Download XII Physics CBSE Sample Paper 2015

ELITE EDUCARE (SIROHI CLASSES)
CLASS XII (2014-15)
(THEORY)
One Paper Time: 3 hrs.
Max Marks: 70
UNITS
Unit I Electrostatics
MARKS
15
Unit II Current Electricity
Unit III Magnetic Effect of Current and Magnetism
16
Unit IV Electromagnetic Induction and Alternating Current
Unit V Electromagnetic Waves
17
Unit VI Optics
Unit VII Dual Nature of Matter
10
Unit VIII Atoms and Nuclei
Unit IX Electronic Devices
12
Unit X Communication Systems
Total
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70
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Sample Question Paper Class XII Physics
(Applicable for March 2015 Examination)
Time Allowed: 3 Hours
Maximum Marks: 70
General Instructions
1.
All questions are compulsory. There are 26 questions in all.
2.
This question paper has five sections: Section A, Section B, Section C, Section D and
Section E.
3.
Section A contains five questions of one mark each, Section B contains five questions of two
marks each, Section C contains twelve questions of three marks each, Section D contains one
value based question of four marks and Section E contains three questions of five marks
each.
4.
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 the three questions of five marks weightage.
You have to attempt only one of the choices in such questions.
5.
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
µo = 4 ×10–7 T m A–1
0
= 8.854 × 10–12 C2 N–1 m–2
1
4 0
 9  109 N m2 C–2
me = 9.1  10–31 kg
mass of neutron = 1.675 × 10–27 kg
mass of proton = 1.673 × 10–27 kg
Avogadro Numbers = 6.023 ×1023 per gram mole
Boltzmann constant = 1.38 × 10–23 JK–1
Section A
( 5×1= 5 Marks)
1) What is the value of the angle between the vectors p and E for which the potential energy of an
electric dipole of dipole moment p , kept in an external electric field E , has maximum value.
2) Name the colours corresponding to the digits 4 and 7 in the colour code scheme for carbon
resistors.
3) State which of the two, the capacitor or an inductor, tends to become a SHORT when the
frequency of the applied alternating voltage has a very high value.
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4) Redraw the diagram given below and mark the position of the centre of curvature of the spherical
mirror used in the given set up.
5) In the given diagram C(t) stands for the carrier wave and m(t) for the signal to be transmitted.
What name do we give to the wave labeled as Cm(t) in the diagram?
Section B
(5×2= 10 Marks)
6) Calculate the value of the unknown potential V for the given potentiometer circuit. The total
length (400 cm) of the potentiometer wire has a resistance 10Ω and the balance point is obtained
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at a length of 240 cm.
7) Name the phenomenon which proves transverse wave nature of light. Give two uses of the
devices whose functioning is based on this phenomenon.
OR
Name the phenomenon which is responsible for bending of light around sharp corners of an obstacle.
Under what conditions does this phenomenon take place? Give one application of this phenomenon in
everyday life.
8) The equivalent wavelength of a moving electron has the same value as that of a photon having an
energy of 6 × 10–17 J.
Calculate the momentum of the electron.
9) The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4 Å. Calculate
the short wavelength limit for Balmer series of hydrogen spectrum.
10) (a) Arrange the following networks in increasing order of the number of computers that may be
present in the network:
Internet ; LAN ; WAN
(b) What is the minimum number of satellites that enables a Global Positioning System (GPS)
receiver to determine one’s longitude/latitude position, i.e., to make a 2D position fix. 1+1=2
Section C
(12×3 = 36 Marks)
11) Eight identical spherical drops, each carrying a charge 1 nC are at a potential of 900 V each. All
these drops combine together to form a single large drop. Calculate the potential of this large
drop. (Assume no wastage of any kind and take the capacitance of a sphere of radius r as
proportional to r).
12) The current flowing in the galvanometer G when the key k2 is kept open is I. On closing the key
k2 , the current in the galvanometer becomes
I
, where n is an integer.
n
Obtain an expression for resistance Rg of the galvanometer in terms of R, S and n. To what form does
this expression reduce when the value of R is very large as compared to S?
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13) The magnitude F of the force between two straight parallel current carrying conductors kept at a
distance d apart in air is given by
F
0 I1I 2
2 d
Where I1 and I2 are the currents flowing through the two wires Use this expression, and the sign
convention that the:
“Force of attraction is assigned a negative sign and Force of repulsion is assigned a positive sign.”
Draw graphs showing dependence of F on
(i)
1 2
when d is kept constant
(ii) d when the product
(iii) d when the product
1 2
is maintained at a constant positive value.
1 2
is maintained at a constant negative value.
OR
The given graphs show the variation of intensity of magnetization I with strength of applied magnetic
field H for two magnetic materials P and Q.
(i)
Identify the materials P and Q.
(ii) For material P, plot the variation of Intensity of Magnetisation with temperature. Justify your
answer.
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14) Find the value of the phase lag/lead between the current and voltage in the given series LCR
circuit. Without making any other change, find the value of the additional capacitor, such that
when ‘suitably joined’ to the capacitor ( C= 2 µF) as shown, would make the power factor of this
circuit unity.
15) Explain how one ‘observes an inconsistency’ when Ampere’s circuital law is applied to the
process of charging a capacitor.
How this ‘contradiction’ gets removed by introducing the concept of an ‘additional current’, known as
the ‘displacement current’?
16) A point object O is kept at a distance of 30 cm from a convex lens of power +4D towards its left.
It is observed that when a convex mirror is kept on the right side at a distance of 50 cm from the
convex lens, the image of the object O formed by the lens-mirror combination coincides with the
object itself. Calculate the focal length of the convex mirror.
17) The arrangement used by Thomas Young to produce an interference pattern is shown in the given
diagram.
Justify why there would be no change in the ‘fringe width’ when the main illuminated slit (S) is
shifted to the position ′ as shown.
18) A given number of atoms No of a radioactive element with a half life T is uniformly distributed in
the blood stream of a
(i)
normal person A having total volume V of blood in the body
(ii)
person B in need of blood transfusion having a volume ′ of blood in the body.
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The number of radioactive atoms per unit volume in the blood streams of the two persons after a time
nT are found to be N1 and N2.
Prove mathematically that the additional volume of blood that needs to be transfused in the body of
 N 2  N1 
V
 N2 
person B equals 
19) A student has to use an appropriate number of
(i)
NAND gates (only) to get the output Y1
(ii)
NOR gates (only) to get the output Y2
From two given inputs A and B as shown in the diagram.
Identify the ‘equivalent gate’ needed in each case. Show how one can connect an appropriate number
of (i) NAND (ii) NOR gates respectively in the two cases to get these ‘equivalent gates’.
20) The data given below gives the photon energy (in eV) for a number of waves whose wavelength
values (in nm) are also given.
Wavelength (in nm)
200
400
600
800
1000
1200
Photon Energy (in eV)
6.216
3.108
2.072
1.554
1.243
1.036
(Without doing any calculation/taking any reading), explain how one can use this data to draw an
appropriate graph to infer
(i)
photon energy corresponding to a wavelength of 100 nm.
(ii)
the wavelength value (in nm) corresponding to a photon energy of 1 eV.
(iii)
velocity of light assuming that the value of Plank’s constant is known.
21) A (sinusoidal) carrier wave
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C(t) = AC sin ωc t
is amplitude modulated by a (sinusoidal) message signal m(t) = Am sin ωm t
Write the equation of the (amplitude) modulated signal.
Use this equation to obtain the values of the frequencies of all the sinusoidal waves present in the
modulated signal.
22) Give reasons for the following:
(i)
The Zener diode is fabricated by heavily doping both the p and n sides of the junction
(ii)
A photodiode, when used as a detector of optical signals is operated under reverse
bias.
(iii)
The band gap of the semiconductor used for fabrication of visible LED’s must at least
be 1.8 eV.
Section D
( 1×4 = 4 Marks)
23) Dimpi’s class was shown a video on effects of magnetic field on a current carrying straight
conductor. She noticed that the force on the straight current carrying conductor becomes zero
when it is oriented parallel to the magnetic field and this force becomes maximum when it is
perpendicular to the field. She shared this interesting information with her grandfather in the
evening. The grandfather could immediately relate it to something similar in real life situations.
He explained it to Dimpi that similar things happen in real life too. When we align and orient our
thinking and actions in an adaptive and accommodating way, our lives become more peaceful and
happy. However, when we adopt an unaccommodating and stubborn attitude, life becomes
troubled and miserable. We should therefore always be careful in our response to different
situations in life and avoid unnecessary conflicts.
Answer the following question based on above information:
a) Express the force acting on a straight current carrying conductor kept in a magnetic field in vector
form. State the rule used to find the direction of this force.
b) Which one value is displayed and conveyed by grandfather as well as Dimpi?
c)
Mention one specific situation from your own life which reflects similar values shown by you
towards your elders.
2+1+1=4
Section E
(3×5 = 15 Marks)
24) a) State the theorem which relates total charge enclosed within a closed surface and the electric
flux passing through it. Prove it for a single point charge.
b) An ‘atom’ was earlier assumed to be a sphere of radius a having a positively charged point nucleus
of charge +Ze at its centre. This nucleus was believed to be surrounded by a uniform density of
negative charge that made the atom neutral as a whole.
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Use this theorem to find the electric field of this ‘atom’ at a distance r (r < a) from the centre of the
atom.
OR
A Dipole is made up of two charges + q and – q separated by a distance 2a.
Derive an expression for the electric field Ee due to this dipole at a point distant r from the centre of
the dipole on the equatorial plane.
Draw the shape of the graph, between | Ee | and r when r >> .
If this dipole were to be put in a uniform external electric field E , obtain an expression for the torque
acting on the dipole.
25) State the law which relates to generation of induced emf in a conductor being moved in a
magnetic field.
Apply this law to obtain an expression for the induced emf when one ‘rod’ of a rectangular conductor
is free to move in a uniform, time independent and ‘normal’ magnetic field.
Apply the concept of the Lorentz (magnetic) force acting on a moving charge to justify the expression
obtained above.
OR
An a.c. voltage V = Vm sin
t is applied across an inductor of inductance L. Apply Kirchoff’s loop
rule to obtain expressions for
(i) the current flowing in the circuit
(ii)
the inductive reactance L Hence find the instantaneous power Pi supplied to the inductor.
Show graphically the variation of Pi with
t.
26) (a) Explain, with the help of a diagram, how is the phenomenon of total internal reflection used in
(i)
an optical fibre
(ii)
a prism that inverts an image without changing its size
(b) A right angled prism made from a material of refractive index
is kept in air. A ray PQ is incident
normally on the side AB of the prism as shown.
Find (in terms of ) the maximum value of
upto which this incident ray necessarily undergoes total
internal reflection at the face AC of the prism.
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2+3=5
Page 9
OR
State Huygen’s principle in wave-optics. How did Huygen ‘explain’ the absence of the backwave?
Use this principle to draw the refracted wave front for a plane wave incident from a denser to a rarer
medium. Hence obtain Snell’s law of refraction.
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SOLUTIONS
Section A
( 5×1= 5 Marks)
1) What is the value of the angle between the vectors p and E for which the potential energy
of an electric dipole of dipole moment p , kept in an external electric field E , has maximum
value.
SOL: P.E. = –
.
= – p E cos
∴ P.E. is maximum when cos
= –1, i.e.
=
(1800)
.
Hence Potential energy P.E.= + p E ( dipole is held anti parallel to the field.)
2) Name the colours corresponding to the digits 4 and 7 in the colour code scheme for carbon
resistors.
SOL: 4 → yellow
,
7 → Violet
3) State which of the two, the capacitor or an inductor, tends to become a SHORT when the
frequency of the applied alternating voltage has a very high value.
SOL: For inductor X L  L.2 f and for capacitor X C 
1
2 fC
when f   X C  0
The capacitor tends to short.
4) Redraw the diagram given below and mark the position of the centre of curvature of the
spherical mirror used in the given set up.
SOL:
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Since image is inverted at the same side hence it is real image hence mirror is concave . since image is
enlarged hence object must be placed between focus and centre of curvature. Centre of curvature is
shown in fig.
5) In the given diagram C(t) stands for the carrier wave and m(t) for the signal to be
transmitted. What name do we give to the wave labeled as Cm(t) in the diagram?
SOL: C(t)= Instantaneous value of carrier ( to be modulated ) signal, m(t) = Instantaneous value of
audio or modulated signal.
C(t) + m(t) = Cm(t)
Cm (t) is the frequency modulated wave.
Section B
(5×2= 10 Marks)
6) Calculate the value of the unknown potential V for the given potentiometer circuit. The total
length (400 cm) of the potentiometer wire has a resistance 10Ω and the balance point is
obtained at a length of 240 cm.
SOL: The current through the potentiometer wire I =
SIROHI CLASSES
E0
3V

= 10–2A
R  R1  290  10  
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∴Potential drop per unit length of the potentiometer wire k =
VAB IR 102 A 10 1


 103V / cm
l
l
400cm
4
Balancing length ( x ) = 240 cm ( given )
V=Kx=
1
103  240  6 102V  60mV
4
7) Name the phenomenon which proves transverse wave nature of light. Give two uses of the
devices whose functioning is based on this phenomenon.
SOL: Polarization
Two Uses: Polaroids can be used in sunglasses, window panes, photographic cameras, 3D movie
cameras.
OR
Name the phenomenon which is responsible for bending of light around sharp corners of an
obstacle. Under what conditions does this phenomenon take place? Give one application of this
phenomenon in everyday life.
SOL: Diffraction; Condition: Size of the obstacle sharpness should be comparable to the wavelength
of the light falling.
Application
The diffraction grating is an important device that makes use of the diffraction of light to produce
spectra. Diffraction is also fundamental in other applications such as x-ray diffraction studies of
crystals and holography.
8) The equivalent wavelength of a moving electron has the same value as that of a photon
having an energy of 6 × 10–17 J.
Calculate the momentum of the electron.
SOL: E = Energy of the photon = h =
∴
=
hc

hc
E
∴Wave length of the moving electron =
∴ Momentum of the electron = p =
h


=
hc
E
hE E 6 1017
 
 2 1025 kgms 1
8
hc c
3 10
9) The short wavelength limit for the Lyman series of the hydrogen spectrum is 913.4 Å.
Calculate the short wavelength limit for Balmer series of hydrogen spectrum.
SOL: Rydberg formula for the wavelengths of spectral lines in hydrogen spectrum is
 1
1 
 R 2  2 
n


 f ni 
1
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The short wavelength limit
for the Lyman series would be
1 1 
 R 2  2   R .

1  
1
1
R
o
913.4 A
The short wavelength limit
B
for the Balmer series, would be
1
1  R
 R 2  2   .
B
2   4
1
B 
o
o
4
 4  913.4 A  3653.6 A
R
10) (a) Arrange the following networks in increasing order of the number of computers that
may be present in the network:
Internet ; LAN ; WAN
(b) What is the minimum number of satellites that enables a Global Positioning System (GPS)
receiver to determine one’s longitude/latitude position, i.e., to make a 2D position fix. 1+1=2
SOL: a) LAN, WAN, Internet
b) Three
Section C
(12×3 = 36 Marks)
11) Eight identical spherical drops, each carrying a charge 1 nC are at a potential of 900 V each.
All these drops combine together to form a single large drop. Calculate the potential of this
large drop. (Assume no wastage of any kind and take the capacitance of a sphere of radius r
as proportional to r).
SOL: Let the radius of each drop be r. The capacitance C of each drop is kr, where k is a constant.
Also q = CV, V = 900 volt
∴ charge on each drop = q = (kr × 900) C
∴ Total charge on all the eight drops = Q = 8 q = 7200 kr
Let R be the radius of the large drop. Then
4 3
4 3
R =8
r
3
3
∴ R = (8)1/3 r = 2r
∴ Capacitance ′ of the large drop = kR = 2kr
∴ Potential of the large drop =
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Q 7200kr

volt = 3600V
C'
2kr
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12) The current flowing in the galvanometer G when the key k2 is kept open is I. On closing the
key k2 , the current in the galvanometer becomes
I
, where n is an integer.
n
Obtain an expression for resistance Rg of the galvanometer in terms of R, S and n. To what
form does this expression reduce when the value of R is very large as compared to S?
SOL: With key K2 open, the current I in the galvanometer is given by I 
E
R  RG
When K2 is closed, the equivalent resistance, say R’, of the parallel combination of S and
by R ' 
is given
SRG
S  RG
The total current, say ′ drawn from the battery would now be I ' 
E
R  R'
This current gets subdivided in the inverse ratio of S and ; Hence the current ″ through G, would
now be given by
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Or n RS + n S
or (n–1) RS = R
or (n–1) RS +
∴
=
= RS + R
+S
– (n–1) S
[ — ( — 1) ]
(n  1) RS
R  (n  1) S
This is the required expression When R >>S , we have
R
(n  1) RS
 (n  1) .
R
13) The magnitude F of the force between two straight parallel current carrying conductors
kept at a distance d apart in air is given by
F
0 I1I 2
2 d
Where I1 and I2 are the currents flowing through the two wires Use this expression, and the
sign convention that the:
“Force of attraction is assigned a negative sign and Force of repulsion is assigned a positive
sign.”
Draw graphs showing dependence of F on
(i)
1 2
when d is kept constant
(ii) d when the product
(iii) d when the product
1 2
is maintained at a constant positive value.
1 2
is maintained at a constant negative value.
SOL: We know that F is an attractive (–ve) force when the currents I1 and I2 are ‘like’ currents i.e.
when the product I1I2 is positive.
Similarly F is a repulsive (+ve) force when the currents I1 and I2 are ‘unlike’ currents, i.e. when the
product I1I2 is negative.
Now F ∝ (I1 I2 ), when d is kept constant and F ∝
1
when I1I2 is kept constant.
d
The required graphs, therefore, have the forms shown below.
OR
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The given graphs show the variation of intensity of magnetization I with strength of applied
magnetic field H for two magnetic materials P and Q.
(iii) Identify the materials P and Q.
(iv) For material P, plot the variation of Intensity of Magnetisation with temperature.
Justify your answer.
SOL: (i) P- Paramagnetic
Q-Ferromagnetic materials
(ii)
14) Find the value of the phase lag/lead between the current and voltage in the given series LCR
circuit. Without making any other change, find the value of the additional capacitor, such
that when ‘suitably joined’ to the capacitor ( C= 2 µF) as shown, would make the power
factor of this circuit unity.
SOL: The current, I, leads the voltage, V, by an angle
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where
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∴ tan
∴
=1
= 450
The power factor becomes unity when = 00
Hence we need to adjust C to a new value C’ where
C’
= XL = 100Ω
Thus, phase angle is 450 with the current LEADING the voltage.
To make power factor as unity we need to have XC also equal to 100 ohms. For this C needs to have a
value of 10 µ F.
We, therefore need to put an additional capacitor of (10-2), i.e, 8 µF in parallel with the given
capacitor.
15) Explain how one ‘observes an inconsistency’ when Ampere’s circuital law is applied to the
process of charging a capacitor.
How this ‘contradiction’ gets removed by introducing the concept of an ‘additional current’,
known as the ‘displacement current’?
SOL: Discussion on the ‘observed inconsistency’
During charging of capacitor current IC would flow through loop-1. But cannot flow through loop-2
be cause loop-2 is in dielectric hence

loop 1
B . dl  0 I C  0 I
But for loop -2

loop  2

B . dl  
loop  2
loop  2
B . dl  0 (0)  0
B . dl
Hence one observe inconsistency in Ampere circuital law.
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Hence concept of displacement current was introduced by Maxwell which removes this inconsistency.
During charging of capacitor, electric flux in between plates of capacitor increases which causes a
displacement current to develop.
d
d
d  
(E )   0 ( EA)   0  . A 
dt
dt
dt   0 
d  q / A  dq
 0 
.A  
 IC
dt   0
 dt
ID  0
Hence I D  I C
Within plates of capacitor during charging of capacitor . Outside plate of capacitor E = 0 , E  0
Hence
d
E   0
dt
 ID  0
So Ampere circuital law was modified to

loop
B.dl  0 ( I C  I D )
Applying this rule

loop  2
So


loop 1
B.dl  0 ( I C  0)  0 I C
B.dl  0 (0  I D )  0 I D  0 I C
loop 1
B.dl  
loop  2
hence IC = Conduction current =
dq
dt
B.dl
Ampere circuital law becomes logically consistent.
Discussion on ‘Removing’ the contradiction through the concept of an additional current, called the
‘displacement current’.
16) A point object O is kept at a distance of 30 cm from a convex lens of power +4D towards its
left. It is observed that when a convex mirror is kept on the right side at a distance of 50 cm
from the convex lens, the image of the object O formed by the lens-mirror combination
coincides with the object itself. Calculate the focal length of the convex mirror.
SOL: Focal length of the convex lens =
1
m
P
Let v be the position of the image, I, of the object formed by the convex lens alone. We then have
1
1
1


v (u ) f
∴ v = 150cm
Hence the distance of the image (formed by the convex lens alone ) from the convex mirror would be
(150-50) cm ,i.e.,100 cm. This distance equals the radius of curvature of the convex mirror.
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∴ Radius of curvature of the convex mirror R = LI – LM = 150–50=100cm
Hence focal length of the convex mirror equals100/2, i.e., 50 cm.
17) The arrangement used by Thomas Young to produce an interference pattern is shown in the
given diagram.
Justify why there would be no change in the ‘fringe width’ when the main illuminated slit (S) is
shifted to the position ′ as shown.
SOL: Initially SS1 = SS2
Two sources do not have initial phase difference .
Fringe width in this case is  
D
.
d
Now s is shifted to S’ towards S1 Because of this two sources have an initial phase difference with S1
leading
S’S2 > S’ S1
Let S’S2 –S’S1 = m 
Total path difference at point P (S ' S2  S2 P)  (S' S1  S ' S1 ) (S2 P  S1P)  m  x
For nth maxima at point P
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m  x  n
x  (n  m)
For this yn 
D
D
D
x  (n  m)  (n  m)
 (n  m) 
d
d
d
For (n–1)th Maxima at point P
m  x  (n  1)
x  (n  m  1)
Corresponding to this point yn 1 
yn1 
D
x
d
D
D
(n  m  1)  (n  m  1)
 (n  m  1) 
d
d
New fringe width = yn  yn1  (n  m)   (n  m  1)   
Hence  '  
No fringe width change take place due to this action.
Calculation of ‘fringe width,
the ‘normal set up’
Calculation of fringe width, ’ in the changed set up Observing that ′ =
18) A given number of atoms No of a radioactive element with a half life T is uniformly
distributed in the blood stream of a
(iii)
normal person A having total volume V of blood in the body
(iv)
person B in need of blood transfusion having a volume ′ of blood in the body.
The number of radioactive atoms per unit volume in the blood streams of the two persons after
a time nT are found to be N1 and N2.
Prove mathematically that the additional volume of blood that needs to be transfused in the
 N 2  N1 
V .
 N2 
body of person B equals 
SOL: Initial number of radioactive atoms, per unit volume, in the blood streams of persons A and B
 N0 
 N0 
 and 
 respectively.
V 
V'
are 
After a time nT (T = Half life), these numbers would get reduced by a factor 2n.
 N0  1
 n
 V 2
N  1
And N2 =  0  n
V'2
N1 V '

N2 V
Hence N1= 
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orV '  V
N1
N2
∴ Additional volume of blood needed by person B is
V – ′ = V— V
N1  N 2  N1 
=
V
N2  N2 
19) A student has to use an appropriate number of
(iii)
NAND gates (only) to get the output Y1
(iv)
NOR gates (only) to get the output Y2
From two given inputs A and B as shown in the diagram.
Identify the ‘equivalent gate’ needed in each case. Show how one can connect an appropriate
number of (i) NAND (ii) NOR gates respectively in the two cases to get these ‘equivalent gates’.
SOL: The equivalent gates in the two cases are the OR gate and the AND gate respectively.
(i) A combination of three NAND gates, connected in the manner shown, would be equivalent to an
OR gate.
(ii) A combination of three NOR gates connected in the manner shown would be equivalent to an
AND gate.
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20) The data given below gives the photon energy (in eV) for a number of waves whose
wavelength values (in nm) are also given.
Wavelength (in nm)
200
400
600
800
1000
1200
Photon Energy (in eV)
6.216
3.108
2.072
1.554
1.243
1.036
(Without doing any calculation/taking any reading), explain how one can use this data to draw
an appropriate graph to infer
(i)
(ii)
(iii)
photon energy corresponding to a wavelength of 100 nm.
the wavelength value (in nm) corresponding to a photon energy of 1 eV.
velocity of light assuming that the value of Plank’s constant is known.
SOL: One can calculate the values of
1
1
and plot a graph between E (photon energy in eV) and


(in nm–1).
The resulting straight line graph can be used to
(i)
read the value of
(ii)
read the value of
(iii) We have E =
, corresponding to
1


1
nm1
100
1
( in nm–1) corresponding to E = 1eV

hc

The slope of the graph (after appropriate adjustment of the units) would equal hc. Since h is known,
one can calculate c.
21) A (sinusoidal) carrier wave
C(t) = AC sin ωc t
is amplitude modulated by a (sinusoidal) message signal m(t) = Am sin ωm t
Write the equation of the (amplitude) modulated signal.
Use this equation to obtain the values of the frequencies of all the sinusoidal waves present
in the modulated signal.
SOL: The equation of the (amplitude) modulated signal is
These are the three sinusoidal waves present in the amplitude modulated signal.
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The frequencies of these three waves are f1 
C
  m
  m
, f2  C
, and f3  C
.
2
2
2
22) Give reasons for the following:
(i)
The Zener diode is fabricated by heavily doping both the p and n sides of the
junction
(ii)
A photodiode, when used as a detector of optical signals is operated under
reverse bias.
(iii)
The band gap of the semiconductor used for fabrication of visible LED’s must at
least be 1.8 eV.
SOL: (i) Heavy doping makes the depletion region very thin. This makes the electric field of the
junction very high, even for a small reverse bias voltage. This in turn helps the Zener diode to act as a
‘voltage regulator’.
(ii) When operated under reverse bias, the photodiode can detect changes in current with changes in
light intensity more easily.
(iii). The photon energy, of visible light photons varies from about 1.8 eV to 3 eV. Hence for
visible LED’s, the semiconductor must have a band gap of 1.8 eV .
Section D
( 1×4 = 4Marks)
23) Dimpi’s class was shown a video on effects of magnetic field on a current carrying straight
conductor. She noticed that the force on the straight current carrying conductor becomes
zero when it is oriented parallel to the magnetic field and this force becomes maximum
when it is perpendicular to the field. She shared this interesting information with her
grandfather in the evening. The grandfather could immediately relate it to something
similar in real life situations. He explained it to Dimpi that similar things happen in real life
too. When we align and orient our thinking and actions in an adaptive and accommodating
way, our lives become more peaceful and happy. However, when we adopt an
unaccommodating and stubborn attitude, life becomes troubled and miserable. We should
therefore always be careful in our response to different situations in life and avoid
unnecessary conflicts.
Answer the following question based on above information:
a) Express the force acting on a straight current carrying conductor kept in a magnetic field in
vector form. State the rule used to find the direction of this force.
b) Which one value is displayed and conveyed by grandfather as well as Dimpi?
c) Mention one specific situation from your own life which reflects similar values shown by you
towards your elders.
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SOL: (i) Expression for the Force :
Let us consider a portion of length l and crosssectional area A of a straight conductor carrying a
current i . Let it be placed at right angles to a
magnetic field B in the plane of the page directed
downwards .
Let n be the number of free electrons per unit volume
of the conductor and vd the drift velocity of the
electrons. From the relation F  qv  B , the magnitude of the force on each electron is evdB, where e
is the (negative) charge on an electron. It however, the conductor makes an angle  with the magnetic
field B measured from the conductor towards the field B , then the magnitude of the force on each
electron is given by
In vector form F  i l  B
Where  is the angle between the vector l and B . The direction of force is given by the vector
l  B where l is taken in the direction of current.
The force is maximum when the conductor is perpendicular to the field (  = 90°). It is zero when the
conductor is parallel to the field (  = 0 ).
The direction in the magnetic field, along which a current-carrying conductor does not experience
any force is called the direction of the magnetic field.
Statement of Flemings’ left hand rule
Fleming's Left-hand Rule : Hold the thumb and first two fingers of the left hand at right angles to
each other with the First finger pointing in the direction of the Field and the seCond finger in the
direction of the Current, then the Thumb points in the direction of the Thrust (Force).
If the wire is not at right angles to the field, the force is smaller and is zero if the wire is parallel to the
field.
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(b)– Adaptation to different situations and flexible and adjustable attitude
- Sharing excitement in classroom learning with family members
(c)– Avoiding unnecessary arguments in conflicting situations in everyday life
Section E
(5×3=15 Marks)
24) a) State the theorem which relates total charge enclosed within a closed surface and the
electric flux passing through it. Prove it for a single point charge.
b) An ‘atom’ was earlier assumed to be a sphere of radius a having a positively charged point
nucleus of charge +Ze at its centre. This nucleus was believed to be surrounded by a uniform
density of negative charge that made the atom neutral as a whole.
Use this theorem to find the electric field of this ‘atom’ at a distance r (r < a) from the centre of
the atom.
SOL: Gauss's theorem
Gauss theorem states that the total flux through a closed surface is 1/
 0 times the net charge
enclosed by the closed surface.
Mathematically, it can be expressed as

1
0
q
   E. d S
1
 Ed S  
E  dS 
q
0
1
0
   EdS cos 
for  =00 , cos   cos 00  1
q
Where  q  q = ( total charge enclosed within the Gaussian surface)
Proof: We prove Gauss's theorem for an isolated positive point charge q. As shown in Fig.,
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suppose the surface S is a sphere of radius r centred on q. Then the surface S is a Gaussian surface.
Electric field at any point on S is
E
1
q
4 0 r 2
This field points radially outward at all points on S. Also, any area element points radially
outwards, so it is parallel to E , i.e.,  =0°.
(b) Set ρ be the uniform density of negative charge. We then have
∴
=
4 a3 
 Ze
3
3Ze
.
4 a 3
Taking a sphere of radius r (centred at the nucleus) as the Gaussian surface, we have
( )×4
2
=
Q
0
Where Q is the net charge enclosed by the Gaussian surface. Now
Substituting this value of Q, we get
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OR
A Dipole is made up of two charges + q and – q separated by a distance 2a.
Derive an expression for the electric field Ee due to this dipole at a point distant r from the
centre of the dipole on the equatorial plane.
Draw the shape of the graph, between | Ee | and r when r >> .
If this dipole were to be put in a uniform external electric field E , obtain an expression for the
torque acting on the dipole.
SOL: Intensity of the Electric Field at a point on the Axis of a Dipole(end-on position) :
Let P be a point on the axis at a distance r metre from the mid-point O of the dipole. We have to
determine the intensity of the electric field at P. E p (net )  ?
Let E1 and E2 be the magnitudes of the intensities of the electric field at P due to the charges
+ q and – q of the dipole respectively. The distance of the point P from the charge + q is (r – l) and
that from the charge – q is (r + l). Therefore
The intensities El and E2 are along the same line in opposite directions. Therefore, the resultant
intensity E at the point P will be equal to their difference and in the direction of dipole axis (since
E1  E2 ). That is,
If r is very large compared to 2l (r > > 2l), then l2 may be neglected in comparison to r. The last
expression may now he written as
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Torque on a Dipole in Uniform Electric Field:
Let an electric dipole consisting of charges + q and –q
and of length 2l placed in an uniform electric field E
, making an angle  with it. It has a dipole moment
of magnitude,
p = q2 l
Force exerted on charge +q F1  qE
(along E )
Force exerted on charge -q F2  qE
(opposit to E )
Hence the net translating force on a dipole in a uniform electric field is zero. But the two equal and
opposite forces act at different points of the dipole. They form a couple which exerts a torque.
Torque = F
 Perpendicular distance between the two forces
  qE  2l sin   (q  2l ) E sin 
  pE sin 
( p  q  2l )
As the direction of torque  is perpendicular to both p and E ,
  p E
  angle between p and E
When the dipole is released, the torque  tends to align the dipole with the field E . When the dipole
gets aligned with E , the torque  becomes zero.
the torque on the dipole will be maximum when the dipole is held perpendicular to E Thus
 max  pE sin 900  pE
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25) State the law which relates to generation of induced emf in a conductor being moved in a
magnetic field.
Apply this law to obtain an expression for the induced emf when one ‘rod’ of a rectangular
conductor is free to move in a uniform, time independent and ‘normal’ magnetic field.
Apply the concept of the Lorentz (magnetic) force acting on a moving charge to justify the
expression obtained above.
SOL: Faraday's laws of electromagnetic induction :
From the experimental observations, Faraday arrived at a conclusion that an emf is induced in a coil
when magnetic flux through the coil changes with time.
the time rate of change of magnetic flux through a circuit induces emf in it.
First law: Whenever magnetic flux linked with a circuit changes , an emf is induces in it and
consequently a current is induced in the circuit ( if the circuit is closed ) and lasts only so long as the
magnetic flux changing .
Second law: The magnitude of the induced emf is directly proportional to the rate of change of
magnetic flux linked with the circuit.
e
d
dt
| e | N
d
dt
d
dt
or
e
or
e  N
d
dt
; 1 volt =
1 weber
1 second
Where  flux linked per turn of coil or circuit , N is the number of turns in the coil.
Motional emf from Lorentz force ( Quantitative energy consideration ):
A conductor has a large number of free electrons. When it moves through a magnetic field, a Lorentz
force acting on the free electrons can set up a current.
Fig. shows a rectangular conductor in which arm PQ is
free to move. It is placed in a uniform magnetic field B,
directed normally into the plane of paper. As the arm
PQ is moved towards left with a speed v, the free
electrons of PQ also move with the same speed towards
left. The electrons experience a magnetic Lorentz force,
F = q v B . According to Fleming's left hand rule, this
force acts in the direction QP and hence the free
electrons will move towards P. A negative charge
accumulates at P and a positive charge at Q. An electric
field E is set up in the conductor from Q to P. This field exerts a force, Fe = qE, on the free electrons.
The accumulation of charges at the two ends continues till these two forces balance each other, i.e.,
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Fm = Fe
Or
qvB=qE
Or
vB=E
The potential difference between the ends Q and P is
V= E l = v B l
Clearly, it is the magnetic force on the moving free electrons that maintains the potential difference
and produces the motional emf,
e=Blv
Current induced in the loop. Let R be the resistance of the movable arm PQ of the rectangular loop
PQRS shown in Fig. Suppose the total resistance of the remaining arms QR, RS and SP is negligible
compared to R. Then the current in the loop will be
I
e Blv

R
R
Force on the movable arm:The conductor PQ of length l and carrying current I experiences a force F
in the perpendicular magnetic field B. The force is given by
This force (due to induced current) acts in the outward direction opposite to the velocity of the arm in
accordance with Lenz's law. Hence to move the arm with a constant velocity v, it should be pulled
with a constant force F.
Power delivered by the external force: The power supplied by the external force to maintain the
motion of the movable arm with constant velocity v is
P  Fv 
B 2l 2 v 2
R
Power dissipated as Joule loss: The power dissipated in the loop as Joule heating loss is
2
B 2l 2 v 2
 Blv 
PJ  I 2 R  
R


R
 R 
Clearly, PJ = P. Thus, the mechanical energy expended to maintain the motion of the movable arm is
first converted into electrical energy (the induced emf) and then to thermal energy.
This is consistent with the law of conservation of energy.
OR
An a.c. voltage V = Vm sin
t is applied across an inductor of inductance L. Apply Kirchoff’s
loop rule to obtain expressions for
(i) the current flowing in the circuit
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(ii)
the inductive reactance L Hence find the instantaneous power P i supplied to the
inductor.
Show graphically the variation of Pi with
t.
SOL: (i)Current flowing in the circuit
AC VOLTAGE APPLIED TO AN INDUCTOR
Consider a coil of self-inductance L connected across an ac supply . Even though the windings of the
coil and the connecting wires have some resistance, it is assumed (for the sake of simplicity) that L is
so great that the resistance of the circuit is negligible. So, if a D.C. is made to flow through the coil L,
the potential difference across its ends would be zero.
Let the instantaneous alternating emf be represented by
V = Vo sin ω t ...(1)
Due to flow of alternating current in the circuit, the magnetic flux linked with
the circuit changes and an induced emf  L
dI
dI
is set up in it. Here
is the
dt
dt
rate of change of current in the circuit. The minus sign indicates that the induced emf acts in a
direction opposite to that of the applied emf. In order to maintain the flow of current, the applied emf
must be equal and opposite to the induced emf.
Using the Kirchhoff’s loop rule , since there is no resistor in the circuit,
V L
dI
=0
dt
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dI
dt
V
dI  dt
L
V sin t
dI  0
dt
L
V L
Integrating both side
V0 sin t
dt
L

dI  
I
V0   cos t  V0
[ cos t ]


L    L
I
V0


sin   t  
L 
2
I
V0


sin   t  
XL
2



I  I 0 sin   t  
2

where X L   L
where
V0
 I0
XL
we find that the current lags the emf by π / 2.
(ii) Finding expression for instantaneous power, Pi, the graph has the form shown.
Pi = VI

 

 (V0 sin t )  I0 sin  t     (V0 sin t )(I0cos t )
2 


V I
V I
 V0 I0 sin t cos t   0 0 (2sin t cos t )   0 0 sin 2t
2
2
Graph ∝ –sin 2t
26) (a) Explain, with the help of a diagram, how is the phenomenon of total internal reflection
used in
(i)
an optical fibre
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(ii)
a prism that inverts an image without changing its size
(b) A right angled prism made from a material of refractive index
is kept in air. A ray PQ is
incident normally on the side AB of the prism as shown.
Find (in terms of ) the maximum value of
upto which this incident ray necessarily undergoes
total internal reflection at the face AC of the prism.
2+3=5
SOL: (a) Explaining the use of the phenomenon of total internal reflection in
(i)
As a Reversing Prism : When the light ray enters the right-angled isoscele's prism normally
at the hypotenuse face, it is totally reflected at each of the shorter faces in turn and emerges antiparallel to its original direction .Thus, it is turned through 180°. The figure also shows the reversal of
image due to reflection by the prism. A pair of such prisms is used to reverse and erect the image in
binoculars.
(ii) Optical Fibre : An optical fibre is a device which transmits light introduced at one end to the
opposite end, with little loss of the light through the sides of the fibre.
Principle: It is based upon the principle of total - internal reflection of light.
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Construction : An optical fibre consists of a cylindrical central core of diameter of the order of a few
micrometers (μ m) made of thousands of long and extremely thin strands of high-quality glass or
quartz. The core is cladded by a material of slightly lower refractive index.(    )
2
1
Light Propagation : When light is incident at a small angle at one end of the optical fibre, it gets
refracted into the core and incident on the core-cladding interface. Now. for a light-ray entering the
fibre, if the angle of incidence at the interface is greater than the critical angle, then the ray undergoes
total internal reflection at the interface. Further, because of the cylindrical symmetry in the fibre
structure, this ray suffers total internal reflection at the lower interface also and is guided through the
core by repeated total internal reflections and finally comes out of the fibre. Even for a bent or twisted
fibre, light guidance can occur through multiple total internal reflections.
(b)
The angle of incidence at the face AC = i
But i = A
(Angle between two lines is the same as the angle between their perpendiculars)
Also θ = π/2 – A = π /2 – i
The minimum value of I, so that there is total internal reflection at the face AC, equals ic where
1


ic = sin 1 
The maximum value of θ corresponding to the minimum value of i (= ic) is therefore,
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1


θ max = π/2 – ic = π/2 – sin 1 
OR
State Huygen’s principle in wave-optics. How did Huygen ‘explain’ the absence of the
backwave?
Use this principle to draw the refracted wave front for a plane wave incident from a denser to a
rarer medium. Hence obtain Snell’s law of refraction.
SOL: Statement of Huygen’s principle :
HUYGENS PRINCIPLE
'Huygens' principle is based on the following assumptions :
1) Each and every point on the given ( primary) wavefront acts as a source of new disturbances,
called “secondary wavelets”, that travels in all directions with the velocity of light in the medium.
These wavelets sending out disturbance in all directions in a similar manner as the original source
of light does.
2)
The new position of the wavefront at any instant (called secondary wavefront) is the envelope (
surface of common tangents ) of the secondary wavelets at that Instant.
Absence of backwave: Huygens argued that the amplitude of the secondary wavelets is maximum in
the forward direction and zero in the backward direction; by making this adhoc assumption, Huygens
could explain the absence of the backwave. However, this adhoc assumption is not satisfactory and
the absence of the backwave is really justified from more rigorous wave theory.
Laws of refraction :
Laws of refraction on the basis of Huygens' wave theory. Consider a plane wavefront AB incident on a
plane surface XY, separating two media 1 and 2. Let v1 and v2 be the velocities of light in the two
media, with v2 < v1.
The wavefront first strikes at point A and then at the successive points towards C. According to
Huygens' principle, from each point on AC, the secondary wavelets start growing in the second
medium with speed v2. Let the disturbance take time t to travel from B to C, then BC = v1 t. During
the time the disturbance from B reaches the point C, the secondary wavelets from point A must have
spread over a hemisphere of radius AD = v2t in the second medium. The tangent plane CD drawn
from point C over this hemisphere of radius v2t will be the new refracted wavefront.
Let the angles of incidence and refraction be i and r respectively.
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From right  ABC, we have
sin BAC  sin i 
BC
AC
From right  ADC, we have
sin DCA  sin r 
AD
AC
sin i BC v1t


sin r AD v2t
Or
sin i v1
 1 2 ( constant)
sin r v2
This proves Snell's law of refraction. The constant 1  2 is called the refractive index of the second
medium with respect to first medium.
Further, since the incident ray SA, the normal AN and the refracted ray AD are respectively
perpendicular to the incident wavefront AB, the dividing surface XY and the refracted wavefront CD
(all perpendicular to the plane of the paper), therefore, they all lie in the plane of the paper, i.e., in the
same plane. This proves another law of refraction.
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