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Grade XII DELHI SET 3
Physics (Theory)
Time allowed: 3 hours]
[Maximum marks:70
General Instructions:
(i)
All questions are compulsory.
(ii)
There are 30 questions in total. Question Nos. 1 to 8 are very short answer type questions and
carry one mark each.
(iii)
Question Nos. 9 to 18 carry two marks each, question 19 to 27 carry three marks each and
question 28 to 30 carry five marks each.
(iii)
There is no overall choice. However, an internal choice has been provided in one question of
two marks; one question of three marks and all three questions of five marks each. You have to
attempt only one of the choice in such questions.
(iv)
Use of calculators is not permitted.
(v)
You may use the following values of physical constants wherever necessary:
c = 3 × 108 ms–1
h = 6.626 × 10 –34 Js
e = 1.602 × 10–19 C
0 = 4 × 10–7 Tm A–1
1
9
2
2

9

1
0
N
m
C
4


0
Mass of electron me = 9.1 × 10–31 kg
Q28. Draw a simple circuit of a CE transistor amplifier. Explain its working. Show that the voltage
βR
 ac L , where ac is the current gain, RL is the load
V=
gain, AV, of the amplifier is given by A
ri
resistance and ri is the input resistance of the transistor. What is the significance of the negative
sign in the expression for the voltage gain?
OR
(a) Draw the circuit diagram of a full wave rectifier using p-n junction diode.
Explain its working and show the output, input waveforms.
(b) Show the output waveforms (Y) for the following inputs A and B of
(i) OR gate (ii) NAND gate
Grade XII DELHI SET 3
Solution:
Circuit diagram of CE transistor amplifier:
Working:
If a small sinusoidal voltage is applied to the input of a CE configuration, the base current and
collector current will also have sinusoidal variations. Because the collector current drives the
load, a large sinusoidal voltage v0 will be observed at the output.
The expression for voltage gain of the transistor in CE configuration is:
v


0
a
cR
L
A
v 
v
r
2
βac → ac current gain
RL → Load resistance
r = RB + ri
ri → Input resistance
RB → Base resistance
Current gain of the transistor will decrease if the base is made thicker because current
Ic
gain,   I
b
If the base of an n-p-n transistor is made thicker, then more and more electrons will recombine
with the p-type material of the base. This results in a decrease in collector current I c .
Furthermore, I b also increases.
Ic
Hence, ac current gain (   I ) decreases.
b
Finding expression for voltage gain of the amplifier:
Applying Kirchhoff’ law to the output loop,
Grade XII DELHI SET 3
VCC = VCE + ICRC
VBB = VBE + IBRB
vi  O
Then, VBB + vi = VBE + IBRB + IB (RB + ri)
 V 
 ri   B E 
  I B V
CE
 v i   I B  R B  ri 
 rIB
 I C ic

 I B ib
It is current gain denoted by Ai.
Change IC due to change in IB causes a change in VCE and the voltage drop across resistor RC,
because VCC is fixed.
∆VCC = VCE + RCIC = 0
 VCE = – RCIC
Change in VCE is the o/p voltage Vo.
 Vo = VCE = – ac RCIB
Voltage gain of amplifier
V V
Av  o  CE
Vi rIB
 ac 
RC
r
Negative sign represents that the o/p voltage is in opposite phase to i/p voltage.
 ac
OR
Grade XII DELHI SET 3
Full wave rectifier:
When the diode rectifies the whole of the AC wave, it is called full wave rectifier.
The figure shows the arrangement for using diode as full wave rectifier. The alternating input signal is
fed to the primary P1P2 of a transformer. The output signal appears across the load resistance RL.
During the positive half of the input signal, suppose P1 and P2 are negative and positive respectively.
This would mean that S1 and S2 are positive and negative respectively. Therefore, the diode D1 is
forward biased and D2 is reverse biased. The flow of current in the load resistance RL is from A to B.
During the negative half of the input signal, S1 and S2 are negative and positive respectively. Therefore,
the diode D1 is reverse biased and D2 is forward biased. The flow of current in the load resistance RL is
from A to B.
(b) Output waveforms (Y)
Grade XII DELHI SET 3
Q 29. Explain the principle of a device that can build up high voltages of the order of a few million
volts.
Draw a schematic diagram and explain the working of this device.
Is there any restriction on the upper limit of the high voltage set up in this machine?
Explain.
OR
(a) Define electric flux. Write its S.I. units.
(b) Using Gauss’s law, prove that the electric field at a point due to a uniformly charged infinite
plane sheet is independent of the distance from it.
(c) How is the field directed if (i) the sheet is positively charged, (ii) negatively charged?
Solution:
Van de Graff generator is the device used for building up high potential differences of the order
of a few million volts.
Such high potential differences are used to accelerate charged particles such as electrons,
protons, ions, etc.
It is based on the principle that charge given to a hollow conductor is transferred to outer
surface and is distributed uniformly over it.
Construction:
Grade XII DELHI SET 3
It consists of a large spherical conducting shell (S) supported over the insulating pillars. A long
narrow belt of insulating material is wound around two pulleys P1 and P2. B1 and B2 are two
sharply pointed metal combs. B1 is called the spray comb and B2 is called the collecting comb.
Working – The spray comb is given a positive potential by high tension source. The positive
charge gets sprayed on the belt.
As the belt moves and reaches the sphere, a negative charge is induced on the sharp ends of
collecting comb B2 and an equal positive charge is induced on the farther end of B2.
This positive charge shifts immediately to the outer surface of S. Due to discharging action of
sharp points of B2, the positive charge on the belt is neutralized. The uncharged belt returns
down and collects the positive charge from B1, which in turn is collected by B2. This is
repeated. Thus, the positive charge on S goes on accumulating. In this way, voltage differences
of as much as 6 or 8 million volts (with respect to the ground) can be built up.
The main limiting factor on the value of high potential is the radii of sphere.
If the electric field just outside the sphere is sufficient for dielectric breakdown of air, no more
charge can be transferred to it.
For a conducting sphere,
Electric field just outside sphere
Q
E
2
4π
0R
and electric potential
Q
V
4π
0R
Thus, E = VR
Now, for E = 3 × 106 V/m
Radius of should be 1 m.
(dielectric breakdown)
Grade XII DELHI SET 3
Thus, the maximum potential of a sphere of radius 1 m would be 3 × 106V.
OR
(a) Electric Flux:It is the number of electric field lines passing through a surface normally.
 
  E A
Where E = Electric Field
A = Area
S.1. Unit of of flux is Nm2C–1
(b)
Consider a uniformly charged infinite plane sheet of charge density .
We have to find electric field E at point as shown in figure.
Now, we construct a Gaussian surface as shown in figure in the form of cylinder.
Applying Gauss’s Law,
    s
   E  ds 
0
 Es  Es  0 
 2Es 
s
0


2 0
E
s
0
Grade XII DELHI SET 3
It shows that electric field is uniform due to charged infinite plane sheet. Also, we can say that
E is independent from distance from the sheet.

(c) i E2
0
Direction of field will be away from the sheet if sheet is positively charged.

(ii) E  2
0
Direction of field will be towards the sheet if sheet is negatively charged.
Q 30. Define magnifying power of a telescope. Write its expression.
A small telescope has an objective lens of focal length 150 cm and an eye piece of focal length
5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the
final image when it is formed 25 cm away from the eye piece.
OR
How is the working of a telescope different from that of a microscope?
The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm
respectively. Find the position of the object relative to the objective in order to obtain an
angular magnification of 30 in normal adjustment.
Solution:
Magnifying power of a telescope is defined as the ratio of the angle subtended at the eye by the
image formed at the least distance of distinct vision to the angle subtended at the eye by the
object lying at infinity, when seen directly.
The formula for magnifying power is,
f 
f 
Magnifying power, m   0 1  e 
fe  D 
where, f0 = Focal length of the objective = 150 cm
fe = Focal length of the eye-piece = 5cm
D = Least distance of distinct vision = 25 cm
Grade XII DELHI SET 3
M 
150
5
 (1  )  36
5
25


tan
M 
tan
M 
tan =
(As angles  and  are small)
Height of object
H 100
1



Distance of object from objective u 3000 30
tanβ
1
( )
30
36
tan =
30
M 
tan 
Height of image
H'

distance of image formation D
Thus,
36  25
H '
 30 cm
30
Negative sign indicates that we get an inverted image.
OR
A microscope is used to look into smaller details like structure of cells etc. On the other hand, a
telescope is used to see larger objects that are very far away like stars, planets etc.
Telescope mainly focuses on collecting the light into the objective lens, which should thus be
large, while the microscope already has a focus and the rest is blurred around it.
There is a big difference is in their magnification factors.
fo
For telescope the angular magnification is given by M  f
e
Where f0 is the focal length of the objective lens and fe is the focal length of the eyepiece.
For microscope the angular magnification is given by
D
M  1
When image is formed at distance of least distinct vision
f0
D
When image is formed at infinity
f0
Where D is the distance of least distinct vision and f0 is the focal length of objective lens.
Magnifying power of compound microscope
M
Grade XII DELHI SET 3
Given :
f o  1.25 cm
f e  5 cm
M  30 (Magnifying power is negative)
We know,
v
D
M  o (1  )
uo
fe
Where, vo  Distance of image from obective
uo  Distance of object from objective
D=Distance of least distinct vision
v
25
30  o (1  )
uo
5
vo  5uo
Using lens formula,
1
1 1
 
fo
uo vo
1
1
1
 
1.25
uo 5uo
uo  1.5 cm
Thus the distance of object from objective is 1.5 cm.