Class XII_All India_Physics_Set-2
9.
In the circuit shown in the figure, identify the equivalent gate of the circuit and
make its truth table.
2
Ans.
Y1  A and Y2  B
 Y  Y1  Y2  Y1.Y2  A.B  A.B
The equivalent gate of the given circuit is AND gate.
Truth table:
A B Y1 Y2 Y
10.
Ans.
1
1
0
0
1
1
0
0
1
0
0
1
1
0
0
0
0
1
1
0
A slab of material of dielectric constant K has the same area as that of the plates
of a parallel plate capacitor but has the thickness d/3, where d is the separation
between the plates. Find out the expression for its capacitance when the slab is
inserted between the plates of the capacitor.
2
Initially when there is vacuum between the two plates, the capacitance of the two parallel
A
plate is, C0  0 , where, A is the area of parallel plates.
d
Suppose, that the capacitor is connected to a battery, an electric field E0 is produced.
Now, if we insert the dielectric slab of thickness t 
d
, the electric field reduces to E.
3
Class XII_All India_Physics_Set-2
Now the gap between plates is divided in two parts, for distance t there is electric field E
and for the remaining distance (d‒ t) the electric field is E0.
If V be the potential difference between the plates of the capacitor, then V = Et + E0(d‒t)
Ed
d  Ed

 2d  d
 E0  d   
 E 0    (E  2E 0 )
3
3 3

 3  3
dE
E
d E
or, V  ( 0  2E 0 )  0 (2K + 1) (As, 0  K)
3 K
3K
E

q
d q
Now, E 0  
V
(2K  1)
0 0 A
3K 0 A
V
We know, C 
11.
Ans.
(t =
d
)
2
3K0 A
q

V d(2K  1)
Draw typical output characteristics of an n-p-n transistor in CE configuration.
Show how these characteristics can be used to determine output resistance.
2
Output characteristics is the plot between collector-emitter voltage (VCE) and the
collector current (IC) at different constant values of base current (IB).
Output resistance is defined as the ratio of the change in collector-emitter voltage (ΔVCE)
to the change in collector current (ΔIC) at a constant base current (IB).
Initially with the increase in VCE the collector current increases almost linearly this is
because the junction is not reverse biased. When the supply is more than required to
reverse bias the base-collector junction, IC increases very little with VCE.
The reciprocal of slope of the linear part of the curve gives the value of output resistance
V
i.e. r0  ( CE ) IB
IC
Class XII_All India_Physics_Set-2
12.
Ans.
A capacitor, made of two parallel plates each of plate area A and separation d, is
being charged by an external ac source. Show that the displacement current inside
the capacitor is the same as the current charging the capacitor.
2
Let the alternating emf charging the plates of capacitor be E = Eo sin ωt
Charge on the capacitor, q = EC = CEo sin ωt
Instantaneous Current is I.
dq d
I 
  CEo sin t   CEo cost  Io cost  where Io  CEo 
dt dt
d E
d E
d q 
d  CEo sin t 
Displacement current, ID  o
 o A
 o A 
  o A 

dt
dt
dt  o A 
dt  o A 
d
(CEo sin t )  CEo cos t  I o cos t
dt
Thus, the displacement current inside the capacitor is the same as the current
charging the capacitor.

13.
Ans.
A parallel beam of light of 600 nm falls on a narrow slit and the resulting
diffraction pattern is observed on a screen 1.2 m away. It is observed that the first
minimum is at a distance of 3 mm from the centre of the screen. Calculate the
width of the slit.
nD
a
Where, D = distance of slit from screen, λ = wavelength of the light, a = width of the slit
For first minimum, n=1
The distance of the nth minimum from the center of the screen is, x n 
Thus, 2.5 103 
14.
Ans.
2
1(1)(500 109 )
 a  2 104 m=0.2 mm
a
In the block diagram of a simple modulator for obtaining an AM signal, shown in
the figure, identify the boxes A and B. Write their function.
2
In the block diagram of modulator, A is Square Law Device and B is Band pass filter.
Band pass filter rejects low and high frequencies and allows a band of frequencies to
pass through.
Square law device is a non linear device. It produces a non-linear output of message and
carrier signals. The output from square law device is,
Class XII_All India_Physics_Set-2
y(t)=Bx(t)+Cx2(t)
Where, B and C are constants and x(t)=message signal (Amsinωmt) + carrier signal
(Acsinωct)
15.
Explain the term ‘drift velocity’ of electrons in conductor. Hence obtain the
expression for the current through a conductor in terms of ‘drift velocity’.
2
OR
Describe briefly, with the help of a circuit diagram, how a potentiometer is used
to determine the internal resistance of a cell.
Ans.
‘Drift velocity’ of electrons in a conductor: Metals contain a large number of free
electrons. These electrons are in continuous random motion. Due to the random motion,
the free electrons collide with positive metal ions with high frequency and undergo
change in direction at each collision. So the average velocity for the electrons in a
conductor is zero.
Now, when this conductor is connected to a source of emf, an electric field is established
in the conductor, such that E = V/L
Where V= potential difference across the conductor and L = length of the conductor.
The electric field exerts an electrostatic force ‘-Ee’ on each free electron in the conductor.
The acceleration of each electron is given by

eE

a
m
Where e = electric charge on electron and m = mass of electron
The negative sign indicates that the force and hence the acceleration is in a direction
opposite to the direction of the electric field. Due to this acceleration, the electrons attain
a velocity in addition to thermal velocity in the direction opposite to that of electric field.
The average velocity of all the free electrons in the conductor is called the drift velocity
of free electrons of the conductor.

eE

vd  
 ...........................(1)
m
Thus, the expression for the drift velocity is
V
Electric field, E  
....................(2)
L
where  = relaxation time between two successive collision.
Let n = number density of electrons in the conductor.
No. of free electrons in the conductor = nAL
Total charge on the conductor, q = nALe
L
Time taken by this charge to cover the length L of the conductor, t 
vd
Class XII_All India_Physics_Set-2
Current I 
q nALe

 vd  nAevd
t
L

 eE 
 e  V    ne2 A 
Using eq(1) and (2), we get that I  nAe       nAe   
  
 V
mL   mL 

 m 
OR
Measurement of internal resistance of a cell using potentiometer:
The cell of emf, E (internal resistance r) is connected across a resistance box (R)
through key K2.
When K2 is open, balance length is obtained at length AN1 = l1
E= Φ l1…………………………………………………………………………….. (1)
When K2 is closed:
Let V be the terminal potential difference of cell and the balance is obtained at
AN2 = l2
∴ V = Φ l2 ……………………………………………………………………………(2)
From equations (1) and (2), we get
E l1

...(3)
V l2
E  I (r  R)
V  IR
E rR
 
...(4)
V
R
From (3) and (4), we get

R  r l1

R
l2
E l1

V l2
l

E 
 r  R   1  r  R  1  1
V

 l2 
We know l1, l2 and R, so we can calculate r.
Class XII_All India_Physics_Set-2
16.
Ans.
A convex lens of focal length f1 is kept in contact with a concave lens of focal
length f2. Find the focal length of the combination.
2
For convex lens, focal length, f = f1 and for concave lens, the focal length, f = -f2
The equivalent focal length of a combination of convex lens and concave lens is given as:
1 1
1
 
F f1  f 2
F
f1 f 2
f 2  f1