1. Studymate Solutions to CBSE Board Examination 2011-2012
Series : SMA/1 Code No. 55/1/1
Candidates must write the Code on
Roll No. the title page of the answer-book.
 Please check that this question paper contains 11 printed pages.
 Code number given on the right hand side of the question paper should be written on the title page of
the answer-book by the candidate.
 Please check that this question paper contains 30 questions.
 Please write down the Serial Number of the questions before attempting it.
 15 minutes time has been allotted to read this question paper. The question paper will be distributed at
10.15 a.m. From 10.15 a.m. to 10.30 a.m., the student will read the question paper only and will not
write any answer on the answer script during this period.
PHYSICS (Theory)
[Time allowed : 3 hours] [Maximum marks : 70]
General Instructuions:
(i) All questions are compulsory.
(ii) There are 30 questions in total. Questions numbered 1 to 8 are very short-answer questions and
carry 1 mark each.
(iii) Questions numbered 9 to 18 are short-answer questions and carry 2 marks each, Questions
numbered 19 to 27 carry 3 marks each and Questions numbered 28 to 30 are long-answer questions
and carry 5 marks each.
(iv) 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 choices in such questions.
(v) Use of calculators is not permitted. However, you may use log tables if necessary.
(vi) You may use the following values of physical constants wherever necessary:
c = 3 × 108 m/s h = 6.63 × 10–34 Js
e = 1.6 × 10–19 C 0 = 4 × 10–7 T mA–1
1
= 9 × 109 Nm2C–2 me = 9.1 × 10–31 kg
40
55/1/1 1 P.T.O.

2. 1. When electrons drift in a metal from lower to higher potential, does it mean that
all the free electrons of the metal are moving in the same direction?
Ans. No. Eventhough, overall motion is from lower to higher potential, individually in
their motion randomness is present.
2. The horizontal component of the earth’s magnetic field at a place is B and angle
of dip is 60°. What is the value of vertical component of earth’s magnetic field at
equator?
Ans. At magnetic equator the angle of dip is zero. So, the vertical component of earth’s
magnetic field is zero (It has been assumed that equator mentioned in question
paper refers to magnetic equator).
3. Show on a graph, the variation of resistivity with temperature for a typical
semiconductor.

Ans.
0 T
4. Why should electrostatic field be zero inside a conductor?
Ans. A conductor has free electrons. As long as electric field is not zero, the free
charge carriers would experience force and drift. In the static situation, the free
charges have so distributed themselves that the electric field is zero everywhere
inside the conductor.
5. Name the physical quantity which remains same for microwaves of wavelength
1 mm and UV radiations of 1600 Å in vacuum.
Ans. Velocity of wave in vacuum remains same for all kinds of E.M. Waves.
6. Under what condition does a biconvex lens of glass having a certain refractive
index act as a plane glass sheet when immersed in a liquid?
Ans. If refractive index of glass equals the refractive index of liquid in which lens is
immersed, the lens behaves as plane glass sheet.
7. Predict a directions of induced currents in metal rings 1 and 2 lying in the same
plane where current I in the wire is increasing steadily.
1
I
2
Ans. In ring 1, the direction of induced current is clockwise.
In ring 2, the direction of induced current is anti-clockwise.
55/1/1 2 P.T.O.

3. 8. State de-Broglie hypothesis.
Ans. De-Broglie hypothesis says that if radiation has a wave-particle nature then
moving particles of matter should also display wave-like properties under suitable
conditions. He reasoned that nature was symmetrical and that the two basic
physical entities – matter and energy, must have symmetrical character.
De-Broglie proposed that the wave length  associated with a particle of momentum
p is given as
h h
 
p mv
where m is the mass of the particle and v its speed.
9. A ray of light, incident on an equilateral glass prism   3  moves parallel to
the base line of the prism inside it. Find the angle of incidence for this ray.
Ans. A = 60°. Since the light moves parallel to the base of prism, prism is under
minimum angle of deviation position.
A
2r = A  r = = 30°
2
sin i
Using  = , we get
sin r
sin i
3
sin30
1
 sin i = 3 sin30  3 
2
3
 sin i =  i = 60°
2
10. Distinguish between ‘Analog and Digital signals’.
OR
Mention the function of any two of the following used in communication system:
(a) Transducer
(b) Repeater
(c) Transmitter
(d) Bandpass Filter
Ans. (a) Analog - continuously varying signal represented generally by a sinusoidal
function. Analog signals are susceptible to noise.
(b) Digital - Are discrete (non-continuous) in nature. Digital signals are relatively
immune to noise.
OR
(a) Transducer - converts one form of energy to another form.
(b) Repeater - picks up the signal from transmitter, amplifies it and re-transmits
it.
(c) Transmitter - sender of information in a communication system.
(d) Bandpass Filter - A band pass filter rejects low and high frequencies and
allows a band of frequencies to pass through as per the requirement.
55/1/1 3 P.T.O.

4. 11. A cell of emf E and internal resistance r is connected to two external resistances
R1 and R2 and a perfect ammeter. The current in the circuit is measured in four
different situations:
(i) without any external resistance in the circuit
(ii) with resistance R1 only
(iii) with R1 and R2 in series combination
(iv) with R1 and R2 in parallel combination
The currents measured in the four cases are 0.42 A, 0.15 A, 1.4 A and 4.2 A, but
not necessarily in that order. Identify the currents corresponding to the four
cases mentioned above.

Ans. (i) i= = 4.2 A
r
Because out of four situations resistance of circuit is minimum in this case.
Hence, current is maximum.

(ii) i= = 1.05 A
r  R1
Effective resistance of circuit is more than situation (i) and (iv) and less
than (iii). So, current is 1.05 A.

(iii) i= = 0.42 A
r  R1  R 2
Effective resistance of circuit is maximum in this case. Hence, current is
0.42 A.
i
(iv) i= = 1.4 A
R1R2
r
R1  R2
In parallel combination of R1 and R2, the effective resistance is less than
either of resistors. So, effective resistance of circuit is more than situation
(i) and less than (ii) and (iii). So, current is 1.4 A.
–5
12. The susceptibility of a magnetic material is –2.6 × 10 . Identify the type of magnetic
material and state its two properties.
Ans. Diamagnetic materials have negative susceptibility. Hence, material is
diamagnetic.
Properties of Diamagnetic materials
(a) Shifts towards a weak field in a non-uniform field.
(b) When placed in a magnetic field, it is weakly magnetised in a direction
opposite to that of the applied field.
13. Two identical circular wires P and Q each of radius R and carrying current ‘I’ are
kept in perpendicular planes such that they have a common centre as shown in
the figure. Find the magnitude and direction of the net magnetic field at the
common centre of the two coils.
55/1/1 4 P.T.O.

5. Q
I
P
I
Q
Ans. Magnetic field due to horizontal loop P at its centre is
I
0i
B1 = in vertically upward direction.
2r
Magnetic field due to vertical loop Q at centre. P
0i I
B2 = in horizontal direction.
2r B1
So, net magnetic field at centre
2 2 B
 i   i 
B= 2
B1    0   0 
2
B2
 2r   2r 
0i  0i
B= 2 
2r 2r
 = 45°
The direction of magnetic field makes an angle 45°
with horizontal. B2
14. When an ideal capacitor is charged by a dc battery, no current flows. However,
when an ac source is used, the current flows continuously. How does one explain
this, based on the concept of displacement current?
Ans. When an ideal capacitor is charged by dc battery, charge flows till the capacitor
gets fully charged. C
V dq
When an ac source is connected then conduction current ic  flows in the
dt
connecting wire. Due to charging current, charge deposited on the plates of the
capacitor changes with time. Changing charge causes electric field between the
d e
plates of capacitor to be varying, giving rise to displacement current id 0 .
dt
[As displacement current is proportional to the rate of flux variation].
C
Between the plates, electric field
 q
E 
0 A 0
q
Electric flux, c  E A  A  A
0
0 d e d  qA  dq
So, id  0    ic
dt dt  A 0  dt
Displacement current brings continuity in the flow of current between the plates
of the capacitor.
55/1/1 5 P.T.O.

6. 15. Draw a plot showing the variation of (i) electric field (E) and (ii) electric potential
(V) with distance r due to a point charge Q.
Ans. Due to point charge,
1 q
electric field E
4 0 r 2
1 q
and electric potential V 
4 0 r
The field and potential can be plotted as shown below:
E V
0 r 0 r
16. Define self-inductance of a coil. Show that magnetic energy required to build up
2
the current I in a coil of self inductance L is given by ½LI .
Ans. Self induction is a property of a coil due to which the coil opposes any change in
the strength of current flowing through it by inducing an e.m.f. in itself.
di
As current i grows in an inductor a back emf of e  L is developed.
dt
In over coming the back emf, work is done.
The work done = dW = edq for a charge dq
di
 W   e dq   L idt
dt
I
  Li di
0
1
 LI 2
2
This is stored as energy in the magnetic field of the inductor.
17. The current in the forward bias is known to be more (~mA) than the current in
the reverse bias (~µA). What is the reason, then, to operate the photodiode in
reverse bias?
Ans. The current in the forward bias is due to majority carriers where as current in
the reverse bias is due to minority carriers. So current in forward bias is more
(~mA) than current in reverse bias (~µA).
On illumination of photodiodes with light, the fractional change in the majority
carriers would be much less than that in minority carriers. It implies fractional
change due to light on minority carrier dominated reverse bias current is more
easily measurable than fractional change in forward bias currnet. So photo-diodes
are operated in reverse bias condition.
18. The metallic rod of ‘L’ length is rotated with angular frequency of ‘’ with one end
hinged at the centre and the other end at the circumference of a circular metallic
ring of radius L, about an axis passing through the centre and perpendicular to
55/1/1 6 P.T.O.

7. the plane of the ring. A constant and uniform magnetic field B parallel to the axis
is present everywhere. Deduce the expression for the emf between the centre
and the metallic ring.
L
Ans.
As the rod is rotated, free electrons in the rod move towards the outer end due to
Lorentz force and get distributed over the ring. Thus, the resulting separation of
charges produces an emf across the ends of the rod. At a certain value of emf,
there is no more flow of electrons and a steady state is reached. The magnitude
of th e em f gen erated across a l gth d r of the rod as it moves at right angles to
en
the magnetic field is given by
d = Bvdr . Hence,
L L B L2
   d    Bv dr   B r dr 
0 0 2
19. The figure shows a series LCR circuit with L = 5.0 H 2, C = 80 µF, R = 40 
connected to a variable frequency 240 V source. Calculate.
R
C
L
(i) The angular frequency of the source which drives the circuit at resonance.
(ii) The current at the resonating frequency.
(iii) The rms potential drop across the capacitor at resonance.
Ans. Given: L = 5.0 H; C = 80 F; R = 40 ; Vrms = 240 V
(i) At resonance, angular frequency
1
w = 2f = 2 
2 LC
1 1
 w= 
LC 5  80  106
55/1/1 7 P.T.O.

8. 1 1
 w=  rad / s
4  104 2  102
 w = 50 rad/s.
(ii) At resonating frequency,
Impedance, Z = R ( XL = XC)
Z = 40 
Vrms 240
 Current, I = 
Z 40
 I=6A
1
(iii) Reactance of capacitor, XC =
wC
1 1
 XC = 6
 
50  80  10 4  103
 Potential drop across capacitor,
VC = I.XC
1
VC = 6 
4  103
VC = 1500 V
20. A rectangular loop of wire of size 4 cm × 10 cm carries a steady current of 2 A. A
straight long wire carrying 5 A current is kept near the loop as shown. If the loop
and the wire are coplanar, find
B 4 cm C
I = 5A 10 cm 2A
1 cm
A D
(i) the torque acting on the loop and
(ii) the magnitude and direction of the force on the loop due to the current
carrying wire.
Ans. (i) Direction of magnetic dipole moment of loop due to current in loop is
perpendicular inside the plane of paper.
Direction of magnetic field at loop due to current flow in straight wire is
also perpendicular inside the plane of paper.
 

 Angle between M and B = 0°
  
 
Torque on loop   M  B
  = MB sin 
  = MB sin 0°
=0
55/1/1 8 P.T.O.

9. 0 I1I 2
(ii) Force between two current carrying wires, F = . l
2 r
B 4 cm C
FAB FCD
I = 5A 10 cm 2A
1 cm
A D
0 5  2
 Force on arm AB, FAB = .  (0.10)
2 1  102
0
FAB =  100N [Attraction – towards the wire]
2
0 52
Force on arm CD, FCD =   (0.10)
2 5  102
0
FCD =  20N [Repulsion – away from the wire]
2
Force on arm BC and AD, FBC = FAD = 0
 Resultant force on loop,
0
F = FAB – FCD = (100  20)
2
–7 –6
 F = 2 × 10 × 80 = 16 × 10 N
[Towards the straight wire i.e. force of attraction]
21. (a) Using Bohr’s second postulate of quantization of orbital angular momentum
shown that the circumference of the electron in the nth orbital state in
hydrogen atom is n times the de Broglie wavelength associated with it.
(b) The electron in a hydrogen atom is initially in the third excited state. What
is the maximum number of spectral lines which can be emitted when it
finally moves to the ground state?
Ans. (a) By Bohr’s postulate of quantization of orbital angular momentum
nh
Angular momentum, mvrn =
2
nh
 rn  ... (i)
2mv
Circumference of the electron in the nth orbital state in hydrogen atom
nh h h
2rn = 2  = n mv  n p ... (ii)
2mv
h
But de-broglie wavelength  = p ... (iii)
 Circumference = n ×  [from equations ii & iii]
55/1/1 9 P.T.O.

10. (b) 3rd excited state
2nd excited state
1st excited state
Ground state
From the above diagram, total number of spectral lines = 6
22. In the figure a long uniform potentiometer wire AB is having a constant potential
gradient along its length. The null points for the two primary cells of emfs 1 and
2 connected in the manner shown are obtained at a distance of 120 cm and 300
cm from the end A. Find (i) 1/2 and (ii) position of null point for the cell 1.
How is the sensitivity of a potentiometer increased?
300 cm
A 120 cm B
1 2
1 2
OR
Using Kirchoff’s rules determine the value of unknown resistance R in the circuit
so that no current flows through 4  resistance. Also find the potential difference
between A and D.
F 1 E D
1 4 R
I 6V
9V B A C
3V
Ans. Let potential gradient of wire = K volt/cm
Case 1: l1 = 120 cm
 E1 – E2 = Kl1 = 120 K ...(i) [ cells are connected in opposite direction]
Case 2: l2 = 300 cm
 E1 + E2 = Kl2 = 300 K ...(ii) [ cells are connected in same direction]
Equation (i) divided (ii)
E1  E 2 120 2
 
E1  E 2 300 5
 5E1 – 5E2 = 2E1 + 2E2
 3E1 = 7E2
E1 7
 
E2 3
55/1/1 10 P.T.O.

11. Let null point for cell E1 is l3.
 E1 = Kl3 ...(iii)
Equation (i) divided (iii)
E1  E 2 120K 120
 
E1 Kl 3 l3
E 2 120
 1 
E1 l3
3 120
 1 
7 l3
4 120
 
7 l3
 l3 = 210 cm
E1 7
(i) 
E2 3
(ii) Position of null point for cell E1 = 210 cm
OR
F I 1 E I D
1 4 R
I 6V I
I I
A
9V B 3V C
Applying Kirchoff’s voltage law in closed loop AFEBA :
9–I×1–I×1–4×0–6=0
 9 – 2I – 6 = 0
 3 = 2I
Applying Kirchoff’s voltage law in closed loop BEDCB
6 + 4 × 0 – IR – 3 = 0
 IR = 3
3 3
 R=  2
I 3
 R=2
Potential difference between A and D = Potential difference between A and E
 V AD = I × 2
= 1.5 × 2
V AD = 3V
23. (i) What characteristic property of nuclear force explains the constancy of binding
energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A <
170?
(ii) Show that the density of nucleus over a wide range of nuclei is constant-
independent of mass number A.
55/1/1 11 P.T.O.

12. Ans. (i) The constancy of the binding energy in the range 30 < A < 170 is a
consequence of the fact that the nuclear force is short-ranged. It will be
under the influence of only some of its neighbours, which come within the
range of the nuclear force. If any other nucleon is at a distance more than
the range of the nuclear force from the particular nucleon it will have no
influence on the binding energy of the nucleon under consideration.
The property that a given nucleon influences only nucleons close to it is
also referred to as saturation property of the nuclear force.
Mass of nucleus
(ii) Nuclear density  =
Volume of nucleus
mA
= 4 [where m = mass of one nucleon, A = mass number]
R 3
3
1/3
R = R0A = Radius of nucleus
mA 3m A
= 
4 4R0 A
3
(R0 A1/3 )3
3
3m
 = 4R 3
0
This expression is independent of mass number A and is constant.
24. Write any two factors which justify the need of modulating a signal.
Draw a diagram showing an amplitude modulated wave by superposing a
modulating signal over a sinusoidal carrier wave.
Ans. Factors for modulating a signal
(i) The energy (strength) of signal wave is low, so it cannot be transmitted
directly to large distance.

(ii) Height of signal antena (transmitter and receiver) 
4
 of signal wave is large
 Height of antenna will also be very large, which is practically not possible
Diagram for AM Modulation
55/1/1 12 P.T.O.

13. 25. Write Einstein’s photoelectric equation. State clearly how this equation is obtained
using the photon picture of electromagnetic radiation.
Write the three salient features observed in photoelectric effect which can be
explained using this equation.
Ans. KEmax = h – 
KEmax  energy of the most energetic photoelectron.
  frequency of incident radiation.
  work function.
h  Planck’s constant.
When a photon falls on an electron, it transfers all of its energy (h) to the
electron. Electron being energized leaves the surface of the metal. But while
leaving the surface energy equal to  is reduced in overcomming the binding
force.
Hence, (KEmax = h – )
h
0 =

The salient features are:
(i) KEmax is independent of the intensity of incident radiation.
(ii) Below a threshold frequency, no photoelectric effect takes place.
(iii) KEmax depends on frequency of radiation.
26. (a) Why are coherent sources necessary to produce a sustained interference
pattern?
(b) In Young’s double slit experiment using monochromatic light of wavelength
, the intensity of light at a point on the screen where path difference is ,
is K units. Find out the intensity of light at a point where path difference is
/3.
Ans. (a) Coherent sources have constant phase difference between, i.e., phase
difference does not change with time. Hence, the intensity distribution on
the screen remains constant and sustained.

(b) I = 4I 0 cos2 ...(i)
2
I0  incident intensity.
I  resultant intensity.
At a point where, path difference = 
2
=    2

Putting in (i)
2
K = 4 I0 cos 
K = 4 I0
55/1/1 13 P.T.O.

14. K
I0 =
4

At a point where, path difference is ,
3
2 2  2
= x = 
  3 3
2  K  2 
I2 = 4 I0 cos = 4  cos
2 4  3
K 1 K
= 4  
4 4 4
27. Use Huygen’s principle to explain the formation of diffraction pattern due to a
single slit illuminated by a monochromatic source of light.
When the width of the slit is made double the original width, how would this
affect the size and intensity of the central diffraction band?
Ans. Let AB be a slit of width ‘a’ and a parallel beam of monochromatic light is incident
on it. According to Huygen’s principle the diffraction pattern is the result of
superposition of a large number of waves, starting from different points of
illuminated slit.
P
A 
Light from
M1 
O C
N
90°
source M2 
B
At the central point C of the screen, the angle  is zero. Hence the waves starting
from all points of slit arrive in the same phase. This gives maximum intensity at
the central point C.
The observation point is now taken at P.
Minima: Now we divide the slit into two equal haves AO and OB, each of width
a
. Now for every point, M1 in AO, there is a corresponding point M2 in OB, such
2
a
that M1M2 = ; Then path difference between waves arriving at P and starting
2
a 
from M1 and M2 will be sin  = .
2 2
a sin  = 
In general,
a sin  = n
Secondary Maxima : Similarly it can be shown that for maxima
 1
a sin    n   
 2
55/1/1 14 P.T.O.

15. The intensity pattern on the screen is shown in the given figure.
I0
–4 –3 –2 – 0  2 3 4
a a a a a a a a
When the width of the slit is made double the original width, the size reduces by
half according to the relation: size  /d. Intensity increases four fold.
28. Explain the principle of a device that can built 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 voltages 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?
Ans. Van-de graff generator is based on the principle the charge given to a hollow
conductor is transferred to the outer surface and uniformly distributed over it.
Potential inside conducting spherical shell of radius R carrying charge Q = constant
1 Q
=
40 R
Now, as shown in figure, let us suppose that in some way we introduce a small
sphere of radius r, carrying some charge q, into the large one, and place it at the
centre. The potential due to this new charge clearly has the following values at
the radii indicated:
55/1/1 15 P.T.O.

16. Potential due to small sphere of radius r carrying charge q
1 q
= at surface of small sphere
40 r
1 q
= at large shell of radius R.
40 R
Taking both charges q and Q into account we have for the total potential V and
the potential difference the values
1 Q q 

40  R R 
V(R) =
 
1 Q q 

40  R r 
V(r) =
 
q 1 1 

40  r R 
V(r) – V(R) =
 
We see that, independent of the amount of charge Q that may have accumulated
on the larger sphere and even if it is positive, the inner sphere is always at a
higher potential: the difference V(r )–V(R) is positive.
Smaller bulb is always at a higher potential than the bigger bulb. Therefore
charge continuously flows from the smaller bulb to the bigger bulb.
Yes, the van-de-graft generator can only be charged upto a limit when the electric
field around it is less than breakdown field of the surrounding air.
OR
(a) Electric flux is defined as the dot product of electric field intensity and area
2
vector. Its SI unit is Nm /C.
(b) Let  be the uniform surface charge density of an infinite plane sheet.
Consider a Gaussian surface to be a rectangular parallelepiped of cross
sectional area A.
Surface charge
z
density 
y
E E
2 x
1
x
x x
Gaussian surface for a uniformly
charged infinite plane sheet
From figure, only the two faces 1 and 2 will contribute to flux. Electric field
lines are parallel to the other faces and they do not contribute to the total
flux.
The net flux through the Gaussian surface is 2EA. The charge enclosed by
the closed surface is  A.
By Gauss’s law,
A 
  
2EA  or E n
0 2 0
Since the expression does not contain ‘x’ electric field is independent of the
distance.
55/1/1 16 P.T.O.

17. (c) (i) The electric field is directed perpendicularly away from the sheet.
(ii) The electric field is directed perpendicularly towards the sheet.
29. Define magnifying power of telescope. Write is 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.
Ans. Magnifying power of an astronomical telescope is defined as the ratio of the
angle subtended at the eye by the final image of the angle subtended at the eye,
by the object directly.
f0  fe 
The expression for magnifying power: m =
fe 1  d 
 
f0
(a) In normal adjustment m = 
fe
f0  fe 
(b) When final image is formed at infinity m =  1  d 
fe  
f0 = 150 cm
fe = 5 cm
1 1 1
Using   for objective.
f v u
1 1 1
 
150 v 3000
1 1 1 20  1
  
v 150 3000 3000
3000
v=
19
3000 1 1
m0 =  
19 3000 19
1 1 1
Using   for eye piece
f v u
1 1 1
 
5 25 u
1 1 1
 
u 25 5
1 5
u=
25
55/1/1 17 P.T.O.

18. 25
u= cm
6
v
me =
u
25
me = 6 = 6
25
 Total magnification = m0 × me = –0.315
hI = 0.315 × 100 = 31.578 cm
OR
Telescope Microscope
1. Resolving power should be higher 1. Resolving power is not so large but
for certain magnification. the magnification should be
higher.
2. Focal length of objective should be 2. Both objective and eye piece
kept larger while eye piece focal should have less focal length for
length should be small for better better magnification.
magnification.
3. Objective should be of large 3. Eye piece should be of large
aperture. aperture.
4. Distance between objective and eye 4. Distance between objective and
piece is adjusted to focus the object eye piece is fixed. For focusing an
at infinity. object distance of objective is
changed.
v D
m=  .
u fe
5
v 25
30 =  
u 5
v
6= 
u
 v = –6u
Applying lens formula for objective
1 1 1
 
v u f
1 1 1 100 4
   
6u u 1.25 125 5
1  6 4

6u 5
24u = –35
35
u= cm
24
u = –1.46 cm
30. Draw a simple circuit of a CE transistor amplifier. Explain the working. Show
ac R L
that the voltage gain, AV, of the amplifier is given by AV = , where ac is the
ri
current gain, R L is the load 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?
55/1/1 18 P.T.O.

19. 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
t1 t2 t3 t4 t5 t6 t7 t8
A
B
Ans. Vcc = VCE + IcRL C
Likewise, the input loop gives IC
C RC
VBB = VBE + IB RB RB B
 V0
When vi is not zero, we get IB
Vi  VBB
E
VCC
VBE + vi = VBE + IB RB + IB (RB + ri) IE
The change in VBE can be related to the input resistance ri and the change in IB.
Hence
vi = IB (RB + ri )
vi = r IB
The change in IB causes a change in Ic. We define a parameter  ac, which is
similar to the dc
I c i
ac =  c
I B ib
which is also known as the ac current gain Ai. Usually ac is close to dc in the
linear region of the output characteristics.
The change in Ic due to a change in IB causes a change in VCE and the voltage drop
across the resistor RL because VCC is fixed.
These changes can be given by
VCC = VCE + RL IC = 0
or VCE = –RL IC
The change in VCE is the output voltage v0. We get
v0 = VCE = –ac RL IB
The voltage gain of the amplifier is
v 0 VCE
Av = 
vi r I B
ac RL
Av = 
r
The negative sign indicates that the output is phase reversed by an angle 180°.
OR
(a) Full wave rectification: The circuit uses two diodes connected to the ends
of a centre tapped transformer. The voltage rectified by the two diodes is
55/1/1 19 P.T.O.

20. half of the secondary voltage i.e., each diode conducts for half cycle of input
but alternately so that net output across load comes as half sinusoids with
positive values only. For positive cycle diode D1 conducts (FB) but D2 is being
out of phase is reverse biased and does not conduct. Thus output across RL
is due to D1 only. In negative cycle of input D1 is R.B. but D2 is F.B. and
conducts as with respect to centre tap point A is negative but B is positive.
Hence output across RL is due to D2.
Centretap
transformer A D1
X
Input Centretap
RL Output
B D2
Y
Fig. Full wave rectification circuit
waveform waveform
at A
t
t
at B
Due to Due to Due to Due to
D1 D2 D1 D2
Output
waveform t
across RL
(b) (i) OR Gate (ii) NAND Gate
A B Y  AB A B A.B Y '  A.B
0 0 0 0 0 0 1
1 0 1 1 0 0 1
0 1 1 0 1 0 1
1 1 1 1 1 1 0
t1 t2 t3 t4 t5 t6 t7 t8
A
B
Y
Y
×·×·×·×·×
55/1/1 20 P.T.O.

21. Studymate Solutions to CBSE Board Examination 2011-2012
Series : SMA/1 Code No. 55/1/2
UNCOMMON QUESTION ONLY
1. Why must electrostatic field be normal to the surface at every point of a charged
conductor. [1]
Ans. Surface of a conductor is an equipotential surface, where the field goes normally.
6. Predict of the direction of induced current in a metal ring when the ring is
moved towards a straight conductor with constant speed v. The conductor is
carrying current I in the direction shown in the figure. [1]
v I
Ans. Clockwise using Fleming’s R.H. rule.
10. Derive the expression for the self inductance of a long solenoid of cross sectional
area A and length l, having n turns per unit length. [2]
Ans. l
 0 NI
Magnetic field at a point inside the solenoid is B =

Where N is the total number of turns of the solenoid and l is its length. B is
constant throughout the length of the solenoid.
Magnetic flux through each turn = B × area of each turn.
N
 1 =  0 I×A where A is the area of each turn.

N
 Total magnetic flux linked with the solenoid =  =  0 IA × N

But from the definition of self inductance (L),   LI .
N 0 N2 A
 LI =  0 IA × N  L = = µ0n2lA. Henry
 
–5
14. The susceptibility of a magnetic material is 2.6 × 10 . Identify the type of magnetic
material and state its two properties. [2]
Ans. Paramagnetic, as susceptibility  is +ve and not so large.
Paramagnetic material
(i) pull most of the field lines inside the material and leave some outside
(ii) Moves towards stronger field in a non-uniform field.
55/1/2 21 P.T.O.

22. 16. Two identical circular loops, P and Q, each of radius r and carrying currents
I and 2I respectively are lying in parallel planes such that they have a common
axis. The direction of current in both the loops is clockwise as seen from O which
is equidistant from the both loops. Find the magnitude of the net magnetic field
at point O. [2]
2r
P O Q
I 2I
2
µo I r
Ans. Field due to loop P = towards P.
2(r 2  r 2 )3/2
µo 2I r 2
Field due to loop Q = towards Q.
2(r 2  r 2 )3/2
µo I r 2
So net field at O = towards Q
2(2r 2 )3/2
µo I r 2 µ0 I
= 2 23/2 r 3  towards Q.
4 2r
20. The figure shows a series LCR circuit with L = 4.0 H 2, C = 100 µF, R = 60 
connected to a variable frequency 240 V source. Calculate.
R
C
L
(i) The angular frequency of the source which drives the circuit at resonance.
(ii) The current at the resonating frequency.
(iii) The rms potential drop across the capacitor at resonance.
1 1 1
Ans. (i)     50 rad/s
LC 4  100  106 2  102
Vrms Vrms 240
(ii) I=    4A
Z R 60
(iii) Vrms across L = 4  X L
1
= 4  2  102  4.0
= 800 volt
21. (a) Why are coherent sources necessary to produce a sustained interference
pattern?
(b) In Young’s double slit experiment using monochromatic light of wavelength
, the intensity of light at a point on the screen where path difference is ,
is K units. Find out the intensity of light at a point where path difference is
/3.
55/1/2 22 P.T.O.

23. Ans. (a) Coherent sources have constant phase difference, i.e., phase difference
does not change with time. Hence, the intensity distribution on the screen
remains constant and sustained.

(b) I = 4I 0 cos2 ...(i)
2
I0  incident intensity.
I  resultant intensity.
At a point where, path difference = 
2
=    2

Putting in (i)
2
K = 4 I0 cos 
K = 4 I0
K
I0 =
4
2
At a point where, path difference is ,
3
2 2 4
=  
 3 3
Putting in (i)
2 2
I2 = 4 I0 cos
3
1 K
I2 = 4 I0 × = I0 =
4 4
22. A rectangular loop of wire of size 2 cm × 5 cm carries a steady current of 1A. A
straight long wire carrying 4 A current is kept near the loop as shown. If the loop
and the wire are coplanar, find
I = 4A
2 cm
1A 1A
5 cm
1cm
(i) the torque acting on the loop and
(ii) the magnitude and direction of the force on the loop due to the current
carrying wire.
Ans. (i) Torque  = 0, as there will be forces on the plane of the rectangular coil and
wire.
µo I 1 I 2
(ii) Forces on the loop = 2 r * length (Newton)
55/1/2 23 P.T.O.

24. 7  41 4 1 
= 2  10  1  102  3  102  × 5 × 10 towards the straight wire
–2
 
5  4
= 2  10   4   × 5 × 10
–2
 3
80 –7
= × 10 newton towards the straight wire
3
27. Name the three different modes of propagation of electromagnetic waves. Explain,
using a proper diagram the mode of propagation used in the frequency range
above 40 MHz. [3]
Ans. Three different modes of propagation of electromagnetic waves are
1. Ground (Surface) wave propagation
2. Sky wave propagation
3. Space wave propagation (LOS communication)
For Frequencies about 40 MHz, space wave propagation is being used as the
ionosphere will not reflect the signals and ground transmission is not possible.
In space wave the transmitter and receiver must be on the line of sight (LOS)
together.
Such wave propagation used for Television broadcast and Satellite communication.
In this propagation the uplink and downlink frequencies are kept different to
avoid the mixingup of the signals.
×·×·×·×·×
55/1/2 24 P.T.O.

25. Studymate Solutions to CBSE Board Examination 2011-2012
Series : SMA/1 Code No. 55/1/3
UNCOMMON QUESTION ONLY
6. Why is electrostatic potential constant throughout the volume of the conductor
and has the same value (as inside) on its surface ? [1]
Ans. E =0 in a conductor makes Q en = 0 and charges stay on the surface. So the
potential is constant and will be the same as on the surface.
9. The relative magnetic permeability of a magnetic material is 800. Identify the
nature of magnetic material and state its two properties. [2]
Ans. µ > > 1 for ferromagnetic
Ferromagnetic material
(i) Pulls all the field lines inside the material
(ii) Moves towards the strongest field in a non-uniform field.
12. Define mutual inductance between two long coaxial solenoids. Find out the
expression for the mutual inductance of inner solenoid of length l having the
radius r1 and the number of turns n1 per unit length due to the second outer
solenoid of same length and n2 number of turns per unit length.
Solenoid S1
Ans.
Solenoid S2
Let l be the length of each solenoid S1 and S2
N1 and N2 be the total number of turns of S1 and S2.
 0 N1I1
Magnetic field in S1 = B1 =

Magnetic flux linked with each turn of S2 = B1 × area of each turn = B1r12
Total magnetic flux linked with S2 = B1r12 N2
 N I   N N r 2 I
  2 =  0 1 1  r12N2 = 0 1 2 1 1
   
But magnetic flux linked with S2 is due to I1
 2  I1 or  2 = M I1
Where M is the mutual inductance of S2 and S1
0 N1N 2 r12 I1
 M I1 =

2
 M = 0 n1n2 r1 l
55/1/3 25 P.T.O.

26. 21. Name the three different modes of propagation of electromagnetic waves. Ex-
plain, using a proper diagram the mode of propagation used in the frequency
range from a few MHz to 40 MHz.
Ans. Three different modes of propagation of electromagnetic waves are
1. Ground (Surface) wave propagation
2. Sky wave propagation
3. Space wave propagation (LOS communication)
The type of propagation used upto a frequency of 40 MHz is sky wave propagation.
The ions in the ionosphere reflects the signal back to the surface of the earth
making long distance communication possible. The maximum frequency f c upto
1/2
which this is used depends on the ion density as fc=9 (Nmax)
23. A rectangular loop of wire of size 2.5 cm × 4 cm carries a steady current of 1A. A
straight long wire carrying 2 A current is kept near the loop as shown. If the loop
and the wire are coplanar, find
I = 2A
2.5 cm
1A 1A
4 cm
2cm
(i) the torque acting on the loop and
(ii) the magnitude and direction of the force on the loop due to the current
carrying wire.
Ans. (i) Torque  = 0, as there will be forces on the plane of the rectangular coil and
wire.
µo I 1 I 2
(ii) Forces on the loop = 2 r * length (Newton)
 21 2 1 
= 2  107   –2
2 2 
× 4 × 10 towards the straight wire
 2  10 4.5  10 
55/1/3 26 P.T.O.

27.  2
= 2  105  1   × 4 × 10
–2
 5
3 24
= 8  107    107
5 5
–7
= 4.8 × 10 newton towards the straight wire
25. The figure shows a series LCR circuit with L = 10.0 H 2, C = 40 µF, R = 60 
connected to a variable frequency 240 V source. Calculate.
R
C
L
(i) The angular frequency of the source which drives the circuit at resonance.
(ii) The current at the resonating frequency.
(iii) The rms potential drop across the capacitor at resonance.
1 1 1
Ans. (i) =   = 50 rad/s
LC 10  40  10 6 2  102
Vrms Vrms 240
(ii) I=    4A
Z R 60
(iii) Vrms across L = 4  XL
1
= 4  10
2  102
= 2000 volt
×·×·×·×·×
55/1/3 27 P.T.O.