PHYSICS 2004 (Delhi Region) PYSP 2004 / Class XII

PYSP 2004 / Class XII
1
PHYSICS 2004
(Delhi Region)
Time Allowed: 3 Hours
M. Marks: 70
 General Instructions:
(a)
All questions are compulsory.
(b)
There are 30 questions in total. Question 1 to 8 carry one mark each, questions 9 to 18 carry two marks
each, questions 19 to 27 carry three marks each and questions 28 to 30 carry five marks each.
(c)
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 give marks. You have to attempt only one of the given
choices in such questions.
(d)
Use of calculators are not permitted.
(e)
You may us the following physical constraints whenever necessary.
8
-1
c
=
3 10 ms
-34
h
=
6.6 10 Js
-19
e
=
1.6 10 C
-7
-1
=
4
10 TmA
o
-12
2
-1
-2
=
8.854 10 C N m
o
1
9
2
-2
=
9 10 N m C
4 o
-31
me
=
9.110 10 kg
-19
1 eV
=
1.6 10 J
-27
1 atomic mass unit (u)
=
1.661 10 kg
1 atomic mass unit (u)
=
931.5 MeV
-27
Mass of neutron mn
=
1.675 10 kg
-23
-1
Boltzmann's constant K
=
1.38 10 J K
23
Avogadro's number N
=
6.023 10 /mole
1.
In the diagram shown below is a circular loop carrying current I. Show the direction of the
magnetic field inside the loop with the help of lines of force.
I
Ans.
The magnetic liens of force of a circular loop carrying current I are shown below:
2.
Ans.
Name the type of modulation scheme preferred for digital communication.
Pulse Code Modulation
3.
Write the nuclear decay process for -decay of
Ans.
4.
32
15 P
32
16 S
32
15 P
0
1e
What is the name given to that part of electromagnetic spectrum which is used for taking
photographs of earth under foggy conditions from great heights?
Assessments by Vriti
PYSP 2004 / Class XII
Ans.
5.
Ans.
6.
2
Infrared rays.
'Heavy water is often used as a moderator in thermal nuclear reactions'. Give reason.
Heavy water is used as a moderator because its mass is nearest to that of a neutron and it has
negligible chances for neutron absorption.
The graph shows the variation of voltage, V across the plates of two capacitors A and B versus
increase of charge, Q stored on them. Which of the two capacitors has higher capacitance? Give
reason for your answer.
V
B
A
Q
Ans.
CA
Q
VA
CB
Q
VB
But VA < VB
CA > C B
so capacitor A has a higher capacitance.
o
7.
A room temperature (27.0 C) the resistance of a heating element is 100 . What is the
temperature of the element if the resistance is found to be 117 , given that the temperature
4 o 1
coefficient of the material of the resistor is 1.70 10 C .
Ans.
Here R27 = 100
= 1.70
10
, Rt = 117
4 o
C ,t=?
Rt = R27[1 + (t
27)]
117 = 101[1 + 1.70
4
10 (t
14
117 = 101 + 17.17
10
117 = 101 = 17.17
10 (t
16 = 17.17
t
27 =
.
1
4
10 (t
16
17.17 10
(t
4
27)]
27)
27)
27)
4
t 27 = 1000
o
t = 1027 C
Assessments by Vriti
PYSP 2004 / Class XII
3
8.
A bulb and a capacitor are connected in series to an a.c. source of variable frequency. How will
the brightness of the bulb change on increasing the frequency of the a.c. source? Give reason.
Ans.
 XC
1
C
1
2
C
1
XC
Therefore, brightness of the bulb will increase.
9.
Explain with the help of diagram the terms
(i) magnetic declination and
(ii) angle of dip at a given place.
Ans.
(i) Magnetic inclination ( ): The angle between the geographic meridian and magnetic meridian
at a place is called magnetic declination at that place.
vertical
declination
Geographic N-S
Horizontal line
BH
Geographical
meridian
dip
BE
BV
Magnetic
meridian

(ii) Angle of dip ( ): The angle made by the earth’s total magnetic field B with the horizontal is
called angle of dip at any place.
10.
Two circular coils, one of radius r and the other of radius R are placed co-axially with their
centers coinciding. For R >> r, obtain an expression for the mutual inductance of the
arrangement.
Ans.
Suppose a current I2 flows through the outer circular coil. Magnetic field at the centre of the coil is
oI2
2R
B2
Field B2 may be considered constant over the cross sectional area of the inner smaller coil.
Hence
r 2B2
1
M
11.
Ans.
r 2I2
2R
o
1
I2
MI2
r2
2R
o
Find the wavelength of electromagnetic wave of frequency 5
applications.
=
c
3 108
m = 0.6
5 1019
-12
10
m = 0.006 Å
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19
10
Hz in free space. Give its two
PYSP 2004 / Class XII
4
This wavelength corresponds to X rays which are used
(i) as a diagnostic tool
(ii) as a treatment for certain forms of cancer.
12.
Draw the graph showing the variation of binding energy per nucleon with mass number. Give the
reason for the decrease of binding energy per nucleon for nuclei with high mass numbers.
Ans.
The graph showing the variation of binding energy per nucleon with mass number.
9.0
Fe56
Binding Energy
per nucleon (MeV )
O16
C12
4
He
7.0
N14
5.0
U238
Li7
3.0
1.0
H2
0
40
80
120
160
200
240
Mass number (A)
13.
Ans.
The binding energy per nucleon for nuclei with high mass numbers is low due to the large
coulomb repulsion between the protons inside these nuclei.
With the help of diagram, show the biasing of a light emitting diode (LED). Give its two
advantages over conventional incandescent lamps.
The diagram is shown below:
n
n
p
Metallised
contact
14.
Ans.
The following are the two advantages over conventional incandescent lamps:
(1) Low operational voltage and less power.
(2) Fast action and no warm up time required.
A compound microscope with an objective of 1.0 cm focal length and an eye-piece of 2.0 cm
focal length has a tube length of 20 cm. Calculate the magnifying power of the microscope, if the
final image is formed at the near point of the eye.
m
mome
L
fo
D
fe
20
1
25
2
250
OR
Assessments by Vriti
PYSP 2004 / Class XII
5
The magnifying power of an astronomical telescope in the normal adjustment position is 100. The
distance between the objective and the eye-piece is 101 cm. Calculate the focal lengths of the
objective and of the eye-piece.
Ans.
Here, M =
fo
= 100
fe
and
fo = 100fe
But fo + fe = 101 or 100fe + fe = 101
15.
fe = 1 cm and fo = 100 cm.
State Gauss's law in electrostatics. Using this theorem, derive the expression for the electric field
intensity at any point outside a uniformly charged thin spherical shell.
Ans.
According to Gauss’ theorem, the electric flux through a closed surface is
1
times the charge q
o
enclosed by the closed surface.
 
q
 E.ds
E
o
S
Consider a Gaussian surface of radius r (>R) outside the spherical shell. By Gauss’s theorem,
 
 E.ds  Eds
E
E
16.
1
4
o
q
E.4 r 2
o
q
r2
Find the total energy stored in the capacitors in the given network.
2 F
6V
2 F
1 F
1 F
2 F
Ans.
2 F and 2 F capacitors are in series, their equivalent capacitance =
2 2
2 2
1 F
Now 1 F and 1 F capacitors are in parallel, so their equivalent capacitance
=1+1=2 F
Then we have 2 F and 2 F capacitors in series, their equivalent capacitance
=
2 2
2 2
1 F
Assessments by Vriti
PYSP 2004 / Class XII
6
Finally, we have 1 F capacitance in parallel with 1 F capacitance, their equivalent capacitance.
C=1+1=2 F=2
10
6
F
Total energy stored =
-6
17.
Ans.
18.
1
CV 2
2
1
(2
2
-6
10 )(6)
2
-5
= 36 10 = 3.6 10 J
What are eddy currents? Discuss briefly any one application of eddy current.
Eddy current: The current induced in the body of a conductor when the amount of magnetic flux
linked with the conductor changes, if called eddy current.
The magnitude of eddy current is
I
Induced emf
Re sis tan ce
I
e
R
d / dt
R
The direction of eddy current is given by Lenz’s law. The eddy current is also called Foucault
current.
Application: Electromagnetic brakes in an example of eddy current. A strong magnetic field is
applied to a metallic drum rotating with the axle connecting the wheels. The large eddy current is
set up in the drum which oppose the motion of the drum. Thus the train stops.
The diagram shows a piece of pure semiconductor, S in series with a variable resistor R, and a
source of constant voltage V. Would you increase or decrease the value of R to keep the reading
of ammeter (A) constant, when semiconductor S is heated? Given reason.
V
S
A
Ans.
19.
Ans.
When the pure semiconductor S is heated, its resistance decreases. So the current in the circuit
increases. To keep the reading of ammeter (A) constant, the value of R has to be increased.
State two conditions to obtain sustained interference of light.
In Young's double slit experiment, using light of wavelength 400 nm, interference fringes of width
X are obtained. The wavelength of light is increased to 600 nm and the separation between the
slits is halved. If one wants the observed fringe width on the screen to the same in the two cases,
find the ratio of the distance between the screen and the plane of the interfering sources in the
two arrangements.
The conditions for obtaining sustained interference of light are
(i) The two light sources should be coherent.
(ii) The two light sources should be narrow and placed close to each other.
Fringe width in first case,
D
d
D 400
d
Fringe width in second case is same.
D' 600
d/ 2
D 400
d
D' 1200
d
D' 1200
d
Assessments by Vriti
PYSP 2004 / Class XII
D
D'
or
20.
7
3
1
3 :1
Give the mass number and atomic number of elements on the right-hand side of the decay
process
220
86 Ru
Po He
The graph shows how the activity of a sample of Radon - 220 changes with time. Use the graph
to determine its half-life. Calculate the value of decay constant of Radon-220.
Ans.
220
86 Ru
216
84 Po
4
2He
Mass number of Po = 216
Atomic number of Po = 84
Mass number of He = 4
Atomic number of He = 2
Data are not provided with the given graph.
21.
Ans.
A circular coil of N turns and radius R, is kept normal to a magnetic field, given by B = B o cos t.
Deduce an expression for emf induced in this coil. State the rule which helps to detect the
direction of induced current.
Flux linked with the N turns of the coil
2
= NBA = N Bo cos t
Induced emf is given by
E
d
dt
2
R = N R Bo cos t
d
cos t
dt
N R2Bo
N R2Bo sin t
The direction of induced current can be determined by Lenz’s law which states that the direction
of induced current is such that it opposes the cause which produces it.
22.
In a series R - C circuit, R = 30 , C = 0.25 F, V = 100 V and = 10,000 radian per second.
Find the current in the circuit and calculate the voltage across the resistor and the capacitor.
Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the
paradox.
Ans.
Here R = 30
Z
1
C2
R2
VRrms
Vrms
Z
IrmsR
10 F, Vrms = 100 V,
0.25 10
100
401.1
= 10,000 rad/s
1
302
2
900 160000
Irms
6
, C = 0.25
160900
6
10000
2
401.1
0.25 A
0.25 30
7.5 V
Assessments by Vriti
PYSP 2004 / Class XII
VCrms
8
Irms
c
Irms XC
0.25
10000 0.25 10
23.
Ans.
Yes, the algebraic sum of voltages across R and C is more than the source voltage. This is
o
because the voltage across C lags behind the voltage across R by 90 .
Define the term 'work function' of a metal. The threshold frequency of a metal is f o. When the light
of frequency 2fo is incident on the metal plate, the maximum velocity of electrons emitted is v 1.
When the frequency of the incident radiation is increased to 5f o, the maximum velocity of
electrons emitted as v2. Find the ratio of v1 to v2.
The minimum amount of energy required to eject an electron from a metal surface (without
imparting it any kinetic energy) is called work function of the metal. As f o is the threshold
frequency, so
W = hfo
From Einstein’s photoelectric equation, the maximum kinetic energy of emitted electron is given
by
1
mv12
2
and
h 2fo
1
mv 22
2
or
24.
Ans.
100V
6
W
h 2fo hfo
h 5fo hfo
1
mv12
2
1
mv 22
2
hfo
4hfo
v1
v2
1: 2
1
2
hfo
4hfo
Draw the circuit diagram of a common-emitter amplifier using n-p-n transistor. Draw the input and
output wave forms of the signal. Write the expression for its voltage gain.
The circuit diagram of a common-emitter amplifier using an n-p-n transistor is shown below:
C
C1
B
Vi
~
E
RB
Input
Output
RC
Vo
VCC
VBB
The voltage gain of a CE amplifier is given by
AV
25.
Ans.
Vo
Vi
ICRC
IBRB
RC
RB
where is the current amplification factor of the transistor.
Write the symbol and truth table of an AND gate. Explain how this gate is realized in practice by
using two diodes.
The symbol of an AND gate is
A
Y
B
Assessments by Vriti
PYSP 2004 / Class XII
9
when A is anded with B, we write Y = A. B.
The truth table is
A
B
Y
0
0
0
0
1
0
1
0
0
1
1
1
Realization of AND gate: For two input AND gate, we use to two diodes D 1, D2 as shown in
figure. The resistor R is connected to positive terminal of a 5 V battery.
26.
Ans.
Operation : (i) When both A and B are grounded, both D1 an D2 get forward biased and conduct.
Output Y will be 0 since it neutralized by 5 V battery sending the current in opposite direction.
(ii) When A is earthed (0) and B is connected to 5 V battery (1), D1 conducts and D2 does not
conduct. The output Y = 0.
(iii) When A is connected to 5 B (1) and B grounded (0), D 2 conducts and D1, does not, output Y
= 0.
(iv) when A and B are both at high (1) level, none of the diodes conduct. The output is equal to
the battery voltage of 5 V. i.e. Y = 1.
With the help of a block diagram, explain the principle of an optical fiber communication system.
Give its two advantages over cable communication system.
Block diagram of an optical communication system is given below:
Transmitter
Input signal
Optical Modulation
Source
Receiver
Demodulation
Optical
Detector
Output
Signal
Principle: The input analog signal is first sampled and modulated using PCM which produces
discrete pulses and gives coded steam, of but (0’s and 1’s) representing the signal,
corresponding to this stream of bits, the optical beam from a source is modulated. This
modulated signal is propagated in medium called optical fibre. At the receiver, an optical detector
detects the signal which in turn, is converted back to electrical signal. This is decoded and
original analog signal is reconstructed.
Advantages of optical communication system over cable communication system are
(i) Wide channel bandwidth and large channel carrying capacity.
(ii) Law transmission loss.
OR
With the help of relevant diagrams, explain the following terms:
Assessments by Vriti
PYSP 2004 / Class XII
Ans.
10
(i) Pulse-position modulation (PPM)
(ii) Pulse-duration modulation (PDM)
(i) In pulse-position modulation, the modulating analog signal is sampled and the timing or
position of the pulse is varied around a fixed value conforming to the signal amplitude at the time
of sampling. The positive going signal sets the pulse ahead and the negative going signal sets
the pulse behind the reference time.
Modulating wave
PPM
(ii) In pulse-duration modulation, the modulating analog signal is sampled and the pulse length or
duration around a fixed value is varied.
Modulating wave
PDM
27.
Ans.
Name the device used for data transmission from one computer to another. Justify the name.
Using this device, draw the block diagram for data communication and explain it briefly.
The device used to interface two digital source is called modem. Both modulation and
Demodulation functions are included in a modem.
The lock diagram for a data communication circuit using MODEMs is given below:
Modem 2
Modem 1
Modulator
Demodulator
Transmitter side
28.
Demodulator
Communication
Channel
Computer
Computer
Modulator
Receiver side
The modem at the transmitting station changes the digital output from a computer to a form
which can be easily set through a communication channel. At the receiving modem, the process
is reversed. In transmitting mode, the modem converts the digital to analog for use in modulating
a carrier signal. At the receiver computer the carrier is demodulated to recover the data.
Derive an expression for the torque acting on a loop of N turns, area A, carrying current i, when
held in a uniform magnetic field B.
With the help of circuit, show how a moving coil galvanometer can be converted into an ammeter
of given range. Write the necessary mathematical formula.
Assessments by Vriti
PYSP 2004 / Class XII
Ans.
11
(i) A rectangular loop efgh is placed in a magnetic field at a certain orientation with it. Using right

 
hand rule we see that there are forces F I l B
on the sides eh and fg, parallel to the axis of
rotation and cause no torque. But there are forces on the sides ef and gh which cause torque.
h
F
B
F
I
N
g
(h, g)
e
m
B
I
F
(e, f)
f
F
(i)
Note: When B field is exactly perpendicular to the plane of the loop there is no turning effect.
Let b be the length of the vertical side and I be the current. These forces are F = BIb each. Let a
be the length of fg and eh. Then the torque as shown in figure (ii) is
Fasin
where
is the smaller angle between B and perpendicular to the area of the loop.
= BIbasin = BIabsin = BIAsin
where A is the area of the loop equal to ab .



The magnetic moment of the loops is m IA , where A is the vector along the outside normal
according to the right hand rule and, therefore, = mB sin .
(ii)
I
Ig
G
I - Ig
S
A galvanometer can measure very small currents of the order of A or so. But when larger
current are desired to be measured, the galvanometer is shunted with a resistance S. The
resistance S allows the current to be by-passed. Suppose, the ammeter range is 0
I, then the
current Ig which is the safe current for full scale deflection passes through the galvanometer
resistance G and (I - Ig) passes through S.
According to Ohm's law
I Ig S
S
Ans.
Ig
I Ig
IgG
G
OR
Write an expression for the force experienced by a charged particle moving in a uniform
magnetic field B. With the help of a diagram, explain the principle and working of a cyclotron.
Show that cyclotron frequency does not depend on the speed of the particle.
A cyclotron is a device developed by Lawrence and Livingstone by which the positively
charged particles likes proton, deutron etc. can be accelerated.
Principle
Assessments by Vriti
PYSP 2004 / Class XII
12
The working of the cyclotron is based on the fact that a positive charged particle can be
accelerated to a sufficiently high energy with the help of smaller values of oscillating electric field.
Construction
It consists of two D-shaped hollow evacuated metal chambers D1 and D2 called the dees. These
dees are placed horizontally with their diametric edges parallel and slightly separated from each
other. The dees are connected to high frequency oscillator which can produce a potential
4
7
difference of the order of 10 volts at frequency 10 Hz. The two dees are enclosed in an
evacuated steel box and are well insulated from it. The box is placed in a strong magnetic field
produced by two pole pieces of strong electromagnets N, S. The magnetic field is perpendicular
to the plane of the dees. P is a place of ionic source or positively charged particle.
N
H.F. Oscillator
~
P
D2
E
W
D1
Target
S
Working and Theory
The positive ion to be accelerated is produced at P. Suppose, at that instant, D1 is at negative
potential and D2 is at positive potential. Therefore, the ion will be accelerated towards D 1. On
reaching inside D1, the ion will be in a field free space. Hence it moves with a constant speed in
D1 say v. But due to perpendicular magnetic field of strength B, the ion will describe a circular
path of radius r(say) in D1, given by
mv 2
r
Bqv
where m and q are the mass and charge of the ion.
r
mv
Bq
Time taken by ion to describe a semicircular path is given by
t
r
v
m
Bq
B q/m
= constant
Thus this time is independent of both the speed of the ion and radius of the circular path. In case
the time during which the positive ion describes a semicircular path is equal to the time during
which half cycle of electric oscillator is completed, then as the ion arrives in the gap between the
two dees, the polarity of the dees is reversed i.e. D 1 becomes positive and D2 negative.
Therefore, the positive ion is accelerated towards D2 and it enters D2 with greater speed which
remains constant in D2. The ion will describe a semicircular path of greater radius due to
perpendicular magnetic field and again will arrive in a gap between the two dees exactly at the
instant, the polarity of the two dees is reversed. Thus, the positive ion will go on accelerating
Assessments by Vriti
PYSP 2004 / Class XII
13
every time it comes into the gap between the dees and will go on describing circular path of
greater and greater radius and finally acquires a sufficiently high energy. The accelerated ion can
be removed out of the dees from window W, by applying the electric field across the deflecting
plates E and F.
Let vo and ro be the maximum velocity and the maximum radius of the circular path followed by
the position ion in cyclotron
Then
mv o2
ro
or
Bqv o
1
mv o2
2
Max. KE =
vo
Bqro
1
m
2
m
Bqro
m
2
B2q2ro2
2m
If T is the time period of oscillating electric field then,
T
2 m
Bq
2t
The cyclotron frequency is given by
1
t
v
30.
It is also known as magnetic resonance frequency.
With the help of a ray diagram, show the transformation of image of a point object by refraction of
light at a spherical surface separating two media of refractive indices n 1 and n2 (n2 > n1)
respectively. Using this diagram, derive the relation
n2
v
Ans.
Bq
2 m
n1
u
n2 n1
R
Write the sign conventions used. What happens to the focal length of convex lens when it is
immersed in water?
AMB is a convex surface separating two media of refractive indices n1 and n2 (n2 > n1). Consider
a point object O placed on the principal axis. A ray ON is incident at N and refracts along NI. The
ray along ON goes straight and meets the previous ray at I. Thus I is the real image of O.
From Snell’s law, n1 sin i = n2 sin r
or
[  sin
n1i = n2r
From NOC, i =
=
as
is very small]
+
From NIC, = r
or r =
n1( + ) = n2(
)
or n1 + n2 = (n2
But
tan
tan
=
=
NP
PI
NP
OP
n1)
NP
OM
[P is close to M]
NP
MI
Assessments by Vriti
PYSP 2004 / Class XII
tan =
n1
NP
PC
NP
MC
NP
NP
n2
OM
MI
n1
OM
or
14
n2
MI
n2 n1
NP
MC
n2 n1
MC
Using Cartesian sign convention,
OM
30.
Ans.
u, MI
v,MC
R
n1
u
n2
v
n2 n1
R
n2
v
n1
u
n2 n1
This is the required relation.
R
OR
Deduce the condition for balance in a Wheatstone Bridge. Using the principle of Wheatstone
Bridge describe the method to determine the specific resistance of a wire in the laboratory. Draw
the circuit diagram and write the formula used.
Write any two important precautions you would observe while performing the experiment.
Wheatstone Bridge is an arrangement of four resistances P, Q, R and S joined together to form a
loop like the four arms of a quadrilateral as shown. It is said to be balanced when P/Q = R/S. In
balanced condition, a cell joined between A and B sends a current i 1 in the branch P and i2 in the
branch R, the same currents i1 flows through Q and i2 through S. There is no current through CD
and hence VC = VD.
C
i1
Q
P
i1
A
B
i2
i2
R
S
i
D
Applying Kirchhoff's law to the loop ACD,
i1P i2R
0
i2R
(1)
i1Q i2S
(2)
i1P
and to the loop CBD,
i1Q i2S
0
Dividing with equations, we get P/Q = R/S.
The figure shows a meter bridge which consists of four resistances (a) P which is unknown (b) Q
which is a resistance known from a resistance box (c) R which is the resistance of the length l of
the meter bridge wire AB (d) S which is the resistance of the length (100 - l) of the meter bridge
wire AB. The wire AB is a uniform resistance wire and 100 cm long. When the potentials V C = VD
there is no current in the galvanometer.
Assessments by Vriti
PYSP 2004 / Class XII
15
P
R
A
C
G
D
Q
S
B
This is a null condition. Point D is called the null point. According to the principle of meter bridge
PQ = R/S =
xl
, where x is the resistance per unit length of the wire.
x 100 l
The unknown resistance, P = Q.
P
Q.
xl
x 100 l
l
100 l
For determination of specific resistance , the first step is to find the resistance. Then use the
formula resistance =
l
. The length l, be measured by a meter scale and area A be measured
A
by the micrometer screw gauge.
Precautions: (i) The resistance Q from the resistance box should be chosen so that the balance
point D comes near to the center of the wire.
(ii) Since the galvanometer G is a sensitive meter, a high protective resistor should be joined in
series with it until a near balance is reached on the wire. When the near balance is reached this
high resistor is shunted or removed and final balance point is found.
Assessments by Vriti