Grade XII Foreign SET 1
Physics (Theory)
[Time allowed: 3 hours]
[Maximum marks:70]
General Instructions:
(i) All questions are compulsory.
(ii) Marks for each question are indicated against it.
(iii) Question number 1 to 8 are very short-answer questions and carry 1 mark each.
(iv) Question number 9 to 18 are short-answer questions and carry 2 marks each.
(v) Question number 19 to 27 are also short-answer questions and carry 3 marks each.
(vi) Question number 28 to 30 are long-answer questions and carry 5 marks each.
(vii) Use Log Tables, if necessary. Use of calculators is not allowed.
Q9.
A conductor of length ‘l’ is connected to a dc source of potential ‘V’. If the length of the
conductor is tripled by gradually stretching it, keeping ‘V’ constant, how will (i) drift speed of
electrons and (ii) resistance of the conductor be affected? Justify your answer.
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Ans:
V = constant
l’ = 3l
V
n el 
Where n is number of electrons
e is charge on electron
l is the length of the conductor
and ρ is the resistivity of conductor.
So, when length is tripled, drift velocity gets one-third.
(i) Drift speed of electrons 
(ii) Resistance of the conductor is given as
l
R=
A
When length is tripled area of cross-section is reduced to
A
3
l'
3l
l
   9   9R
A
A'
A
3
Thus, new resistance will be 9 times the original.
Hence, R  
Q10. Two students ‘X’ and ‘Y’ perform an experiment on potentiometer separately using the circuit
given below:
Grade XII Foreign SET 1
Ans:
Keeping other parameters unchanged, how will the position of the null point be affected if
(i) ‘X’ increases the value of resistance R in the set-up by keeping the key K1 closed and the
Key K2 opens?
(ii) ‘Y’ decreases the value of resistance S in the set-up, while the key K2 remains open and
they K1 closed?
Justify.
(i) K1  closed, K2  Open
Suppose null point occurs at l,
Apply KVL in smaller loop,
 – l = 0
…..(1)
Where,  = potential drop per unit length in potentiometer wire
 = l
l 


As X increase the value of resistance R, So, current in the circuit (wire) decrease. Hence,  will
be increases. Then l will decrease.
We can say, as X increases the value of R, null point decrease.
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(ii) K2  open, K1  closed.
Then the circuit will be same as shown earlier.
We see that resistance S is not involved in the circuit because K2 is open.
So, from equation (1)
 = l
l 


Here,  does not depends on the value of resistance S.
So,  null point is not affected by decreasing the value of resistance S.

Q11. A particle of charge ‘q’ and mass ‘m’ is moving with velocity V . It is subjected to a uniform

magnetic field B directed perpendicular to its velocity. Show that it describes a circular path.
Write the expression for its radius.
2
Ans.
A charge ‘q’ projected perpendicular to the uniform magnetic field ‘B’ with velocity ‘v’. The
perpendicular force F  qv  B acts like a centripetal force perpendicular to the magnetic field.
Then the path followed by charge is circular as shown in the figure below.
The Lorentz magnetic force acts as centripetal force thus,
mv 2
qvB =
r
mv
r=
qB
Here, r = radius of the circular path followed by charge projected perpendicular to the uniform
magnetic field.
Q12. Calculate the quality factor of a series LCR circuit with L = 2.0 H, C = 2F and R = 10 .
Mention the significance of quality factor in LCR circuit.
Ans.
Given,
Grade XII Foreign SET 1
L = 2.0 H
C = 2µF = 2 × 10–6 F
R = 10 Ω
1 L
Now, Q-factor =
R C
1
2
10 2 106
1
1

 2  100
3
10 10
10
Quality factor is also defined as
Energy Stored
Q = 2 f 
Power loss
So higher the value of Q means the energy loss is at lower rate relative to energy stored i.e. the
oscillations will die slowly and damping would be less.

Q13. Explain briefly how electromagnetic waves are produced by an oscillating charge. How is the
frequency of the em waves produced related to that of the oscillating charge?
2
Ans.
An oscillating charge is considered as the accelerating charge. This produces on oscillating
electric field in space, which produces an oscillating magnetic field, which again produces
oscillating electric field and so on. These oscillating electric and magnetic field thus keep on
regenerating each other as the wave propagates through the space.
Thus, the frequency of the electromagnetic waves naturally equals the frequency of oscillation
of the charge
Q14. In a given sample, two radioisotopes, A and B, are initially present in the ration of 1 : 4. The
half lives of A and B are respectively 100 years and 50 years. Find the time after which the
amounts of A and B become equal.
2
Ans.
Let NA be the concentration of A after tA time and NB be the concentration of B after tB time.
So, NA = N0e–AtA
NB = 4N0e–BtB (as N0B = 4N0A)
Now half-life of A is 100 years and B is 50 years.
l n2
l n2
So A 
and B 
100
50
Dividing we get
A 1

or B  2A
B 2
Now let after t years NA = NB
Grade XII Foreign SET 1
So
N A e  At

N B 4e  Bt
NA  NB
4e  Bt  e  At
4e
 A  B t
l n 4     A  2A  t
 B  2A 
l n 4  A t
t
l n4
100
l n2
l n2 

 A 

100 

 200 years
Q15. Figure shows a block diagram of a transmitter identify the boxes ‘X’ and ‘Y’ and write their
functions.
2
Ans.
Modulator: Since the frequency range of signal is quiet low and it is associated with very small
amount of energy, it dies out very soon if transmitted as such. So, it is modulated by mixing
with very high frequency waves called carrier waves. This is done by modulator power.
Amplifier: Since the signal gets weaken after travelling through long distances it cannot be
transmitted as such. Thus, we use a power amplifier to provide it necessary power before
feeding the signal to the transmitting antenna.
Grade XII Foreign SET 1
Q16. Trace the path of a ray of light passing through a glass prism (ABC) as shown in the figure. If
the refractive index of glass is 3 , find out the value of the angle of emergence from the prism.
Ans.
Refractive index of glass ng  3
Since i = 0
At the interface AC, we have according to shell’s law
sin i ng

sin r na
But sin i = sin 0 = 0
Grade XII Foreign SET 1
Thus sin r 
na sin i
0
ng
Hence r = 0
This ray pass unrefracted at AC interface and reaches AB interface. Here we can see angle of
incidence becomes 30°.
Thus, applying Snell’s law
sin 30 na 1
 
sin e ng
3
sin e  3  sin 30 
3
2
Thus e = 60°
Hence, angle of emergence is 60°.
Q17. Write two characteristic features to distinguish between n-type and p-type semiconductors.
2
OR
How does a light emitting diode (LED) work? Give two advantages of LED’s over the
conventional incandescent lamps.
Ans.
In n-type semiconductor, the semiconductor is doped with pentavalent impurity. In it the
electrons are majority carriers and holes are minority carrier or ne >> nh [ne – number density of
electrons, nh – number density of holes]
In energy band diagram of n-type semiconductor the donor energy level ED is slightly below
the bottom of EC conduction band and thus the electron can move to conduction band thus the
electron can move to conduction band with even small supply of energy,
In p-type semiconductor, the semiconductor is doped with trivalent impurity. In this the holes
are the majority carries and electrons are the minority carriers i.e nh >> ne
In energy-band diagram of p-type, the accepter energy level in slightly above the top of valence
band EV.
Thus even with small supply of energy electron from valence band can jump to level EA and
ionize the acceptor negatively.
OR
When we apply sufficient voltage to LED, electron move across the junction into the p-region
and get attracted to the holes there. Thus, electrons and holes recombine. During each
Grade XII Foreign SET 1
recombination, electric potential energy is converted into electromagnetic energy and a photon
of light with a characteristic frequency is emitted, this is how LED works.
Advantage of LEDs over incandescent lamps
(1) Since LEDs do not have a filament that can burn out, they last longer.
(2) They do not get hot during use.
Q18. A short bar magnet of magnetic moment 0.9 J/T is placed with its axis at 30° to a uniform
magnetic field. It experiences a torque of 0.063 J.
(i) Calculate the magnitude of the magnetic field.
(ii) In which orientation will the bar magnet be in stable equilibrium in the magnetic field?
Ans.
(i) Magnetic moment M = 0.9 J/T
 = 0.063 J,  = 30°
We know  = M × B
= MB sin 
0.063 = 0.9 × B × sin 30°
2×0.063
B=
= 0.14T
0.9
 
(ii) Stable equilibrium is position of minimum energy. Since U =  M.B
U = – M B cos 
Where, U is the energy stored or P.E. of the magnet inside magnetic field B.
So, when  =0, U = – MB is the minimum energy.
Thus, when M and B are parallel to each other bar magnet is in stable equilibrium.
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