SCHEME OF EVALUATION FOR PHYSICS I PREPARATORY QUESTION PAPER
Q. No
1
2
3
ANSWERS
Magnitude of the magnetic field 𝒅𝒃 =
4
5
6
7
8
𝜇𝑜 𝐼(𝒅𝒍×𝒓)
4𝜇
𝜇𝑜 is permeability of free space
I is current in the current element
dl is length vector
r is the displacement vector
Which material is used in the core of the transformer?
Soft iron
State Lenz's law.
The polarity of induced emf is such that it tends to produce a current which opposes the
change in magnetic flux that produces it.
If the object is placed within the focus of concave mirror what is the nature of the
image?
Virtual, erect, enlarged
Define one electron volt.
One electron volt is the energy gained by an electron when it has been accelerated by a
potential difference of 1 volt.
The ground state energy of hydrogen atom is -13.6 eV. What is its kinetic energy in
the ground state?
We know that kinetic energy K is
10
11
12
13
𝑒2
8𝜋𝜀𝑜 𝑟
1
1
1
𝑟3
This means total energy of E of electron is -13.6 eV. i.e. −
9
Marks
If two electrons are removed from a body, what is the charge on the body?
q =ne=2 × 1.6 × 10−19 𝐶
The colour coding on the resistor is black brown red and silver. What is the
resistance of the carbon resistor?
Black =0, brown=1, red=102, silver=10%
So the resistance is 100𝛺 ±10% tolerance
Write the expression for Biot-savart's law in vector form?
𝑒2
1
1
1
1
8𝜋𝜀𝑜 𝑟
which is actually − (−
1
𝑒2
8𝜋𝜀𝑜 𝑟
)
=› K= -( -13.6 eV)= 13.6 eV
Which property of the nuclear force makes the nuclear force between the two
nucleons to vanish beyond few femtometers?
Short range nature of nuclear force
What is an electrical transducer?
It is a device that converts some physical variables into corresponding variations in
electrical signal at its output.
Mention any two properties electric field lines?
1) Field lines start from positive charges and end at negative charges. If there is a single
charge the may start or end at infinity
2) In a charge free region electric field lines can be taken to be continuous curves without
any breaks
3) Two field lines can never cross each other. If they did the field at the point of
intersection will not have a unique direction which is absurd
4) Electrostatic field lines do not form any closed loops
Define drift velocity and mobility?
Drift velocity: It is an average velocity with which free electrons drift in metallic wire
when potential difference is applied across it ends.
Mobility: It is the magnitude of drift velocity per unit electric field
The magnetic flux threading a coil changes from 12 × 10−3 𝑊𝑏 to 6 × 10−3 𝑊𝑏 in
0.001s. Calculate the induced emf?
Here ∅1 = 12 × 10−3 𝑊𝑏, ∅2 = 6 × 10−3 𝑊𝑏, 𝑡 = 0.001𝑠 = 10−3 𝑠, 𝑒 =?
1
1
Among
the four
any
two
carry 2
marks
1
1
1
𝑒=
−𝑑∅
𝑑𝑡
−(∅2 −∅1 )
14
15
16
17
18
−(6×10−3 −12×10−3 )
=
= 6V
𝑑𝑡
10−3
Define declination and dip?
Declination: angle made by the magnetic meridian at a point with geographic meridian
OR angle between the true geographic north and the north shown by a compass needle.
Dip: it is the angle that the total magnetic field of earth makes with the surface of earth.
Write the expression for Ampere-Maxwell law? A metal target bombarded with high
energy electron produce which electromagnetic radiation?
𝑑∅𝑒
∮ 𝑩. 𝑑𝒍 = 𝜇𝑜 𝑖𝑐 + 𝜇𝑜 𝜀𝑜
𝑑𝑡
X-rays
Write the Cartesian sign conventions in Ray-optics.
1) All the distances are measured from the pole of the mirror or the optical centre of lens.
2) The distances measured in the same direction as incident light are taken as positive and
those measured in the direction opposite to the direction of incident light are taken as
negative.
3) The heights measured upwards with respect to x-axis and normal to the principal axis of
mirror or lens are taken as positive and the heights measured downward are taken as
negative.
Which diode is used for regulating the supply voltage? Why?
Zener
Because zener voltage remains constant, even though the current through the zener diode
varies over a wide range.
Write the block diagram for receiver.
1
1
1
1
1
Any
two
1
1
1
1
1
2
Receiving antenna
Output
Amplifier
19
IF stage
detector
amplifier
Arrive at the relation between field and potential using two closely spaced
equipotential surfaces.
Figure and explanation :
Consider two closely spaced equipotential surfaces A and B with potential values V and V
+ϨV, where ϨV is the change in V in the direction of the electric field E. let P be a point
on the surface B. Ϩl is the perpendicular distance of the surface a form P. imagine that a
unit positive charge is moved along this perpendicular from the surface B to surface A
against the E field. The work done in this process is |E| Ϩl
This work equals the pd VA-VB .
Thus ,
|E| Ϩl=V-(V +ϨV)= -ϨV
1
1
i.e.
|E|=-ϨV/ Ϩl
Since ϨV is negative ϨV=-| ϨV|. We can rewrite as
|E|=-ϨV/ Ϩl= |ϨV|/Ϩl
This is relation between field and potential
20
Obtain the condition for selection of charged particle of specific velocity out of the
beam using crossed electric and magnetic fields.
1
A charged particle moving with velocity v in the presence of both electric and magnetic
field experiences a force given by
𝑭 = 𝑞 (𝐸 + 𝒗 × 𝑩) = 𝑭𝑬 + 𝑭𝑩
Consider the simple case in which electric and magnetic fields are perpendicular to each
other and also perpendicular to the velocity of the particle
1
1
E
FE
B
V
FB
𝒗 = 𝑣𝒊, 𝑬 = 𝐸𝒋, 𝑩 = 𝐵𝒌
=›𝑭𝐸 = 𝑞𝑬 = 𝑞𝐸𝒋,
𝑭𝑩 = 𝑞(𝒗 × 𝑩) = 𝑞(𝑣𝒊 × 𝐵𝒌) = −𝑞𝐵𝒋
There electric and magnetic forces are in opposite directions as shown in the figure
Suppose we adjust value of e and b such that magnitude of two forces are equal then total
force on the charge is zero and it will move in the field undeflected .
this happens when
|𝐹𝐸 | = |𝐹𝐵 |
𝑞𝐸 = 𝑞𝑣𝐵
𝐸
𝑣=
𝐵
this condition can be used to select charged particle of a particular velocity out of a beam
containing charges moving with the different velocities
21
Write any three differences between the diamagnetic and paramagnetic materials.
Diamagnetic
paramagnetic
Any
other
valid diff
could
award
mark
1) diamagnetic substances are those which
have tendency to move from stronger to the
weaker part of the external magnetic field
1) ) paramagnetic substances are those
which have tendency to move from weaker
to the stronger part of the external magnetic
field
2)paramagnetic substances posses a
permanent magnetic moment of their own
1
2)diamagnetic substances are the one in
which resultant magnetic moment in an
atom is zero
3)they do not obey the curie law
22
1
1
3) they obey the curie law
Derive expression for self inductance of a long solenoid.
Let l be the length, A be the area, n be the number of turn per unit length of the solenoid
and I be the current in the solenoid. Total no. of turns in the solenoid is N which is nl
The magnetic field due to current I in the solenoid is
𝐵 = 𝜇𝑜 𝑛𝐼
Total magnetic flux linked with the solenoid is
∅ = 𝑁𝐵𝐴
∅ = 𝑁𝜇𝑜 𝑛𝐼𝐴
Or
1
Also we know that
1
∅ = 𝐿𝐼
𝑁𝜇𝑜 𝑛𝐼𝐴 = 𝐿𝐼
=›
𝐿 = 𝑁𝜇𝑜 𝑛𝐴
1
Or
𝐿 = 𝜇𝑜 𝑛2 𝐴𝑙
Since N =nl
When the solenoid is filled with a material of relative permeability 𝜇𝑟 then,
𝐿 = 𝜇𝑜 𝜇𝑟 𝑛2 𝐴𝑙
23
What are the sources of power losses in practical transformers?
24
1)hysteresis loss
2)loss due to flux leakage
3)loss due to resistance of the winding
4)loss due to eddy currents
Using Huygens principle derive Snell's law for refraction of plane wavefront moving
from rarer medium to denser medium?
Relevant figure with explanation:
𝐵𝐶 𝑣1 𝜏
sin 𝑖 =
=
𝐴𝐶 𝐴𝐶
3
2
𝐴𝐸 𝑣2 𝜏
=
𝐴𝐶 𝐴𝐶
sin 𝑖 𝑣1
=
𝑠𝑖𝑛𝑟 𝑣2
sin 𝑟 =
We know that
𝑐
𝑣1
𝑐
𝑛2 =
𝑣2
𝑛1 =
𝑛1 sin 𝑖 = 𝑛2 sin 𝑟
Hence snell’s law
25
Define ionisation energy. How much amount of energy is required to free an electron
from the first excited state of hydrogen atom?
The minimum energy required to free an electron from the ground state of hydrogen atom
is called the ionisation energy
The total energy of electron in first excited state is
i.e.
−13.6 𝑒𝑉
22
=
−13.6 𝑒𝑉
4
−13.6 𝑒𝑉
𝑛2
where n =2
1
= −3.4 𝑒𝑉
So the energy required to free the electron is 3.4 𝑒𝑉
26
1
1
How to get NOR gate? Write the circuit symbol and truth table for NOR gate?
A NOT operation applied after OR gate gives a NOT-OR(NOR) gate
1
Logic symbol
1
A
Y
B
Truth table
1
Input
27
Output
A
B
Y
0
0
1
0
1
0
1
0
0
1
1
0
State gauss law in electro-statics and hence arrive at the expression for field due to an
infinitely long straight uniformly charged wire.
Electric flux through a closed surface =q/𝜀𝑜
1
Figure and explanation
1
Flux through the Gaussian surface = flux through the curved cylindrical part of the
surface= 𝐸 × 2𝜋𝑟𝑙
1
Since surface includes a charge 𝑞 = 𝜆𝑙 according gauss’s law flux =
𝜆𝑙
𝜀𝑜
𝜆
2𝜋𝜀𝑜 𝑟
𝒏 in vector form
𝐸=
𝑬=
28
𝜆
2𝜋𝜀𝑜 𝑟
1
State Kirchhoff's rules. Get the balance condition using null deflection of
galavanometer in Wheatstone bridge.
Kirchhoff’s junction rule: at any junction, the sum of the currents entering the junction ids
equal to the sum of currents leaving the junction
1
Kirchhoff’s loop rule: the algebraic sum of changes in potential around any closed loop
involving resistors and cells in the loop is zero
1
Figure with explanation
Applying junction rule and getting I1=I3 and I2=I4
1
1
Applying loop rule
−𝐼1 𝑅1 + 0 + 𝐼2 𝑅2 = 0
1
𝐼4 𝑅4 + 0 − 𝐼3 𝑅3 = 0
Solving
𝑅2
𝑅
= 4 the balance condition
𝑅1
29
1
1
𝑅3
Show that parallel currents attract each other. Using the expression for force
between parallel currents define ampere.
Figure with explanation :
1
Magnitude of the magnetic field due to current in long conductor a is
𝜇𝑂 𝐼𝑎
𝐵𝑎 =
2𝜋𝑑
Force on long conductor b due to magnetic field of a is
𝜇𝑂 𝐼𝑎 𝐼𝑏 𝐿
𝑭𝑏𝑎 = 𝐼𝑏 𝐿𝐵𝑎 =
2𝜋𝑑
1
It can be shown that
1
𝑭𝑏𝑎 = −𝑭𝑎𝑏
Hence parallel currents attract each other
𝜇𝑂 𝐼𝑎 𝐼𝑏 𝐿
𝐹𝑏𝑎 =
2𝜋𝑑
𝜇 𝐼 𝐼
𝑓𝑏𝑎 = 𝑂 𝑎 𝑏 force per unit length
1
2𝜋𝑑
1
Using this expression we can define ampere
30
The ampere is the value of that steady current which, when maintained in each of the two
very long, straight, parallel conductors of negligible cross section, and placed one meter
apart in vacuum, would produce on each of these conductors a force equal to 2 × 10−7 N
per meter of length.
Using Young’s double slit experiment obtain the expression for distance between two
consecutive bright or dark fringes.
Figure and explanation:
We know that for an arbitrary point on the screen to corresponds to maximum we must
have
Equation S2P-S1P=n 𝜆 n=0,1,2,3….
𝑑 2
𝑑 2
(s2 p)2 − (𝑠1 𝑝)2 = [𝐷2 + (𝑥 + ) ] − [𝐷 2 + (𝑥 − ) ] = 2𝑥𝑑
2
2
Now S2P+S1P=2D
1
1
1
1
s2 p − 𝑠1 𝑝 ≈
𝑛𝜆 ≈
𝑥𝑛 𝑑
𝐷
𝑥𝑛 ≈
𝑛𝜆𝐷
𝑑
𝑥𝑑
𝐷
𝛽 = 𝑥𝑛+1 − 𝑥𝑛 =
𝜆𝐷
𝑑
1
Is the distance between two consecutive bright fringes.
31
Write Einstein's photo electric equation and using the equation account for the
observations of photo electric effect.
𝐾𝑚𝑎𝑥 = ℎ𝜗 − ∅𝑜
32
1
1) According to Einstein's photo electric equation, 𝐾𝑚𝑎𝑥 depends linearly on 𝜗and it is
independent of intensity of radiation in agreement with observation.
1
2) since 𝐾𝑚𝑎𝑥 must be non negative equation implies that photoelectric emission is possible
∅
only if ℎ𝜗 > ∅𝑜 or 𝜗 > 𝜗𝑜 where 𝜗𝑜 = 𝑜
ℎ
The above equation shows that the greater the work function ∅𝑜 , the higher the minimum
or threshold frequency 𝜗𝑜 needed to emit photo electron.
Thus there exists a threshold frequency 𝜗𝑜 for the metal surface, below which no
photoelectric emission is possible, no matter how intense the incident radiation may be or
how long it falls on the surface.
3) In this picture, intensity of radiation as noted above is proportional to the number of
energy quanta per unit area per unit time. The greater the no. of energy quanta available,
the greater is the no. of electron absorbing the energy quanta and greater, therefore is the
no. of electron coming out of the metal.
4) In Einstein’s picture, the basic elementary process involved in photoelectric effect is the
absorption of a light quantum by an electron. This process is instantaneous. Thus whatever
may be the intensity i.e., the number of quantum of radiation per unit area per unit time,
photoelectric emission is instantaneous. A low intensity does not mean delay in emission
since the basic elementary process is the same. Intensity only determines how many
electrons are able to participate in the elementary process.
What is rectifier? Discuss the working of half-wave rectifier?
1
1
1
33
An alternating voltage is applied across the diode the current flows only in that part of the
cycle when the diode is forward biased. this property is used to rectify alternating voltages
and the circuit used for this purpose is called rectifier
Circuit diagram
1
Working
3
A 900pF capacitor is charged by 100V battery. How much electrostatic energy is
stored by the capacitor? The capacitor is disconnected from the battery and
connected to another 900pF capacitor. How much is the electrostatic energy stored in
the system?
Here 𝐶 = 900𝑝𝐹 = 900 × 10−12 𝐹 = 9 × 10−10 𝐹, 𝑉 = 100𝑉
1
34
Energy stored 𝑈1 = 𝐶𝑉 2 = 4.5 × 10−6 𝐽
2
On connecting to another 900 pF capacitor, charge is shared equally.
1
1
𝑄 ′ = 𝑄 𝑎𝑛𝑑 𝑉 ′ = 𝑉
2
2
1 ′ ′
1
1
Total energy of the system 𝑈2 = 2 × 2 𝑄 𝑉 = 2 𝑄 × 2 𝑉 = 2.25 × 10−6 𝐽
2
Let R be the external resistance in series with the cell of emf ɛ and internal resistance r.
ɛ
The current in the circuit is 𝐼 =
𝑅+𝑟
First case: I=0.5A, R=12 Ω
ɛ
ɛ
𝐼=
= 0.5 =
= 6.0 + 0.5𝑟.........(1)
𝑅+𝑟
12+𝑟
Second case: I=0.25A, R=25 Ω
ɛ
ɛ
𝐼=
= 0.25 =
= 6.25 + 0.25𝑟……..(2)
2
From 1 and 2
r=1 Ω
ɛ=6.5V
2
36
1
2
A cell of emf “ɛ” and internal resistance “r” gives a current of 0.5A with an external
resistance of 12Ω and a current of 0.25A with an external resistance of 25 Ω.
Calculate: a) Internal resistance of the cell and b) emf of the cell.
𝑅+𝑟
35
1
1
25+𝑟
A resistor of 200 Ω and a capacitor of 15.0μF are connected in series to a 220V, 50Hz
a.c source Calculate: a) The current in the circuit b) The voltage (rms) across the
resistor and the capacitor.
Here 𝑅 = 200Ω, C = 15μF = 15.0 × 10−6 F, V = 220V, ϑ = 50Hz.
In order to calculate current we need the impedance
𝑍 = √𝑅2 + 𝑋𝑐2
𝑍 = √𝑅2 + (2𝜋𝜗𝐶)2
𝑍 = 291.5Ω
2
The current in the circuit is
I=V/Z=0.755A
Since current is same throughout the circuit
VR=IR=151V
VC=IXC=160.3V
1
2
At what distance form a convex mirror of focal length 2.5m should a boy stand so
that his image has a height equal to half his original height? The principal axis is
perpendicular to the height.
−𝑣
1
Here 𝑚 = = (the magnification is positive as the image is virtual)
𝑢
2
−𝑢
𝑣=
2
1
1
1
And + =
𝑢
𝑣
𝑓
1
1
1
+
=
𝑢 −𝑢/2
2.5
𝑢 = −2.5 𝑚
2
2
37
He must stand at a distance of 2.5m in front of mirror
Determine the amount of 210
84𝑃𝑜 necessary to provide a source of alpha particles of
10mCi strength. Half life of 210
84𝑃𝑜 is 138days.
1
Activity 𝐴 = 𝜆𝑁
Here 𝐴 = 10𝑚𝐶𝑖 = 10 × 10−3 𝐶𝑖 = 3.7 × 1010 × 10 × 10−3 = 37 × 107 𝑑𝑖𝑠/𝑠
1
𝑇1/2 = 138 × 24 × 60 × 60 = 11923200𝑠
0.693
𝜆=
= 5.812 × 10−8 /𝑠
𝑇1
1
2
𝐴
𝑁 = = 6.37 × 1015 𝑎𝑡𝑜𝑚𝑠
𝜆
1
1 mole of polonium contains 6.023 × 1023 𝑎𝑡𝑜𝑚𝑠 and 1mole of polonium =210g
i.e. 210g of Po-----------6.023 × 1023 𝑎𝑡𝑜𝑚𝑠
? g of Po---------6.37 × 1015 𝑎𝑡𝑜𝑚𝑠
1
So amount of polonium in order to provide a source of strength 10mCi is
1
210×6.37×1015
6.023×1023
= 2.22μg