1. Studymate Solutions to CBSE Board Examination 2012-2013
Series : SKS/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 15 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 29 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 29 questions in total. Questions numbered 1 to 8 are very short-answer questions and carry 1 mark
each.
(iii) Questions numbered 9 to 16 carry two marks each, Question Nos. 17 to 25 carries three marks each and Question
Nos. 27 to 29 carry five 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) Question No. 26 is value based question carries four marks.
(vi) 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
40
me = 9.1 × 10–31 kg
Mass of the Neutron = 1.675 × 10–27 kg
Mass of the Proton = 1.673 × 10–27 kg
55/1/1 1 P.T.O.

2. 1. What are permanent magnets? Give one example.
Ans. Magnets which possess higher retentivity and coercivity and retain the magnetic field for
longer period are called permanent magnets. Example. Alnico, Nipermag, Steel.
2. What is the geometrical shape of equipotential surfaces due to a single isolated charge?
1
Ans. Concentric spheres with the gap between them not being uniform as V 
r
3. Which of the following waves can be polarized
(a) Heat waves (b) Sound waves?
Ans. Heat waves can be polarised as they are transverse by nature.
4. A capacitor has been charged by a dc source. What are the magnitudes of conduction and
displacement currents, when it is fully charged?
Ans. On full charging the source will maintain the potential across the plates. Displacement current
and conduction current will be zero.
5. Write the relationship between angle of incidence ‘i’, angle of prism ‘A’ and angle of minimum
deviation for a triangular prism.
A + δm
Ans. Angle of incidence i =
2
6. The given graph shows the variation of photo-electric current (I) versus applied voltage (V) for
two different photosensitive materials and for two different intensities of the incident radiation.
Identify the pairs of curves that correspond to different materials but same intensity of incident
radiation.
I
1
3
2
4
V
Ans. The pairs (2,4) and (1,3) are having the same intensity and different metal.
7. A 10 V battery of negligible internal resistance is connected across a 200 V battery and a
resistance of 38  as shown in the figure. Find the value of the current in circuit.
10V
38 200V
Ans. Applying Kirchoff’s rule, we get
200 – I(38) – 10 = 0
190 95
 I=   5A
38 19
55/1/1 2 P.T.O.

3. 8. The emf of a cell is always greater than its terminal voltage. Why? Give reasons.
Ans. Emf is the p.d. when no current is drawn. When current is drawn, there will be potential drop
across the internal resistance of the cell. So, terminal voltage will be less than the emf.
9. (a) Write the necessary conditions for the phenomenon of total internal refletion of occur.
(b) Write the relation between the refractive index and critical angle for a given pair of
optical media.
Ans. (a) (i) The light should traverse from denser to rarer media.
(ii) The angle of incidence should be more than the critical angle.
d 1
(b)  where d and r are the refractive index of the denser and the rarer media.
r sin ic
10. State Lenz’s Law.
A metallic rod held horizontally along east-west direction, is allowed to fall under gravity. Will
there be an emf induced at its ends? Justify your answer.
Ans. According to Lenz’s Law, “the emf produced due to the change in magnetic flux will oppose the
cause of the emf.”
Yes, there will be an emf as the horizontal component of field of earth, velocity of the motion
of the rod and the length of the rod are all perpendicular to each other.
11. A convex lens of focal length 25 cm is placed coaxially in contact with a concave lens of focal
length 20 cm. Determine the power of the combination. Will the system be converging or
diverging in nature?
Ans. f1 = 25 cm f2 = –20 cm
1 1 
Peq = P1 + P2 = 100  f  f 
 1 2 
 1 1   20  25 
Peq = 100     100   = –1 Dioptre
 25 20   500 
The system will be a diverging lens as if has negative power.
12. An ammeter of resistance 0.80  can measure current upto 1.0 A.
(a) What must be the value of shunt resistance to enable the ammeter to measure current
upto 5.0 A?
(b) What is the combined resistance of the ammeter and the shunt?
Ans. (a) Rg = 0.80 , Ig = 1.0 A, I = 5A
Ig R g 1  0.8
We know that, Shunt resistance (S) =  = 0.2 
I  Ig 4
RgS 0.80  0.2
(b) The combined resistance will be R =  = 0.160 
Rg  S 1.00
13. In the given circuit diagram, a voltmeter ‘V’ is connected across a lamp ‘L’. How would (a) the
brightness of the lamp and (b) voltmeter reading ‘V’ be affected, if the value of resistance ‘R’
is decreased? Justify your answer.
55/1/1 3 P.T.O.

4. V L
+
9V
–
R
Ic
Ans. As R is decreased, Ib increases. So, Ic increases as  = . With increase in Ic, power dissipated
Ib
by bulb is more. So, voltmeter reading and brighness of the bulb, both increase.

14. (a) An em wave is travelling in a medium with a velocity v  viˆ . Draw a sketch showing
the propagation of the em wave, indicating the direction of the oscillating electric and
magnetic fields.
(b) How are the magnitudes of the electric and magnetic fields related to the velocity of
the em wave?
Ans. (a) Propagation direction is X.
Ey
E
X
Bz B
Electric field  Along y
Magnetic field  Along z
(b) If Em and Bm are the magnitude of the electric and magnetic fields, then, the velocity of
Em
e.m. wave is given by C = B
m
15. Block diagram of a receiver is shown in the figure:
Receiving Antenna
Received Signal Output
Amplifier X Detector Y
(a) Identify ‘X’ and ‘Y’.
(b) Write their functions.
Ans. (a) X – Intermediate frequency amplifier
Y – Audio frquency amplifier
(b) The intermediate frquency amplifier brings down the frequency to a lower value. Audio
amplifier amplifies the final demodulated output.
16. Explain, with the help of a circuit diagram, the working of a photo-diode. Write briefly how it
is used to detect the optical signals.
55/1/1 4 P.T.O.

5. OR
Mention the important considerations required while fabricating a p-n junction diode to be
used as a Light Emitting Diode (LED). What should be the order of band gap of an LED if it is
required to emit light in the visible range?
Ans. It is a p-n junction fabricated with a transparent window to allow light photons to fall on it.
These photons generate electron hole pairs upon absorption. If the juction is reverse biased
using an electrical circuit, these electron hole pair move in opposite directions so as to produce
current in the circuit. This current is very small and is detected by the microammeter placed
in the circuit. The magnitude of the photocurrent depends on the intensity of incident light
(photocurrent is proportional to incident light intensity). It is easier to observe the change in
the current with change in the light intensity. This property is used in directing optical
signal.
I
I4>I3>I2>I1 (m A)
R. B.
p n
A I1 V
I2
I3
I4 I ( A)
OR
Ans. The important considerations required while fabricating a p-n junction diode to be used as a
Light Emitting Diode (LED) are
(i) the light emitting efficiency is maximum.
(ii) care should be taken that high reverse voltages do not appear across them.
The semiconductor used for fabrication of visible LEDs must at least have a band gap of
1.8 eV (spectral range of visible light is from about 0.4 m to 0.7 m, i.e., from about 3 eV to
1.8 eV).
17. Write three important factors which justify the need of modulating a message signal. Show
diagrammatically how an amplitude modulated wave is obtained when a modulating signal
is superimposed on a carrier wave.
Ans. Need for Modulation
(a) Size of antenna: For transmitting a signal, the antenna must have a size h = /4. For
e.m wave of frequency 20 kHz, l = 15 km. Such a long antenna is not practical. So,
there is a need of translating the low frequency signal into high radio frequencies.
2
l 
(b) Power radiated: Is proportional to   . l – antenna length to have good power radiation,

high frequency (low wavelength) transmission is needed.
(c) If many transmitters are transmitting message signals simultaneously, all these
signals will get mixed up and it will be difficult to distinguish them. Possible solution is
using communication at high frequencies and allotting a band of frequencies to each
message signal for its transmission.
55/1/1 5 P.T.O.

6. Diagrammatic representation of Amplitude Modulation
In A.M., the amplitude of carrier wave is changed in accordance with information signal.
+Ac Carrier wave
c(t) c(t) = Ac sin ct
0 t
–Ac
Am
Message signal
0 t m(t) = Am sin mt
m(t)
–Am
A
cm(t) 0 t
Amplitude modulated wave
18. A capacitor of unknown capacitance is connected across a battery of V volts. The charge
stored in it is 360 C. When potential across the capacitor is reduced by 120 V, the charge
stored in it becomes 120 C.
Calculate:
(a) The potential V and the unknown capacitance C.
(b) What will be the charge stored in the capacitor, if the voltage applied had increased by
120 V?
OR
A hollow cylindrical box of length 1 m and area of cross-section 25 cm2 is placed in a three
dimensional coordinate system as shown in the figure. The electric field in the region is

given by E  50xiˆ , where E is in NC–1 and x is in metres. Find
(a) Net flux through the cylinder.
(b) Charge enclosed by the cylinder.
Y
O X
1m
Z
Ans. (a) Let the capacitance be C V
so charge Q1 = cV
or 360 C = cV ...(i)
In second case Q1 C
Q2 = c(V – 120)
or 120 C = c(V – 120) ...(ii)
55/1/1 6 P.T.O.

7. From equation (i) and (ii), (V – 120)
360 C cV

120 C c  V  120 
V
3= Q2 C
V  120
or 3V – 360 = V
or 2V = 360
V = 180 volt
From equation (i),
360 C = C × 180
360  C
C= = 2 f
180V
C = 2 f
1
(b) Energy stored V = cV2
2
1
U= × 2 × 10–6 × (180 + 120)2
2
1
=  2  10 6  300  300
2
U = 9 × 10–2 J
OR
 
Ans. (a) Flux acoss curved surface 2 = E.A  0 ( = 90°)
 
Flux acoss first surface 1 = E.A = EA cos 0°
1 = (50 × 2) × 25 × 10–4
= 100 × 25 × 10–4
1 = 25 × 10–2 Vm
 
Flux across third surface 3 = E.A = EA cos 180° ( = 180°)
= (50 × 1) × 25 × 10–4 (–1) Y
–4
= –1250 × 10
II
= –12.50 × 10–2 Vm III I
O X
So, net flux  = 1 + 2 + 3 1m
= 0 + 25 × 10–2 – 12.50 × 10–2 Z
–2
 = 12.50 × 10 Vm
q en
(b) From Gauss Law  = 
0
q en
or 12.50 × 10–2 = 
0
qen = 0 × 12.50 × 10–2
= 8.85 × 10–12 × 12.50 × 10–2
qen = 110.625 × 10–14 C
55/1/1 7 P.T.O.

8. 19. (a) In a typical nuclear reaction, e.g.
2
1
H + 21H  3
He + n + 3.27 MeV
2
although number of nucleons is conserved, yet energy is released. How? Explain.
(b) Show that nuclear density in a given nucleus is independent of mass number A.
Ans. (a) In all types of nuclear reaction, the law of conservation of number of nucleons is followed.
But during reaction, the mass of final product is found to be slightly less than the sum
of the masses of reactant components. This difference in masses is called mass defect.
So, as per mass energy relation E = (m)c2, energy is released.
In the given reaction, the sum of the masses of two deutrons is more than the mass of
helium and neutron. Energy equivalent of mass defect is released.
(b) Let m = mass of one nucleon
A = mass number
R = radius of nucleus
mass of nucleus
Density (d) =
volume
m.A mA
= 
4
 
3
R 3 4 1
 R0 A 3
3 3
3m A
=
4 R 3 A
0
3m
d=
4 R 3
0
We can see that density is independent of mass number.
20. (a) Why photoelectric effect can not be explained on the basis of wave nature of light? Give
reasons.
(b) Write the basic features of photon picture of electromagnetic radiation on which
Einstein’s photoelectric equation is based.
Ans. (a) (i) According to wave nature, the energy carried by light is measured in terms of
intensity. So, when light of higher intensity falls on metal surfce, it will impart
more energy to electron. Hence, electrons will eject out with more kinetic energy.
But experimentally, kinetic energy of photoelectrons is independent of intensity
of light.
(ii) According to wave theory, threshold frequency should not exist. Light of all
frequencies should emit electron provided intensity of light is sufficient for
electrons to eject.
(iii) According to wave theory, photoelectric effect should not be instantaneous. Energy
of wave cannot be transferred to a particular electron but will be distributed to all
the electrons present in the illuminated portion. Hence, there has to be a time
lag between incident of radiation and emission of electrons.
(b) Basic features of photon picture of electromagnetic radiation:
(i) Radiation behaves as if it is made of particles like photons.
Each photon has energy E (= h) and momentum p    .
h
(ii)  
 
55/1/1 8 P.T.O.

9. (iii) Intensity of radiation can be understood in terms of number of photons falling per
second on the surface. Photon energy depends only on frequency and is
independent of intensity.
(iv) Photoelectric effect can be understood as the result of the one to one collision
between an electron and a photon.
(v) When a photon of frequency () is incident on a metal surface, a part of its energy
is used in overcoming the work function and other part is used in imparting
kinetic energy so,
1 2
h  0  mVmax
2
21. A metallic rod of light ‘l’ is rotated with a freqneucy  with one end hinged at the centre and
the other end at the circumference of a circular metallic ring of radius r, about an axis
passing through the centre and perpendicular to the plane of the ring. A constant uniform
magnetic field B parallel to the axis is present every where. Using Lorentz force, explain how
emf is induced between the centre and the metallic ring and hence obtain the expression for
it.
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 the emf generated across a length dr of the
rod as it moves at right angles to the magnetic field is given by
d = Bv dr
Hence,
r r
B r 2 B  2  r 2
 
  d  Bvdr  B r dr 
0

0
2

2
  Br 2
22. Output characteristics of an n-p-n transistor in CE configuration is shown in the figure.
50µA
6
40µA
5
4 30µA
3.5
20µA
3
10µA
2
IB = 0
1
0
2 4 6 8 10 12 14 16 18
Determine:
(a) dynamic output resistance
(b) dc current gain and
(c) ac current gain at an operating point VCE = 10 V, when IB = 30 A.
55/1/1 9 P.T.O.

10. Ans. (a) Dynamic output resistance
 VCE 
r0 =  
 IC IB
 12  8 
r0 =  3 
  3.6  3.4  10  30  A
r0 = 20 k
(b) For IB = 30 A, the IC = 3.5 A
3.5  10 3
dc = 6 = 0.12 × 103
30  10
dc = 120
 I 
(c) ac =  C 
 IB  VCE
For IB = 30 A, IC = 3.5 mA and for IB = 40 A, IC = 4.6 mA
  4.6  3.5   103 
ac =  6 
  40  30   10  VCE  10V
1.1 
=   103 
 10 
ac = 110
23. Using Bohr’s postulates, obtain the expression for the total energy of the electron in the
stationary states of the hydrogen atom. Hence draw the energy level diagram showing how
the line spectra corresponding to Balmer series occur due to transition between energy
levels.
Ans. The basic postulates of Bohr’s theory are :
(a) The central core of every atom is called nucleus and it carries only positive charge and
almost the entire mass of the atom. Negatively charged electrons revolve around it with
a centripetal force which is provided by the electrostatic force of attraction between
electron and nucleus.
mv2
F=
r
where m and v are mass and velocity of electron moving in orbit of radius r. .....(1)
(b) Electron revolve in discrete non-radiating orbits (stationary orbits) with angular
h
momentum equal to integral multiple of (h is Planck’s constant)
2
nh
i.e., mvr = .....(2)
2
where mvr = angular momentum of e–
(c) Radiation of energy takes place when electron jumps from one orbit to another and the
amount of energy radiated is equal to the difference in energies of the two permitted
orbits i.e.,
hv = E2 – E1 .....(3)
where v = frequency of radiation emitted
nh nh
We know, mvr = (from eqn 2)  v = ......(4)
2 2mr
55/1/1 10 P.T.O.

11. Also, electrostatic force between electron (e–) and nucleus of charge (+Ze) is
1 |Ze||e|
F= ......(5)
40 r2
mv2 1 Ze2
From (1) and (5),  ......(6)
r 40 r2
1 Ze2
or r=
40 mv2
Substitute from (4),
1 Ze2 42m2r2
r= 
40 m n2h2
(40 )n2h2
r=
42mZe2
0 n2 h2
 r=
 me2 .Z
For H, Z = 1
(40 ) n2 h2
 r= ......(7)
42 me2
 r  n2
Since energy of moving electron is its kinetic energy, hence from (6),
mv2 1 Ze2

r 40 r2
1 Ze2
 mv2 =
40 r2
1 1  1 Ze2 
K. E. = mv2    ......(8)
2 2  40 r 
 
Potential energy = Potential × Charge
 1 Ze 
Since potential of electron is due to nucleus  i.e., 
 40 r 
1 Ze 1 Ze2
 P.E. =  (e) = .....(9)
40 r 40 r
 Total energy of electron = KE + PE
1  Ze2  1 Ze2
E=  
2  40 r  40 r
 
n=5
1  Ze 2  n=4
 E=   .....(10)
40  2r



n=3
n=2
22mZ 2e4
Put (7) in (10) we get, E = n=1
(40 )2 n2h2
55/1/1 11 P.T.O.

12. For H-atom, substitute standard values,
13.6
We get, E = eV
n2
1 1 
When electron jumps from outer to second orbit then 2  R  2  2  where k = 3, 4,
2 k 
13.6
5....... and energy of electron is E2 = = – 3.4 eV. This is Balmer Series.
22
24. (a) In what ways is diffraction from each slit related to the interference pattern in a double
slit experiment?
(b) Two wavelengths of sodium light 590 nm and 596 nm are used, in turn, to study the
diffraction taking place at a single slit of aperture 2 × 10–4 m. The distance between the
slit and the screen is 1.5 m. Calculate the separation between the positions of the first
maxima of the diffraction pattern obtained in the two cases.
Ans. (a) When there are only a few sources, say two interfering sources, then the result is
usually called interference, but if there is a large number of them, it seems that the
word diffraction is more often used.
In the double-slit experiment, we must note that the pattern on the screen is actually
a superposition of single-slit diffraction from each slit. This is shown in figure.
The actual double-slit interference pattern. The envelope shows the single slit diffraction.
(b) d = 2 × 10–4 m
D = 1.5 m
1 = 590 × 10–9 m
2 = 596 × 10–9 m
For first maxima,
31
d sin  =
2
For small angle (sin   tan  )
31
So, d tan  =
2
y1 31
d. 
D 2
31D
or, y1 =
2d
For second wavelength
3 2D
y2 =
2d
3D
So, separation y =  2  1 
2d
55/1/1 12 P.T.O.

13. 3  1.5
y =  596  590   10 9
2  2  10 4
y = 6.75 × 10–5 m
25. In a series LCR circuit connected to an ac source of variable frequency and voltage  = vm sin
t, draw a plot showing the variation of current (I) with angular frequency () for two different
values of resistance R1 and R2 (R1 > R2). Write the condition under which the phenomenon of
resonance occurs. For which value of the resistance out of the two curves, a sharper resonance
is produced? Define Q-factor of the circuit and give its significance.
Ans. (a) Condition for resonance to occur is I
X L = XC
R2
(b) Sharper resonance is produced for R2
(c) Quality factor is the ratio of voltage drop across the R1
inductor or capacitor to the voltage drop across
resistor. 
0
(d) Significance  it shows the sharpness of the circuit.
Sharper the curve is more sensitive the circuit will be.
26. While travelling back to his residence in the car, Dr. Pathak was caught up in a thunderstorm.
It became very dark. He stopped driving the car and waited for thunderstorm to stop. Suddenly
he noticed a child walking alone on the road. He asked the boy to come inside the car till the
thunderstorm stopped. Dr. Pathak dropped the boy at his residence. The boy insisted that Dr.
Pathak should meet his parents. The parents expressed their gratitude to Dr. Pathak for his
concern for safety of the child.
Answer the following questions based on the above information:
(a) Why is it safer to sit inside a car during a thunderstorm?
(b) Which two values are displayed by Dr. Pathak in his actions?
(c) Which values are reflected in parents’ response to Dr. Pathak?
(d) Give an example of a similar action on your part in the past from everyday life.
Ans. (a) The body of the car provides a electrostatic shield. This prevents the lightning strike
on the people in the car.
(b) Concern for others, values human life,.
(c) Obliged with the gesture of Dr. Pathak, highly thankful.
(d) I saw an old couple stranded in the street due to puncture of their vehicle tyre. I guided
them to a mechanic, got the tyre rectified and guided them home.
27. (a) Draw a ray diagram showing the image formation by a compound microscope. Hence
obtain expression for total magnification when the image is formed at infinity.
(b) Distinguish between myopia and hypermetropia. Show diagrammatically how these
defects can be corrected.
OR
(a) State Huygen’s principle. Using this principle draw a diagram to show how a plane
wave front incident at the interface of the two media gets refracted when it propagates
from a rarer to a denser medium. Hence verify Snell’s law of refraction.
55/1/1 13 P.T.O.

14. (b) When monochromatic light travels from a rarer to a denser medium, explain the
following giving reasons:
(i) Is the frequency of reflected and refracted light same as the frequency of incident
light?
(ii) Does the decrease in speed imply a reduction in the energy carried by light wave?
Ans. (a) Compound Microscope: Compound microscope is used for observing highly magnified
images of tiny objects. It makes use of two converging lenses. The objective lens is of
very small focal length and short aperture and the eye-piece is of small focal length and
large aperture. The image formed by the objective lens becomes the object for the
eyepiece. The eyepiece acting as a magnifying glass produces enlarged virtual image
of it
fe
u f0
A
h  B’ Eyepiece
B 
B” h’ E
Objective
A’
d
A”
v0  d
Magnifying power of a Compund microscope is given by (i), m  1  
|u0 | fe 
u0 = object distance for objective lens
v0 = image distance for objective lens
f0 = focal length of objective lens
fe = focal length of eyepiece.
Since the object lies very close to f0, the image formed by the objective lens is very
close to the focal length (fe) of the eyepiece.
 u0  f0 , v0  L where L is the length of the microscope tube.
L  d
Magnifying Power, m  1   .
f0 |  fe 
When image is formed at infinity
v L d
m0  0  me 
u0 f0 ; fe
L d
 Magnifying powerm m = m0 × me  m = 
f0 fe
The magnifying power of compound microscope is negative hence the image formed is
inverted.
(b) The light from a distant object arriving at the eye-lens may get converged at a point in
front of the retina. This type of defect is called nearsightedness or myopia. This means
that the eye is producing too much convergence in the incident beam. To compensate
this, we interpose a concave lens between the eye and the object, with the diverging
effect desired to get the image focussed on the retina.
If the eye-lens focusses the incoming light at a point behind the retina, a convergent
55/1/1 14 P.T.O.

15. lens is needed to compensate for the defect in vision. This defect is called farsightedness
or hypermetropia.
Corrected Myopic eye. Corrected Hypermetropic eye.
OR
Ans. (a) HUYGENS PRINCIPLE
(i) Each point of the wavefront is a source of a secondary disturbance and secondary
wavelets emanating from these points spread out in all directions with the speed
of the wave.
(ii) The new position of the wavefront at a later time can be found using the product vt
where v is the speed and t is the time.
(iii) Energy propagates in the direction perpendicular to the wavefront.
C
A
C'
Secondary
A' A C wavelents
Primary
wavefront
New Position of
wavefront
B' B D D'
B
D
REFRACTION OF PLANE WAVES USING HUYGENS PRINCIPLE
Refraction of a plane wave:– Let v1 and v2 be the speed of light in medium 1 and medium
2. Suppose a plane wavefront AB is incident on PP' at an angle i.
Incident wavefront
B
A' v1
v1 i
medium 1 i
P A r C P'
v2
r
medium 2
Refracted
E wavefront
v2<v1
Let t be the time taken by the wavefront to travel distance BC in medium 1. Thus, BC =
v1t.
55/1/1 15 P.T.O.

16. In the same time the wavefront travels distance AE in medium 2. Thus, AE = v2t and CE
would represent the refracted wavefront.
BC v1t AE v 2t
In ABC , sin i   ; In AEC , sin r  
AC AC AC AC
where i and r are the angles of incidence of refraction.
sin i v1
  ...(i)
sin r v 2
c c
Now, n1  v and n 2  v
1 2
n 2 v1 v1
 
n1 v 2 or n21 = v 2 ...(ii)
sin i
From (i) and (ii),  n 21
sin r
This is the Snell's law of refraction.
(b) (i) Frequency is the property of source. Hence, frequency does not change when light
is reflected or refracted
(ii) No, decrease in speed does not imply reduction in energy carried by light wave.
This is because the frequency does not change and according to the formula E =
h, energy will be independent of speed.
28. (a) State the working principle of a potentiometer. With the help of the circuit diagram
explain how a potentiometer is used to compare the emf’s of two primary cells. Obtain
the required expression used for comparing the emfs.
(b) Write two possible causes for one sided deflection in a potentiometer experiment.
OR
(a) State Kirchhoff’s rules for an electric network. Using Kirchhoff’s rules. obtain the balance
condition in terms of the resistances of four arms of Wheatstone bridge.
(b) In the meterbridge experimental set up, shown in the figure, the null point ‘D’ is obtained
at a distance of 40 cm from end A of the meterbridge wire. If a resistance of 10  is
connected in series with R1, null point is obtained at AD = 60 cm. Calculate the values
of R1 and R2.
R1 R2
B
G
D
A C
K
Ans. (a) Principle: The fall of potential across any portion of the potentiometer wire is directly
proportional to the length of that portion provided the wire is of uniform cross–sectional
area and a constant current is flowing through it.
55/1/1 16 P.T.O.

17. – +
A
Rh K
A J J B
E1
3
G
2
E2
 Working:
A – The area of cross-section of the wire.
r – Specific resistance of the material of the wire
V – Potential drop across the portion of the wire of length l.
R – Resistance
I – Current flowing through the wire.
I 
V  IR   k
A
 V  l where I and A are constants.
V
 k  potential gradient or falls of potential per unit length of the wire.

To find the e.m.f of a cell the following circuit is used.
Let the null point is found at J and AJ = l.
 The potential drop across the potentiometer wire AJ is equal to the unknown
e.m.f E.
 E = kl where k is the potential gradient of the potentiometer wire. Knowing k and
measuring l from the attached scale E can be calculated.
– +
– + A
A Rh K
Rh K
A J J B
A l J B
E1 1
3 G
E
G 2
E2
Comparison of e.m.f of two cells
 When the key is put between 1 and 3 terminals, cell of e.m.f E1 is connected in the
circuit and the corresponding balancing length is found to be l1.
 E1 = kl1 [where k = Potential gradient of the potentiometer wire]
 When the key is put between 2 and 3 terminals, cell of e.m.f E2 is connected in the
circuit and the corresponding balancing length is found to be l2.
55/1/1 17 P.T.O.

18.  E2 = kl2
E1 1
 
E2  2
Measuring l1 and l2 from the attached scale the ratio of E1 and E2 can be calculated.
(b) Causes for one sided deflection
(i) Potential difference between the ends of the potentiometer wire is less than the
emf of the cell in the secondary circuit.
(ii) The positive side of the driving cell is connected to the negative terminal of the
cells in the secondary circuit.
OR
Ans.(a) Kirchhoff’s Rules:
 Junction Rule (  I = 0). The sum of the charge/currents entering any junction must
equal the sum of the charge/currents leaving that junction.
 This is a statement of Conservation of Charge.
 Loop Rule (  V = 0). The sum of the potential differences across all the elements around
any closed circuit loop must be zero.
 This is a statement of Conservation of Energy.
 Wheatstone Bridge Principle: Wheatstone bridge is a very accurate and precise method
of determining an unknown resistance.
B
(I1 – Ig)
P Q
I1 Ig
(I – I1) G C
A
R S
I
I – I1 + Ig I
D
Applying loop rule to mesh ADBA , we get
–(I – I1)R + IgG + I1P = 0 …(1)
On applying loop rule to mesh DCBD, we get
–(I – I1 + Ig)S + (I1 – Ig)Q – IgG = 0 …(2)
The value of R is so adjusted that the points B and D are at the same potential so that the
current Ig through the galvanometer is zero. In this position the bridge is said to be balanced.
When Ig = 0, eqns, (1) and (2) reduce to the form
I1P = (I – I1)R
and I1Q = (I – I1)S
Dividing one by the other
P R
 (Condition for the balance of bridge)
Q S
55/1/1 18 P.T.O.

19. R1 40
(b)  ...(i)
R 2 60
R1  10 40

R2 60
Dividing
R1  10 40 40 4
  
R1  10 60 60 9
9R1 = 4R1 + 40
5R1 = 40
R1 = 8 
Using (i) we get,
60
R2 =
40
60
R1 =  8 = 12 
40
29. (a) Derive the expression for the torque on a rectangular current carrying loop suspended
in a uniform magnetic field.
(b) A proton and a deuteron having equal momenta enter in a region of uniform magnetic
field at right angle to the direction of the field. Depict their trajectories in the field.
OR
(a) A small compass needle of magnetic moment ‘m’ is free to turn about an axis
perpendicular to the direction of uniform magnetic field ‘B’. The moment of inertia of
the needle about the axis is ‘I’. The needle is slightly disturbed from its stable position
and then released. Prove that it executes simple harmonic motion. Hence deduce the
expression for its time period.
(b) A compass needle, free to turn in a vertical plane orients itself with its axis vertical at
a certain place on the earth. Find out the values of (i) horizontal component of earth’s
magnetic field and (ii) angle of dip at the place.
Ans. (a) Torque experienced by a Current Loop (Rectangular) in a uniform Magnetic Field:
Let  be the angle between the plane of the loop and the FSP
direction of the magnetic field. The axis of rotation of the
coil is perpendicular to the magnetic field. b S
   
FSP  I(b  B)
I
| FSP | = I b B sin  P FRS
  
FQR  I(b  B) × B
| FQR | = I b B sin 
  l
Forces FSP and FQR are equal in magnitude but opposite
in direction and they cancel out each other. Moreover FPQ
I R
they act along the same line of action (axis) and hence do 
not produce torque.
  
FQR  I(l  B) Q
FQR
| FPQ | = I l B sin 90° = I l B
55/1/1 19 P.T.O.

20.   
FRS  I(l  B)
| FRs | = I l B sin 90° = I l B
 
Forces FPQ and FRS being equal in magnitude but opposite in direction cancel out
each other and do not produce any translational motion. But they act along different
lines of action and hence produce torque about the axis of the coil.
FRS
Torque experienced by the coil is
 = FPQ × PN (in magnitude) × S
b
 = I l B (b cos )  B

 = I lb B cos   I
P N
 = I A B cos  (A = lb) n
 = N I A B cos  (where N is the no. of turns) FPQ
If  is the angle between the normal to the coil and
the direction of the magnetic field, then
I R
 +  = 90° i.e.  = 90° – 
B
So, = I A B cos (90° - ) 
Q n
 = N I A B sin 
mv p
(b) Since r   and the charge on both the proton and deutron are the same, the
qB qB
particles will follow a circular path with their radius in the ratio 1: 1
OR
Ans. (a) This is done by placing a small compass needle of known magnetic moment m and
moment of inertia I and allowing it to oscillate in the magnetic field. This arrangement
is shown in figure.
The torque on the needle is,
=m×B
In magnitude  = mB sin 
Here  is restoring torque and  is the angle between m and B.
Therefore, in equilibrium
d2 
I  mBsin 
dt2
Negative sign with mB sin  implies that restoring torque is in opposition to deflecting
torque.
This represents a simple harmonic motion. The square of the angular frequency is 2
= mB/I and the time period is,
I
T  2
mB
(b) Since, the compass needle is oriented vertically,
(i) Horizontal component of earth’s field will be zero.
BV
(ii) Angle of dip will be 90° as tan  = B .
H
×·×·×·×·×
55/1/1 20 P.T.O.

21. Studymate Solutions to CBSE Board Examination 2012-2013
Series : SKS/1 Code No. 55/1/2
UNCOMMON QUESTION ONLY
1. A cell of emf ‘E’ and internal resistance ‘r’ draws a current ‘I’. Write the relation between
terminal voltage ‘V’ in terms of E, I and r.
Ans. E = V+Ir where r is the internal reistance.
2. Which of the following substances are diamagnetic?
Ans. Bi and Cu
3. A heating element is marked 210 V, 630 W. What is the value of the current drawn by the
element when connected to a 210 V dc source?
Ans. Current drawn = I = P/V = 630/210 = 3 A.
10. An ammeter of resistance 1 W can measure current upto 1.0 A (i) What must be the value of
the shunt resistance to enable the ammeter to measure upto 5.0A? (ii) What is the combined
resistance of the ammeter and the shunt?
Ig G
Ans. S  I  I
g
[Here, Ig represents the current for full scale deflection in the ammeter of resistance 1.]
[G  resistance of ammeter that measures 1 Amp]
Ig = 1A
G = 1
I = 5A
(1) (1)
S  0.25
(5  1)
Shunt resistance S = 0.25 
GS (1)0.25
Req 
G  S 1  0.25
0.25 1
Req =   0.2 
1.25 5
13. In the given circuit diagram, a voltmeter ‘V’ is connected across a lamp ‘L’. How would (a) the
brightness of the lamp and (b) voltmeter reading ‘V’ be affected, if the value of resistance ‘R’
is increased? Justify your answer.
V L
+
9V
–
R
55/1/1 21 P.T.O.

22. Ic
Ans. As R is increased, Ib decreases. So, Ic decreases as  = . With decrease in Ic, power dissipated
Ib
by bulb is less. So, voltmeter reading and brightness of the bulb, both decrease.
14. A convex lens of focal length 20 cm is placed coaxially in contact with a concave lens of focal
length 25 cm. Determine the power of the combination. Will the system be converging of
diverging in nature?
Ans. f1 = + 20; f2 = – 25 cm
P = P1 + P2
1 1 25  20 5
   
20 25 500 500
1 1
 cm1   100m 1  1D
100 100
The system will behave as a converging lens because its power is positive.
19. Using Bohr’s postulates, obtain the expression for (i) kinetic energy and (ii) potential energy
of the electron in stationary state of hydrogen atom.
Draw the energy level diagram showing how the transitions between energy levels result in
the appearance of Lyman series.
Ans. The basic postulates of Bohr’s theory are :
(a) The central core of every atom is called nucleus and it carries only positive charge and
almost the entire mass of the atom. Negatively charged electrons revolve around it with
a centripetal force which is provided by the electrostatic force of attraction between
electron and nucleus.
mv2
F=
r
where m and v are mass and velocity of electron moving in orbit of radius r. .....(1)
(b) Electron revolve in discrete non-radiating orbits (stationary orbits) with angular
h
momentum equal to integral multiple of (h is Planck’s constant)
2
nh
i.e., mvr = .....(2)
2
where mvr = angular momentum of e–
(c) Radiation of energy takes place when electron jumps from one orbit to another and the
amount of energy radiated is equal to the difference in energies of the two permitted
orbits i.e.,
hv = E2 – E1 .....(3)
where v = frequency of radiation emitted
nh nh
We know, mvr = (from eqn 2)  v = ......(4)
2 2mr
Also, electrostatic force between electron (e–) and nucleus of charge (+Ze) is
1 |Ze||e|
F= ......(5)
40 r2
mv2 1 Ze2
From (1) and (5),  ......(6)
r 40 r2
55/1/2 22 P.T.O.

23. 1 Ze2
or r=
40 mv2
Substitute from (4),
1 Ze2 42m2r2
r= 
40 m n2h2
(40 )n2h2
r=
42mZe2
0 n2h2
 r=
 me2 .Z
For H, Z = 1
(40 ) n2h2
 r= ......(7)
42 me2
 r  n2
Since energy of moving electron is its kinetic energy, hence from (6),
mv2 1 Ze2

r 40 r2
1 Ze2
 mv2 =
40 r2
1 1  1 Ze2 
K. E. = mv2    ......(8)
2 2  40 r 
 
Potential energy = Potential × Charge
 1 Ze 
Since potential of electron is due to nucleus  i.e., 
 40 r 
1 Ze 1 Ze2
 P.E. =  (e) = n=5 .....(9)
40 r 40 r
n=4
 Total energy of electron = KE + PE
n=3
For H-atom, substitute standard values,
n=2
13.6 n=1
We get, E = eV
n2
1 1 
When electron jumps from outer to first orbit then 1  R  2  2  where k = 2, 3, 4,
1 k 
13.6
5....... and energy of electron is E1 = = –13.6 eV. This is Lymann Series.
12
22. Figure shows a rectangular loop conducting PQRS in which the arm PQ is free to move. A
uniform magnetic field acts in the direction perpendicular to the plane of the loop. Arm PQ is
moved with a velocity v towards the arm RS. Assuming that the arms QR, RS and SP have
negligible resistances and the moving arm PQ has the resistance r, obtain the expression for
(i) the current in the loop (ii) the force and (iii) the power required to move the arm PQ.
55/1/2 23 P.T.O.

24. × × × × × × × × ×
× × S× × × × × × ×
× × × × × × × × P×
× × × × × × × × ×
× × × × ×v × × × ×
× × × × × × × × Q×
× × R× × × × × × ×
× × × × × × × × ×
× × × × × × × × ×
Ans. As the arm PQ = l is moved to the left as shown with a velocity v, the flux linked will decreased
d B
as B = Blx = Blvt. Therefore, the emf induced will be e =   Blv
dt
e
(a) The current flowing in the loop will be I =
r
e
(b) The force experienced will be F = I(l × B) = Bl
r
e2 e 2 B 2l 2v 2
(c) The power required to move the arm will be I2r = r  
r2 r r
23. Distinguish between ‘sky waves’ and ‘space waves’ modes of propagation in communication
system.
(a) Why is sky wave mode propagation restricted to frequencies upto 40 MHz?
(b) Give two examples where space wave mode of propagation is used.
Ans. Sky Wave Propagation
• Waves after being emitted are reflected by the atmosphere to reach the receiver.
• Ionosphere reflects waves within the region (1710 kHz to 40 Mhz).
• The phenomenon of bending of em waves so that they are diverted towards the earth is
similar to total internal reflection in optics.
Space wave propagation (LOS communication)
• Space waves (54Mhz – 4.2 Ghz) travel in a straight line.
• Hence, the transmitter and receiver must be on the line of sight (LOS) together.
(a) 40 Mhz is critical frequency. Beyond this frequency, rays penetrate and escape the
ionosphere.
(b) Space waves are used for television broadcast and satellite communication.
×·×·×·×·×
55/1/2 24 P.T.O.

25. Studymate Solutions to CBSE Board Examination 2012-2013
Series : SKS/1 Code No. 55/1/3
UNCOMMON QUESTION ONLY
6. A 5V battery of negligible internal resistance is connected across a 200 V battery and a
resistance of 39  as shown in figure. Find the value of the current.
5V
39 200V
Ans. Let, current in circuit = I
Effective E.M.F of circuit E = 200 – 5
= 195V
Total resistance of circuit,
R = 0 + 39 = 39
E 195
I==   5A
R 39
7. Which of the following substances are para-magnetic?
Bi, Al, Cu, Ca, Pb, Ni
Ans. Al, Ca
8. A heating element is marked 210 V, 630 W. Find the resistance of the element when connected
to a 210 V dc source.
Ans. V = 210 V, P = 630 W
R=?
V2
P=
R
(210)2
630 =
R
210  210
R=  70
630
R = 70 
9. An ammeter of resistance 06W can measure current upto 1.0 A. Calculate (i) The shunt
resistance required to enable the ammeter to measure current upto 5.0 A (ii) The combined
resistance of the ammeter and the shunt.
Ans. IA = 1.0A, I = 5.0 A,
RA = 0.6
 IA 
(i) rs =  I  I   R A
 A 
 1 
=    0.6
 5 1
55/1/1 25 P.T.O.

26. 1
=  0.6
4
rs = 0.15 
R A  rs 0.6  0.15
(ii) Reff = R  r  0.6  0.15
A s
0.6  0.15
Reff = = 0.12 
0.75
15. A convex lens of focal length 30 cm is placed coaxially in contact with a concave lens of focal
length 40 cm. Determine the power of combination. Will the system be converging or diverging
in nature?
Ans. f1 = + 30 cm, f2 = –40 cm
P = P1 + P2
1 1
= f  f
1 2
1 1 10 10
= 0.30  (0.40)  3  4
 4  3  10
P = 10    0.83D
 12  12
System will be converging in nature.
18. A capacitor of unknown capacitance is connected across a battery of V volts. The charge
stored in it is 300 µC. When potential across the capacitor is reduced by 100 V, the charge
stored in it becomes 100 µC. Calculate the potential V and the unknown capacitance. What
will be the charge stored in the capacitor if the voltage applied had increased by 100V?
OR
A hollow cylindrical box of length 0.5 m and area of cross-section 20 cm2 is placed in a three
dimensional coordinate system as shown in the figure. The electric field in the region is

ˆ
given by E  20x i , where E is NC–1 and x is in metres. Find.
(i) Net flux through the cylinder.
(ii) Charge enclosed in the cylinder.
Y
O X
0.5 m
Z
Ans. C = ?, battery voltage, V1 = V volts
Q1 = 300 µC, Q2 = 100 µC, V2 = (V – 100)
Q1 = CV1
300 × 10–6 = C × V ...(i)
Q2 = CV2
100 × 10–6 = C × (V – 100) ...(ii)
55/1/3 26 P.T.O.

27. Eqn. (ii) ÷ Eqn. (i)
1 V  100

3 V
V = 3V – 300
2V = 300
V = 150 volts
From (i)
300 × 10–6 = C × 150
C = 2 µF
V 3 = V + 100 = 250 volts
Q3 = CV3 = 2 × 250 = 500 µC
V = 150 volts, C = 2µF and Q3 = 500 µC
OR
L = 0.5 m (box length)
Y
A = 20 cm2 = 20 × 10–4 m2

ˆ
E  20x i II
(i) Net flux through the cylinder, O I III
X
0.5 m
 = 1 + 2 + 3
Z
= E1. A1 cos180° + E2A2 cos90° + E3A3 cos0°
= (20 × 0.5) × (20 × 10–4) (–1) + 0 + (20 × 1.0) × (20 × 10–4) × (1)
= –2 × 10–2 + 4 × 10–2
= 2 × 10–2
= 0.02 V-m
1
(ii) By Gauss theorem,   Qen
0
1
0.02 = 8.85  10 12  Qen
Qen = 1.772 × 10–13 C
21. (a) In a nuclear reaction
3 3 4 1 1
2 He 2He  2He  1H  1H  12.86 MeV, though the number of nucleons is conserved on
both sides of the reaction, yet the energy is released. How? Explain.
(b) Draw a plot of potential energy between a pair of nucleons as a function of their
separation. Mark the regions where potential energy is (i) positive and (ii) negative.
Ans. (a) 2 He 3  2He 3  2 He 4  1H1  1H1  12.86 MeV
though the number of nucleons is conserved on both sides of the reaction, but total
mass of products is less then total mass of reactants. Energy is released due to this
mass defect.
E = m × C2
55/1/1 27 P.T.O.

28. Y
+P.E
Separation
(b) X
–P.E
r0
25. (a) Using Bohr’s postulate, obtain the expression for total energy of the electron in the nth
orbit of hydrogen atom.
(b) What is the significance of negative sign in the expression for the energy?
(c) Draw the energy level diagram showing how the line spectra corresponding to Paschen
series occur due to transition between energy levels.
Ans. (a) See solution question no. 23 set-1
(b) Negative sign of total energy signifies that if equal amount of energy is provided to
revolving electron it will move out of atom. Hence it denotes binding energy of electron
in the orbit.
n = 6
(c) n = 5
n = 4
n = 3
n = 2
n = 1
×·×·×·×·×
55/1/3 28 P.T.O.