistance R Ω, when connected in
series have an effective resistance X Ω, and when the resistors are connected
in parallel their effective resistance is Y Ω. Find the relation between R, X and
Y.
12 A stream of electrons travelling with a speed ‘v’ m s-1 at right angles to a
uniform magnetic field B is deflected in a circular path of radius ‘r’. Prove
that (e/m) = (v/rB)
13 A rectangular coil of ‘N’ turns and area of cross section ‘A’ is held in a time –
varying magnetic field given by B = Bo Sin ωt, with the plane of the coil
normal to the magnetic field. Deduce an expression for the e m f induced in
the coil.
14 An electric lamp, which runs at 80 V d c and consumes 10 A is connected to a
100 V, 50 Hz a c mains. Calculate the inductance of the choke required.
15 Which constituent radiation of the electromagnetic spectrum is used:
(i)
in radar
(ii)
to photograph internal parts of human body
(iii)
for taking photographs of the sky during the night and foggy
conditions
(iv)
emitted during a nuclear decay.
16 Two nearby narrow slits are illuminated by a single monochromatic source.
Name the pattern obtained on the screen. One of the slits is now completely
covered. What is the name of the pattern now obtained on the screen. Write
any one difference between the patterns obtained in the two cases.
17 What do the terms ‘depletion region’ and ‘barrier potential’ mean for a p-n
junction?
18 Show that the range of transmission `d’ of a T.V. tower of height `h’ is given
by the relation d = √(2Rh) , where `R’ is the radius of the Earth.
19 Define the term resistivity and write its S I unit. Derive the expression for
resistivity of a conductor in terms of number density of electrons and
relaxation time.
20 An a c voltage E = Eo Sin ωt is applied across a pure capacitor of capacitance
C. Show mathematically that the current flowing through it leads the applied
voltage by a phase angle of π/2.
OR
Show that the average value of the alternating voltage E = Eo Sin ωt, over the
first half cycle is 2IO/π
21 You are given three lenses having powers P and aperture A as follows;
P1 = 6 D
A1 = 3 cm
P2 = 3D
A2 = 15 cm
P3 = 12 D
A3 = 1.5 cm
Which two of these will you select to construct (i) telescope (ii) microscope?
State the basis for your answer in each case.
22 What is the effect on the interference fringes in a Young’s double-slit
experiment due to each of the following operations:
(a) the screen is moved away from the plane of the slits;
(b) the (monochromatic) source is replaced by another (monochromatic)
source of shorter wavelength;
(c) the separation between the two slits is increased.
Give reason for all.
23 Draw the graph showing the variation of binding energy per nucleon with
mass number of different atomic nuclei.
Calculate binding energy/nucleon of 26 Fe 56 nucleus.
[Given: Mass of 26 Fe 56 = 55.934939 u, mass of proton = 1.007825 u,
mass of neutron = 1.008665 u, 1 u = 931 MeV/c2]
24 Define the term modulation. Name three different types of modulation used
for a message signal using a sinusoidal continuous carrier wave. Explain the
meaning of any one of these
25 Red light, however bright it is, can not produce the emission of electrons
from a clean zinc surface. But even weak ultra violet radiation can do so.
Why?
X – rays of wavelength `λ’ fall on a photosensitive surface, emitting electrons.
Assuming that the work function of the surface is negligible, prove that the
de – Broglie wavelength of electrons emitted will be [h λ/2mc]1/2
26 The energy levels of an atom are shown below. Which of the transitions will
result in the emission of a photon of wavelength 275 nm?
A
B
0 eV
C D
-2 eV
-4.5 eV
-10 eV
27 Draw a circuit diagram for use of a n-p-n transistor as an amplifier in
common emitter configuration.
The input resistance of a transistor is 1000 Ω . On changing its base current
by 10μA the collector current increases by 2mA. If a load resistance of 5kΩ is
used in the circuit, calculate; (i) current gain and (ii) voltage gain of the
amplifier.
I
28 Derive an expression for the capacity of a parallel plate capacitor with a
dielectric slab of thickness, less than the plate separation, placed between
the plates.
A parallel plate capacitor is to be designed with a voltage rating of 1 kV using
a material of dielectric constant 3 and dielectric strength about 107 Vm-1. For
safety we would like the field never to exceed say 10 % of dipole strength.
What minimum area of the plates is required to have a capacitance of 50 pF?
OR
What is an electric dipole? Deduce an expression for the torque acting on an
electric dipole placed in a uniform magnetic field. Hence define dipole
moment.An electric dipole of length 2 cm is placed with its axis making an
angle of 60o to a uniform electric field of 105 NC-1. If it experiences a torque
of 8√3 Nm, calculate the; (i)
magnitude of charge on the dipole.
(ii)
potential energy of the dipole.
29 Explain the principle and working of a cyclotron, with the help of a labelled
diagram.
A cyclotron’s oscillator frequency is 10 MHz. What should be the operating
magnetic field for accelerating protons? If the radius of its dees is 60 cm,
what is the kinetic energy of the proton beam produced by the accelerator?
Express your answer in MeV. (Charge of proton = 1.6 x 10-19 C, mass of
proton = 1.67x10-27 kg, 1 MeV = 1.6 x 10-13 J)
OR
With the help of a neat labelled diagram, explain the principle and
construction of a moving coil galvanometer. Define the terms current
sensitivity and voltage sensitivity of a galvanometer.
In a galvanometer there is a deflection of 10 div per mA. The coil resistance
of the galvanometer is 78 Ω. If a shunt of 2Ω is connected to the
galvanometer and there are 75 divisions in all on the scale of the
galvanometer, calculate the maximum current which the galvanometer can
read.
30 Draw a ray diagram to show the formation of the image of a point object
placed in a medium of refractive index n1 on the principal axis of a convex
spherical surface of radius of curvature R and refractive index n2. Using the
diagram derive the relation;
n2
n1
n2 - n1
v
u
R
A converging lens of focal length 50 cm is placed co axially in contact with
another lens of unknown focal length. If the combination behaves like a
diverging lens of focal length 50 cm, find the power and nature of the second
lens.
OR
Derive the lens maker’s formula in case of a double convex lens. State the
assumptions made and conventions of signs used.
A converging lens has a focal length of 20 cm in air. It is made of a material of
refractive index 1.6. If it is immersed in a liquid of refractive index 1.3, what
will be its new focal length?
SAMPLE PAPER (Level II)
PHYSICS FOR CLASS XII
Qn.
No.
Value points
Marks
Total
Marks
1
q1 q2 > 0 – repulsive
½
q1 q2 < 0 – attractive
½
1
2
Circles with same radius oriented in opposite directions
1
1
3
Phase difference = 0
1
1
4
Velocity
1
1
5
θ = tan-1 √3 = 60o
½
r = 180o – (90o + 60o) = 30o
½
1
6
Metal X is having higher threshold wave length
1
1
7
Energy gap, Eg = hc/λm
1
1
8
0 to t1 and t3 to t4
½+½
1
9
Diagram or assumption of distance
½
Kq q/x2 = K Q 2q?(d – x)2
½
Solution
½
x = d/(√2 + 1)
½
OR
K x 4q/x – K 2q/(1 – x) = 0
½
Calculation
1
2
10
11
12
13
14
15
Result x = 2/3 m
½
E1 + E2 = K x 351
½
E1 – E2 = K x 70.2
½
Calculation
½
E1/E2 = 1.57
½
X=nR
½
Y = R/n
½
R2 = X Y
½
R = √(X Y)
½
mv2/r = e v B
1
e/m = v/rB
1
Flux Φ = N A Bo Sin ωt
½
E = - (dΦ/dt)
½
E = - NABoω Cosωt
1
R = 8Ω and Z = 10Ω
½
Z = √(R2 + X2)
½
X2 = 36 and X = 6 Ω
½
L = 6/ 100π H
½
(i)
(ii)
(iii)
(iv)
microwaves or radio waves
X – rays
I R radiations
γ - rays
2
2
2
2
2
2
½
½
½
½
16
17
18
Interference pattern
½
Diffraction pattern
½
Any one difference
1
Explanation of depletion region
1
Explanation of potential barrier
1
Diagram
½
2
2
2
Derivation of the equation
19
20
1½
Definition
½
Unit
½
Derivation of euation ρ = m/n e2 τ
2
Diagram or p d across the capacitor q/c = Eo Sin ωt
½
I = dq/dt
½
Derivation
1
Result I = Io Cos ωt
½
Statement Current leads the voltage as Cos ωt = Sin(ωt + π/2)
½
2
3
3
OR
21
22
Average value Eav = [∫Edt] /[∫dt]
1
Further step and derivation
2
(i) Telescope P2 and P3
1
(ii)Microscope P3 and P1
1
Correct reasoning
1
(a) fringe width increases
(b) fringe width decreases
(c)
fringe width decreases
Suitable reasons using relevant equation
½
3
3
½
½
1½
23
Diagram with approximate shape
Binding energy calculation
24
25
1
1½
Result B E per nucleon = 8.79 MeV
½
Definition
½
Types of modulations(three)
3
3
1½
Explanation of any one
1
Threshold frequency of zinc is more than frequency of red light and less than that of U V rays
1
Energy of photon E = hc/λ
½
de – Broglie wavelength λ’ = h/√(2mE)
½
3
26
λ’ = h/ (2mhc/λ) and final result
1
Energy of the photon E = hc/λ
½
Calculation of Energy of the photon
27
28
3
1½
Result E = 4.5 eV
½
Identification of transition B
½
Correct circuit diagram with proper biasing for the junctions
1
Current gain β = ΔIC/ΔIB
½
Result β = 200
½
Voltage gain AV = β RL/Ri
½
Result AV = 1000
½
Diagram or equivalent description
1
Derivation of equation for capacity with dielectric slab
2
Electric field E = 106 Vm-1
½
d = V/E = 10-3 m
½
Area A = C d/ Kεo
½
Result A = 1.88 x 10-3 m2
½
3
3
5
OR
29
Definition of dipole
½
Derivation of equation for torque
2
Definition for dipole moment
½
Torque τ = 2a q E Sinθ
½
Charge q = 8 x 10-3 C
½
Potential energy = 2a q E Cosθ = 8J
1
Diagram
1
Principle
1
Explanation of working
1
Frequency ν = qB/2πm
½
Calculation of B and result B = 0.65 T
½
5
Energy of proton = q 2 B2 R2/2 m
½
Result e = 6.8 MeV
½
5
OR
Diagram of M C G
1
Construction
1
Principle
½
Definitions Current sensitivity
½
Voltage sensitivity
½
Conversion formula I/IG = 1 + G/S
½
IG = 7.5 mA
½
Calculation of I
30
Result I = 300 mA
½
Diagarm
1
Derivation of the equation
2
Resulting power,P = P1 + P2
½
Power of the second lens P2 = - 4 D
1
Nature of the lens – Diverging
½
5
5
OR
Diagram
1
Derivation of lens maker’s formula
1½
Sign conventions and assumptions
½
Statement of lens maker’s formula
½
Substitution and further steps
1
New focal length f2 = 52 cm
½
5
QUESTION PAPERS L-2 - 5
QUESTION PAPER WITH HIGHER DIFFICULTY LEVEL - II
1. Which physical quantity has S I unit
(i) C- m
(ii) V-m
2.
3.
4.
5.
Graphically show how the resistivity of carbon varies with temperature?
In a LCR circuit if VL = VC find the power factor of LCR circuit.
Why potentiometer is preferred over voltmeter to measure emf of a cell?
Which one of the following will experience maximum force, when projected with same velocity v
perpendicular to the magnetic field B (a) α – particle(b) β – particle
6. For what value of angular diffraction minima is not obtained on the screen of a single slit diffraction
set up?
7. A glass is immersed in water. How is power of lens affected? Justify your answer.
8. The figure shows the variation of photocurrent with anode potential for a photosensitive surface fro
three different radiations. I1, I2 and I3 are intensities and υ1 υ2 υ3 be the frequencies for the curves 1,
2 and 3 respectively.
Figure
9.
Write the expression for the magnitude of force per unit length between two infinitely long
parallel straight current carrying conductors. Hence, define S I unit of current.
10. State the principle of an ac generator. Write an expression for the maximum emf produced in it.
11. Three capacitance C1= 10 μF C2 = C3 = 5 μF are connected to a battery as shown in the figure. Find
charges on the three capacitors.
12. A semiconductor has equal electron and hole concentration of 6×108 m-3. On doping with a certain
impurity, electron concentration increases to 8×1012 m-3
(i) Calculate new hole concentration.
(ii) Draw new energy level diagram.
Or,
Draw energy level diagram for an
(a) Unbiased p – n junction diode.
(b) Forward biased p – n junction diode and explain how the barrier potential changes?
13. Why modulation of message signals is necessary?
14. Electric field inside a sphere varies with distance as E = Ar. Find the total charge enclosed within the
sphere. If A = 3000Vm-2 and r = 30 cm, where r is the radius of sphere.
15. Potential difference across terminal of a cell were measured (in volts) against different currents (in
ampere) flowing through the cell. A graph was drawn which was a straight line ABC. Using the data
given in the graph, determine (i) emf and (ii) internal resistance of the cell.
Figure
16. A copper coil L wound on a soft iron core and a lamp B are connected to a battery E through a
tapping key K. When the key is closed, the lamp glows dimly. But when the key is suddenly opened,
the lamp flashes for an instant to much greater brightness. Explain
Figure
17. Find the position of the image formed by the lens shown in the figure.
Figure
18. A square wire of size 3 cm is placed 25 cm away from a concave mirror of focal length 10 cm. What
is the area enclosed by the image of the wire? (The centre of the wire is on the axis of the mirror
with its two sides normal to the axis).
19. A nucleus makes a transition from permitted energy level to another level of lower energy. Name
the region of electromagnetic spectrum to which the emitted photon belongs. What is the order of
its energy in electron volt? Write four characteristics of nuclear forces.
20. Explain with the help of circuit diagram, how we can detect amplitude modulated wave?
21. The half life of C–14 is 5700 years. What does it mean? Two radioactive nuclei X and Y initially
contain an equal number of atoms. Their half lives are 1 hours and 2 hours respectively. Calculate
the ratio of their rates of disintegration after two hours.
22. Why are de – Broglie waves associated with a moving football not visible?
23. Electromagnetic waves with wavelength(i) λ1 are used to treat muscular strain.
(ii) λ2 are used by a FM radio station for broadcasting
(iii) λ3 are used to detect fracture in bones
(iv) λ4 are absorbed by the ozone layer of the atmosphere.
Identify and name the part of electromagnetic spectrum to which these radiation belong
24. Derive an expression for fringe width in a Young’s double slit experiment.
25. A capacitor C is connected to a Bulb to an ac source as shown in the figure alongside. Explain how
the brightness of bulb changes does.
(i)
(ii)
(iii)
Frequency of ac source is decreased.
A dielectric slab is placed between plates of the capacitor.
An inductor of reactance XL = XC is placed in the same circuit.
26. A thin uniform wire AB of length 1 m and unknown resistance X. if value of resistance R is 12 W and
balancing length AC = 40 cm, then
(i) Value of unknown resistance X.
(ii) How the balancing length changes in resistance R is increased?
(iii) Shift in balancing length if R and X are interchanged.
27. Two dipoles A and B of very small dimensions are kept in the field of +Q charge as shown. Explain
the motion of dipole A and B with reason.
28. What are the two main considerations that have to be kept in mind while designing of an
astronomical telescope?
Obtain an expression for angular magnifying power and the length of a tube of an astronomical
telescope in its normal adjustment position.
An astronomical telescope having an objective of focal length 2 m and an eye piece of focal
length 1cm is used to observe a pair of star with actual angular separation of 0.75 what would be
their observed angular separation as seen through the telescope?
Or,
What are the two ways of adjusting the position of eye piece by observing the final image in
compound microscope? Which of these is usually preferred and why? Obtain expression for the
magnifying power of a compound microscope. Hence, explain why
(i) we prefer both the objective and eye piece to have small focal length.
(ii) We regard the lengths of microscope tube to be nearly equal to the separation between the focal
points of its objective and its eye piece?
(iii) Calculate the magnification obtained by an compound microscope having an objective of focal
length 1.5 cm and eye piece of focal length 2.5 cm and tube length of 30 cm.
29. Draw the circuit diagrams of the circuit used to study the out put characteristics of npn transistor in
the CE configuration. Give the shape of these characteristics and use them to define the
(i) output resistance
(ii) Current amplification factor
Or
With the help of suitable diagrams explain the working of transistor as an oscillator.
30. State Biot Savart law use it to obtain the magnetic field at an external point distance x from the
centre of circular coil of radius ‘a’ carrying current ‘I’. hence compare the magnitudes of the
magnetic field of the coil at its centre and an axial point for which x =√3a
Or
How will dia, para and ferromagnetic materials behave when kept in non – uniform external
magnetic field? Give two example of each of these materials. Name two main characteristics of
ferromagnetic materials which help us to decide its suitability for making
(i) Permanent magnet
(ii) Electromagnet
SOLUTIONS
1. (i) Dipole moment (ii) Electric flux
2.
3. Power factor = cos φ = 1.
4. When voltmeter is connected across terminals of the cell it draws current from it . hence it
measures terminal potential difference across the cell which is always less than the emf of
the cell.
While potentiometer is based on null deflection method and does not draw any current
from the cell. Hence it measures the emf of the cell.
5.
F = qvBsinθ F α q
For α – particle q = 2e for β – particle q = -e
Hence Fα > Fβ
6. a sinθ = nλ , n = 1 sinθ = λ/a
if θ = π /2 first minima is found at angular position of π /2.
Therefore λ ≥ a, minima is not obtained on the screen.
7. On immersing the lens its focal length increases hence its power decreases.
8. υ1 = υ2 as they have same stopping potential. I2 = I3 they have same saturated current.
9. Force per unit length; F/l = (μ0 I1 I2) / 2πr
Force is attractive if currents are in the same direction and repulsive if currents are in the
opposite direction.
10. Principle of ac Generator: - is based on electromagnetic induction i. e. Whenever magnetic
flux through a coil changes with time, e.mf in induced in it.
E0 = NBAω
11. C2 and C3 are parallel C23 = 5+5= 10μF. C1 and C23 are equal hence the p.d. across them is
same.
V = 6/2 = 3 V
Charge of C1, q1 = C1V =10×3×10-6 = 30 μC similarly q2 = q3 =15 μC
12. (i) n2 i = ne nh
nh = (6×108 m-3)2/ 8×1012 m-3 = 4.5×104m-3
(ii) n – type semiconductor.
Or
Unbiased pn junction - NCERT Text Book page no. 479 fig. 14.11 (b)
Forward Biased - NCERT Text Book page no. 480 fig. 14.13 (b)
13. (i) for efficient transmission length of antenna must be at least equal to ¼ of the message
signal. For low frequency signal it is not possible.
(ii) Energy radiated by antenna for frequency below 15 kHz is practically zero. So message
below this frequency are converted to higher frequency signal.’
(iii) To avoid mixing of signals simultaneously beamed from different sources. The message
signals are needed to be modulated.
14. Electric Flux through a sphere
∫E. ds = ∫Eds cos0 = ∫Eds
φ= E∫ds
φ =E 4πr2 = Ar 4πr2 = 4πAr3
φ = q / ε0 = 4πAr3
q = 4π ε0Ar3 = 9×10-9 C.
15. Emf = intercept of Y – axis = 1.4 V (When no current drawn from the cell.)
For point A ,
V = E – ir = 1.4 – 0.04 r
0.2 = 0.04 r
r = 5 ohm.
16. When key is suddenly opened, current through the inductor decreases. Decreasing current
induces emf in the coil. An induced current also flows through the lamp in addition do the
current send by the battery E. Hence, lamp flashes with greater brightness.
17. f = 10 cm , u = -30 cm v = ?
1/f = 1/v – 1/u
1/v = 1/f + 1/ u
1/v = 1/10 – 1/ 30 = 1/15
v = 15 cm
Second lens is concave lens.
18. u = -25 cm , f = -10 cm, O = 3
1/f = 1/v + 1/u
v = -50/3 cm.
m = - v/u = I/ O
I = -2 cm
Area enclosed by the image 2 × 2 = 4 cm2
19. (i) emitted photon belong to γ rays.
(ii)
Energy is of the order of MeV.
(iii)
Four characteristics of nuclear forces.
(a) They are independent of charges
(b) They do not obey inverse square law.
(c) They are non central forces.
(d) They short range forces.
20. To detect an amplitude modulated signal, we need antenna the signals are then amplified.
Then signal fed to detector. This circuit can read the analog block diagram of detector. This
circuit can read the envelop.
Block diagram of detector for amplitude modulated signal. – NCERT Text Book Page no. 527
fig 15.13.
21. The half life mean time in which half of present number of a radioactive nuclei of the given
substance. Here C -14 remain undecayed.
No. Of nuclei of X after 2 Hrs
N = No (½ )n ; n = t / T1/2
22. de – Broglie wave length for a moving object is given as
λ = h/ mv , mv = h / λ
EK = (mv)2/ 2m = (h / λ)2/ 2m =
Energy of photon E P= hc/ λ
EK/ EP = h/2m λc
23. (i) infrared
(ii)
radio and microwave
(iii)
X – rays
(iv)
U V rays.
λ2 > λ1> λ4 >:λ3
24. Let S1 and S2 be two coherent sources. The disturbances due to them reach at point P with a
certain path differences is
S2P – S1P = Δ x
Δ x = d sin θ (θ is very small)
tan θ = y/D (θ is very small)
Δ x/d = y / D
Δ x = yd / D
For maxima, Δ x = n λ
yn = n λD/d
β = yn+1 – yn = D λ/d
Fig. NCERT Text Book Page No. 363 Fig. 10.12 (b)
25. (i) When frequency is decreased XC = 1/ωC increase. Hence, current in the circuit decrease.
As a result brightness of bulb will decrease.
(ii) When dielectric slab is placed between the plate of capacitor, Capacitance increases then
reactance will decrease. Hence, brightness of bulb will increase.
(iii) Inductor is place in series, net impedance of circuit becomes R as XL = XC. Brightness of
bulb becomes maximum.
26. (i) R/AC = X/ CB
X = 12×60/40 = 18 ohm
(ii) If R is increase AC will also increase.
(iii) If R and X are interchanged then AC and CB lengths get interchanged. Shift is equal to 20
cm.
27. For dipole A and B Net force on the dipole is zero as the size of the dipole is very small and
charge on dipole is zero. Torque on dipole A
τ = pE sin θ , (θ = 0)
τ=0
For dipole B
τ = pE , (as θ = 900)
28. Objective lens should be selected as
(i)
To collect as much light as possible.
(ii)
To have high resolving power.
Figure: - NCERT Text Book page No. 342 Fig. 9.32
M = β/α;
β = A’B’ / B’E
α = A’ B’ / OB’
M = O B’ / B’E
M = fo / fe = 2/10-2 = 200.
L = fo+ fe
Observed angular separation = α M = 0.75’ ×200 = 2030’
Or,
Two ways to adjust the position of eye piece
(i)
Final image is formed at the distance of distinct vision.
(ii)
Final image is formed at infinity.
Second adjustment is preferred as the observer is relaxed while observing the image.
Ray Diagram: NCERT Text Book Page No. 340, Fig 9.31
M = β/α;
M = mo×me
mo = h’/ h = L/ fo
me = 1 + D/ fe
So, M = L/ fo (1 + D/ fe)
M = L/ fo (D/ fe) = 200.
29. Circuit Diagram: - NCERT Text Book page no. 493 Fig 14.29.
Output characteristics – As Shown in NCERT Text Book Page No. 494 Fig. 14.30 (b).
Output Resistance: -Defined as the ratio of change in collector emitter voltage to the
change in collector current at a constant base current.
R0 = (ΔVCE/ΔIC) at constant Ib.
Current amplification factor: β = (ΔIC/ ΔIB)
Or,
Circuit Diagram: -NCERT Text Book page no. 500, Fig. 14.33 (b) and (c).
Suppose Switch S1 is put on to apply proper bias for the first time. Obviously collector
current flow in the transistor. The current flow through the coil T2. The current does not reach
full amplitude instantaneously but increase X to Y as shown in figure. The inductive coupling
between coil T1 and T2 . Now causes a current to flow in the emitter circuit. As a result of this
positive feed back this current also increased X’ To Y’ as shown in figure. The current in T2
connected in the collector circuit acquires the value Y when the transistor becomes saturated.
This means the maximum collector current is flowing and can increase no further. Since there is
no further change in collector current without continues feedback the emitter current become
to fall. Consequently collector current decrease from Y to Z as shown in figure. However a
decrease of collector current causes the magnetic field to decay around the coil T2. thus T1 is
now seeing a decaying field in T2 this cause further decrease in emitter current till it reaches Z’
when the transistor is cut off the both Ie and Ic cease to flow . Therefore, transistor reverted
back to its original state. The whole process is now repeats itself.
30. Magnetic field at an axial coil due to a current carrying loop :-
dB = μ0 I dl / 4πr2
Components of magnetic field dB along Y – axis = dB cosθ
Components of magnetic field dB along X – axis dB sinθ
Vertical components are equal and opposite and hence cancels each other and
net magnetic field B = ∫dB sinθ
B = ∫μ0 I dl / 4πr2 (a/r)
B = μ0I2π a2/(4πr3)
B = μ0 Ia2/ 2r3.
r2 = a2 + X2
B = μ0Ia2/ 2 (a2 + X2)3/2
At X = √3 a,
B = μ0I/ 8a
B0/B = 4.
Or.
When kept in a non uniform magnetic field
(i)
Diamagnetic substance has a tendency to move from a stronger region to a weaker
region of field.
Ex. Sb, Bi.
(ii)
Paramagnetic substance has a tendency to move from a weaker region to a stronger
region of field.
Ex. Al, O2.
(iii)
ferroamagnetic substance has a tendency to move from a weaker region to a stronger
region of field.
Fe and Ni.
Two characteristics of Ferromagnetic Substance for making
Permanent magnet: -Large retentivity and high coercivity.
Electromagnet: - Large retentivity and low coercivity