Download KV No. 1 (Paper-I)

K.V. No. 1 (Paper-I)
Max. Marks : 70
Sub : Physics
Class : XII
Time : 3 Hrs
General Instructions:1. All questions are compulsory
2. There are 29 question in total Q. No. 1 to 8 are very short question and carry
one mark each.
3. Q. No. 9 to 16 carry two marks each, Q. No. 17 to 25 carry three marks each,
question, 26 is value based question carry four marks and 27 to 29 carry 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 three questions
of five marks each. You have to attempt only one of the given choices in such
questions.
5. Use of calculators is not permitted. However, you may use log tables if
necessary.
6. You may use the following values of physical constants wherever necessary.
c  3  10 8 m / s; h  6.6  10 34 Js; e  1.6  10 19 C;  0  4x10 7 TmA 1
1
4 0
 9  10 9 Nm 2 / C 2 m e  9.1x10 31 kg
1.
A 500 c charge is at the centre of square of side 10cm. Find the work done in
moving a charge of 10C between two diagonally opposite points on the
square.
2.
A wire of resistivity  is stretched to four times its length. What will be its
new resistivity?
3.
A silver wire has a resistance of 2.1 at 27.50C, and a resistance of 2.7 at
1000C. Determine the temperature co-efficient of resistivity of silver.
4.
An electron is moving with a velocity of 107m/s, enters uniform magnetic
field of 1T. The direction of the velocity of election is parallel to the magnetic
field. What would be its trajectory in this field?
5.
What is the name given to the curve, the tangent to which at any point gives
the direction of magnetic field at that point? Can two such curves intersect
each other? Justify you answer.
6.
An electric motor running on a 50V DC supply draws 12A current. Find the
resistance of its windings.
7.
What is meant by modulation?
8.
A carrier wave of peak voltage 12V is used to transmit a message signal.
What should be the peak voltage of modulating signal in order to have a
modulation index of 75%?
9.
Write the expression for electrostatic potential energy of a dipole, of dipole


moment Pe in uniform electric filed and E . Draw a graph showing the value


of this at different angles between Pe and E and indicate the positions of
unstable and stable equilibrium.
10. Two identical cells each of emf E and internal resistance ‘r’ are connected in
parallel to an external resistance R. Find an expression for the total current
flowing in the circuit as shown in figure.
11. An ac generator has coil of 50 turns and area 2.5m2 rotating at an angular
speed of 60 rad S-1 in a uniform magnetic field of B=0.3T between two pole
pieces. The resistance of the windings is 500.
(i)
What is the maximum current drawn from the generator?
(ii)
What is the flux through the coil when the current is Zero? What is the
flux when the current is maximum?
12.
How is resolving power of a microscope affected when
(i)
(ii)
The wavelength of illuminating radiations is decreased?
The diameter of the objective lens is decreased? Justify your answer.
13. The following table gives the values of work function for a few photosensitive
metals:
S. No.
1
Na
Metal
Work function (ev)
1.92
2
K
2.15
3
Mo
4.17
If each of these metals is exposed to radiations of wavelength 300nm, which
of them will not emit photo electrons and why?
Or
By how much would the stopping potential for a given photosensitive surface
go up if the frequency of the incident radiations were to be increased from
4x1015 H 2 to 8x1015 H 2 ?
Given h  6.4x10 34 Js e  1.6x10 19 C and c  3x108 m / s.
14. Draw of graph to show the variation of stopping potential with frequency of
radiation incident on a metal plate. How can the value of Planck’s constant be
determined from this graph?
15. Indentify the logic gate ‘1’ and ‘2’ in the logic circuit given. Also obtain the
truth table for the final output for all possible combination of the input A and
B.
16. Give any two difference between a half wave rectifier and a full wave
rectifier.
17. When two known resistances R and S are connected in the left and right gaps
of a meter bridge, the balance point is found at a distance l1 from the zero end
of metre bridge wire. An unknown resistance X is now connected in parallel
to the resistance S and the balance point is now found at a distance l 2 from the
zero end of the metre bridge wire as shown in figure. Obtain a formula for X
in terms of l1 x l2 and S.
R
18.
A galvanometer coil has a resistance of 30 and meter shows full scale
deflection for a current of 2.0mA. Calculate the value of resistance required
to convert it into an ammeter of range O to 1A. Also, calculate the resistance
of ammeter.
19.
A thick copper wire carrying a current of 10A is bent into semicircular arc of
radius 7.0cm as shown in fig (a). State the direction and calculate the
magnitude of field at the centre of the arc. How would your answer change if
the same wire is bent into semicircular arc of the same radius but in opposite
way as shown in fig (b)
Or
A closely wound solenoid of 2000 turns and area of cross-section 1.6x10 4 m 2
carrying current of 4.0A is suspended through its centre allowing it to turn in
a horizontal plane.
(a) What is the magnetic moment associated with the solenoid?
(b) What is the force and torque on the solenoid if a uniform horizontal
magnetic field of 7.5x10 2 T is set up at an angle of 30 0 with the axis of the
solenoid?
20.
Write the order of frequency range and one use of each of the following
electromagnetic radiations.
(i) Microwaves
(ii) Ultraviolet rays
(iii) Gamma rays
21.
Which two main consideration are kept in mind while designing the
objective of an astronomical telescope? Obtain an expression for the angular
magnifying power and the length of the tube of an astronomical telescope in
its ‘normal adjustment’ position.
22. Define the term Binding energy of a nucleus. Complete B.E per nucleon for
56
26 Fe .
Given mass of portion = 1.0078254amu
mass of neutron = 1.00866549amu
mass of 56
26 Fe nucleus = 55.9349394amu
and 1amu = 931.5 Mev.
 13.6
eV . Calculate the
n2
energy required to take an electron from ground state to the second excited
state.
23.
The energy of an electron in orbit is given by En 
24.
Draw a labelled circuit diagram to show the used of Zener diode as a voltage
regulator.
25.
(a)
What does the term LOS communication mean? Name the types of
waves that are used for this communication
(b)
Draw the block diagram of a communication system.
26.
A student X sitting in the last row in the class have difficulty in reading the
matter written on the black board. The teacher requests the student sitting in
the front row to volunteer for exchange of seat to help X. Student Y sitting
in the front row volunteers herself to exchange the seat with X. Now X was
able to read the matter on black board clearly and Y also did not face any
difficulty. The teacher also advised X to get his eyes checked up.
Answer the following questions based on the above information.
(i)
Which eye defect of vision X is suffering from?
(ii)
Which values is the teacher displaying through her actions and
advice?
(iii)
(iv)
27.
Which values is Y displaying through volunteering?
In what possible ways can X respond and reciprocate the help
rendered by Y?
Define the term electric dipole moment and magnetic dipole moment. Derive
an expression for electric field intensity at a point on equitorial line of a
electric dipole.
Or
With the help of a labelled diagram explain the working of Van de Graff
generator.
28.
A spherical surface of radius of curvature R, separates a rarer and a denser
medium as shown in the figure.
Complete the path of the incident ray of light, showing the formation of a
real image. Hence derive the relation connecting the object distance, image
distance, radius of curvature and refractive indices 1 and 2 of the two
media.
Or
The figure, drawn here shows a modified Young’s double slit experimental
set up. If SS2 – SS1 = /4.
(i)
State the conditions for constructive & destructive interference.
(ii)
Obtain an expression for the fringe width.
(iii)
Locate the position of the central fringe.
29.
(a)
An alternating voltage V= Vm Sin wt applied to a series LCR ckt
drives a current given by I = Im Sin (wt +). Deduce an expression for
the average power dissipated
(c)
For a circuit used for transporting electric power a low power factor
implies large power loss in transmission line explain?
Or
(a)
Current in a circuit falls from 5A to 0A in 0.1s if the average e.mf of
200V is induced what is the self inductance of the circuit?
(b)
An ac of 0.2A RMS and frequency 100/2 Hz flows in a LRC ckt with
L=0.15H, R=20, C= 500F.
Calculate the ac voltage (i) across each component (ii) across R&L
together (iii) across L&C together (iv) The total voltage across L, C
and R. What power is dissipated in each component?
SOLUTIONS (Paper-I)
Max. Marks : 70
Sub : Physics
Class : XII
1.
Zero, as the two points are at same potential.
2.
Resistivity depends upon the nature of material and does not depend upon
the dimension of the material. So, resistivity will be unchanged.
(1)
R 2  R 1 (1  T ), we get 
R 2  R1
2.7  2.1

R 1 T
2.1x (100  27.5)
(1)
3.
Using
4.
Force on the electron in magnetic field will be
F = q VB Sin O0 = O
So there is no force on electron hence it will move along a straight line.
5.
The name of the curve is “magnetic lines of force”. Two magnetic lines of
force never intersect with each other because if they do so then two tangents
drawn at the same point of intersection will give two directions of magnetic
field at same point, which is impossible.
(1)
6.
Given V  50V and I  12A
0.6

 .0039 0 C 1
2.1(72.5)
Re sis tan ce of motor  R 
V 50

 4.18 
I 12
(1)
(1)
7.
The phenomenon of superposing low frequency modulating signal on high
frequency carrier wave in accordance to the same characteristics of
modulating signal is called modulation.
8.
Using  
Em
we get E c  12 V, E m  ?   75%
Ec
E m  E c 
9.
(1)
75
x12  9V
100
The electrostatic potential energy Up of a
dipole of dipole moment Pe in uniform



electric field E is up = Pe. E = pe E cosIt
varies since soidally with  s follows
(1)
10.
Total e.m.f =E
Total Resistance = R + r/2
 Current I 
11.
(i)
Eo = NABW
I0 
(ii)
12.
2E
2R  r
E o NABW 50x0.3x 2.5x60


 4.5A
R
R
500
The flux is maximum when current is zero. The flux
the current is maximum.
is zero when
The expression for R.P. of a microscope i.e.
R.P. 
2 Sin 

Where  is R.I. of the medium between the object
and the lower surface of the objective lens and ‘’
is the angle subtended by a radius of the objective
on one of the points of the object obviously.
(i)
13.
Obviously R.P. is increased (ii) R.P. is
decreased because  is decreased.
hc
For photoelectric effect to occur h  o or  o

hc 12375 evA 0

 4.125 eV
We calculate roughly
0

3000 A
Obviously Mo will not emit photoelectrons when 30nm light is shined on it.
Or
U sin g h  h 0  eVs
and h1  h 0  eVs1
h ( 1   )
 v 1s  v s
e
Putting values Vs1  Vs 
6.4x10 34 Js
(8x1015  4x1015 )S 1  16V
19
1.6x10 C
14.
The variation of stopping potential with the
frequency of radiation incident on a metal
plate is a straight line AB as shown in figure.
Take two points C and D on the graph. Note
down the corresponding frequency of
radiations (  1 and  2) and stopping
potential (V1 and V2)
Then, eV  h  o
1
1
and eV  h  o
2
2
15.
e( V  V )  h (    )
2
1
2
1
e( V  V )
Thus planck’s constant can be determined.
2
1 ,
or h 
 
2
1
Logic gate ‘1’ is OR gate and logic gate ‘2’ is NOT gate.
Truth table
A
0
0
1
1
16.
17.
B
0
1
0
1
Y
1
0
0
0
(i)
In half wave rectifier, the output voltage is unidirectional intermittent
and varying voltage.
(ii)
In half wave rectifier, the frequency of output signal in same as that of
input signal.
In full wave rectifier, the frequency of output voltage signal is double
than that of input signal.
When R and S are connected to the left and right gaps of metrebridge and
bridge is balanced at length l1 from zero end
l
R
1
Then

(i)
S (100  l )
1
When unknown resistance X is connected in parallel to S, then effective
resistance in right gap is S1 
SX
SX
(ii )
Now balance point is obtained at length l2 and So.
l1
R

1
S
(l00  l1 )
Putting the value of S11 we have
R (S  X)
l2

SX
(100  l 2 )
(iii )
Dividing (iii) by (i) we get
SX
l2
100  l1

x
X
(100  l 2 )
l1
Or
S
l (100  l 2 )
1  2
X
l 2 (100  l 2 )
S 100 l 2  100 l1  l1l 2 100 (l 2  l1 )


X
l 2 (100  l 2 )
l1 (100  l 2 )
(100  l 2 )l1
xS
100 l 2  l1 
Here, current through the galvanometer IG = 2mA = 2x10-3 A
Resistance of galvanometer G = 30. Maximum value of correct = 1A
I xG 0.002 x 30
R G

 0.06
As,
I  I G 1  0.002
X 
18.
R  6x10 10 
19.
Here I = 10A
Length to the semicircle L 
2r
 r  x 7x10 2 m
2
The magnetic field at the centre of the semicircle will be given as
 IL 
o Idl sin 90 
B  2 2  dervied from dB 

4r 
4
r2

7
o I
o I oI 4x10 x10




 4.5x10 5 T.
2
4 r
4 r
4r
4x 7x10
Now if the same wire is bent into semicircular arc of the same radius but in
opposite way as shown in fig (b) the direction of magnetic field at the centre
will be inverted while magnitude will remain constant.
Or
(a)
(b)
20.
21.
As magnetic moment M = NIA
= 2000 x 4 x 1.6x10-4Am2
= 1.28 Am2
Net force F = 0
But torque T = MB sin
= 1.28x7.5x10-2 sin 300
= 4.8 x 10-2 N-m
(1½)
(1½)
(i)
Microwaves having frequency range 3x1011 Hz to 1x109 Hz. It is
used in radar system.
(ii)
UV rays having frequency range 5x1017 H 2 5x1017 to 8x1014 Hz. It is
used in checking of forged documents.
(iii)
Gamma rays having frequency range of 5x1022 Hz to 3x1019 Hz. it is
used studying the nucleus.
The objective of telescope must have (i) high R.P. (ii) large light gathering
power. Thus it must have a large diameter.
In normal adjustment as shown the magnifying power is given by M =   .
The final image is formed at infinity.
AB
AB

EA
OB
AB

EA  OB  fo
M  
 AB
EA fe
OB
 
The length of the tube is fo + fe
22.
As mass defect M = Zmp + (A-Z)mn-M
and energy corresponding to M is E = MC2 E = M x 931.5MeV.
Here M
= 26xmp + (56-26)mn – 55.935
= (26x1.0079+30x1.0087) – 55.9350
= (26.2054+30.2610) – 55.9350
= 56.4664-55.9350
= 0.5314 amu.
Now energy corresponding to this mass defect
E = 0.5314 x 93.15 MeV = 494.9991 = 495 MeV.
23.
The energy of nth orbit of the H-atom is given by
13.6
eV
n2
 13.6   13.6 
 E 2  E1 


( 2) 2  12 
 3.4  13.6  10.2eV.
 En  
Now energy required to take an electron from ground state to second
excited state.
24.
Here the zener diode is joined in reverse bias to functioning d.c. input
voltage through a resistance R of suitable value depending upon the zener
voltage and power rating of zenerdiode used. The constant output voltage is
taken across a load resistance RL connected in parallel with zener diode.
When the input d.c. voltage increases beyond a certain limit, the voltage
across zener diode becomes constant equal to zener break down voltage, but
the current through the zener diode circuit rises sharply as the dynamic
resistance of zener diode becomes almost zero after zener break down
voltage. Due to which there is an increase in voltage drop across R since R L
is connected in parallel so the voltage across RL remains same as that of
zener break down voltage.
25.
(a)
LOS means line of sight communication. The two type of waves used
for LOS communication are (i) space waves (ii) waves from
transmitting stations near the ground to flying air craft.
Transmitted
signal
Received
signal
Message Signal
Information
source
Transmitter
Channel
Receiver
User of
information
Noise
26.
(i)
(ii)
(iii)
(iv)
Myopia/short sightedness
Concern for students, helpfulness, Duty/role as a guide counseling.
Empathy, Helpfulness, Cooperation.
Friendship, Sharing and caring, thanks and grate fullness.
27.
Definition
P = qx2l
M = mx2l
E
1
P
4eV0 ( r 2  a 2 ) 32
Or
Construction
Diagram will labeled
Working
-
28.
i   
r   
sin i  2

sin r  1
Re lation
 2 1  2  1


v
u
R
2
3
Or
Initial path difference between S1 and S2 = SS2 – SS1 =
  x ; x  S P  S P  path difference between disturbance from S1 and
1
2
1
4
2

S2 at point P. =
yd
D
 total path difference between the two disturbances at P
 yd
x T  x1  x 2  
4 D
For constructive interference
  yd 
X T      n where n  0,1,2 ......
4 D 
yd 
1
 n  n  
(i)
D 
4
For destructive int erference

  yd 
X T      2n  1
2
4 D 
ynd 
3 
  2n  
D 
2 2
D
d
The position y o of Central fringe is obtained by putting n  0 in  n (i)
  fringe width  y n 1  y n 
d
ud
[- ve sign shows that the central fringe is obtained at a point below the
central point O.]
yo  
29.
(a)
Let at any instant the current and voltage in an LCR series AC ckt is
given by
I= I0 sin wt
V = V0 sin (wt+)
The instantaneous power is given by
P = VI = Vo sin (wt +)/ Io sin wt
Vo I o
2 sin wt. sin ( wt  
2
Vl
P  VI  o o Cos  Cos(2 wt  
2
 2 sin A sin B  cos(A  B)  cos(A  B)
P
(i)
Work done for a very small time interval dt is given by
dW = Pdt
dW = IVdt
 Total W.D. over Ta complete cycle is given by
T
W   VIdt  T
0
T
W
But Pav 

T
T
(b)
 VIdt
0
T
Vl
I 0 0 0
Pav
[Cos  cos(2 wt  )]dt
T 2
T
V I T

 o o   cos dt   Cos( 2 wt  )dt
2T  0
0

VI
 o o cos [ t ]T0  0
2T
VI
 o o cos T
2T
V I
 o o cos 
2 2
 Vrms Irms Cos
Power factor cos = R/Z
R – Resistance
Z- Impedance
Low power factor (cos ) implies lower ohmic resistance and higher power
loss as Pav  1/R in transmission.
Or
(a)
(b)
(i)
dI
e
200
or L 

 40 H.
dt
dI dt 5
0.1
We make use of phasor diagram
e l
100 

VL  I RMS X L  0.2x  2x
 0.15  3V
2 

VR  I RMS R  0.2x 20  4V
VC  I RMS X c  I RMS
X
L
 L  2L
1
1
1
1 

 0.2x
 4V  X C 


100
C
C

2

fc
6


2x
x500 x10
2
(ii)
VLR  VR2  VL2  4 2  32  5V
(iii)
VLC  V:L  VC  4  3  IV
(iv)
VLCR VR2  ( VL  VC ) 2  4 2  12  17  4.1V
(v)
PL  o; Pc  O PR  I 2RMS R  0.2 x20  0.8W
2
K.V. No. 1 (Paper-II)
Max. Marks : 70
Sub : Physics
Class : XII
Time : 3 Hrs
General Instructions:1. All questions are compulsory
2. There are 29 question in total Q. No. 1 to 8 are very short question and carry
one mark each.
3. Q. No. 9 to 16 carry two marks each, Q. No. 17 to 25 carry three marks each,
question, 26 is value based question carrying four marks and 27 to 29 carry
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 three questions
of five marks each. You have to attempt only one of the given choices in such
questions.
5. Use of calculators is not permitted. However, you may use log tables it
necessary.
6. You may use the following values of physical constants wherever necessary.
1
8
 34
19
7
1
9
2
2
c  3  10 m / s; h  6.6  10 Js; e  1.6  10 C;   4x10 Tm A
 9  10 Nm / C ;
0
4C
0
 31
m e  9.1x10 kg
1.
Why are alloys used for making standard resistance coils?
2.
How does focal length of a lens change when red light is replaced by blue
light?
3.
Name the series of hydrogen spectrum lying in the infrared region.
4.
Would sky waves be suitable for transmission of TV singnals of 60 MHz
freqneucy?
5.
What is the colour code for a resistor of resistance 3.5K with 15% tolerance.
6.
Which physical quantity has the unit Wb m-2? Is it a scalar or a vector
quantity?
7.
A solenoid is connected to a battery so that a steady current flows through it.
If an iron core is inserted into the solenoid, will the current increase or
decrease? Explain.
8.
Why is ground wave transmission of signals restricted to a frequency of 1500
kHz?
9.
A uniform field E exists between two charged plates as shown
in figure. What would be work done in moving a charge q along
the closed rectangular path ABCDA?

10. What is an ideal diode? Draw the output waveform across the load resistor R,
if the input waveform is an shown in the figure.
11. Find the ratio of intensities of two points P and Q on a screen in Young’s
double slit experiment when waves from sources S1 and S2 have phase
difference of (i) 00 and (ii)

respectively.
2
12. Two conductors are made of the same material and have the same length.
Conductor A is solid wire of diameter 1mm. Conductor B is a hollow tube of
outer diameter 2mm and inner diameter 1mm. Find the ratio of resistance RA
to RB.
13. The current sensitivity of a moving coil galvanometer increases by 20% when
its resistance is increased by a factor 2. Calculate by what factor the voltage
sensitivity changes.
14. A coil of 0.01 H inductance and I  resistance is connected to 200 V, 50 Hz
ac supply. Find the impedance of the circuit and time lag between maximum
alternating voltage and current.
15. With the help of an example, explain, how the neutron to proton ratio changes
during alpha decay of a nucleus.
16. Tow long parallel wires are hanging freely. If they are connected to a battery
(i) in series, (ii) in parallel, what would be the effect on their positions?
17. Obtain equivalent capacitance of the following network. For a 300 V supply,
determine the charge and voltage across each capacitor.
18. Which of the following waves can be polarized?
(i) X-rays (ii) Sound waves. Give reasons.
Two polaroids are used to study polarization. One of them (the polarizer) is
kept fixed and the other (the analyser) is initially kept with its axis parallel to
the poalriser. The analyser is then rotated through angels of 45 0, 900 and 1800
in turn. How would the intensity of light coming out of anlayser be affected
for these angles of rotation, as compared to the initial intensity and why?
19. Illustrate the basic elements required for transmitting and receiving an audio
signal with the help of a block diagram.
20. Two cells of voltage 10 V and 2 V and internal
resistance 10  and 5  respectively, are connected
in parallel with the positive end of 10 V battery
connected to negative pole of 2 V battery as shown
in figure. Find the effective voltage and effective
resistance of the combination.
21. A source contains two phosphorus radionuclides
33
15
and P(T1
2
 25.3
days ) . Initially 10% of the decays come from
must wait until 90% do so?
32
15
P (T1  14.3 days )
33
15
P . How long one
2
22. The two plates of a parallel plate capacitor are 4 mm apart. A slab of dielectric
constant 3 and thickness 3 mm is introduced between the plates with its faces
parallel to them. the distance between the plates is so adjusted that the
rd
2
capacitance of the capacitor becomes   of its original value. What is the
 3
new distance between the plates?
Or
Define the term electric potential due to a point charge. Find the electric
potential at the centre of a square of side
2m , having charges
100 C,50 C,20 C and  60 C at the four corners of the square.
23. How would you establish an instantaneous displacement current of 2.0 A in
the space between the two parallel plates of 1 F capacitor?
24. The figure shows a rectangular current carrying loop,
placed 2 cm away from a long straight, current
carrying conductor. What is the direction and
magnitude of the net force acting on the loop?
25. Two lenses of powers + 15 D and -5 D are in contact with each other forming
a combination lens.
(a)
What is the focal length of this combination.
(b)
An object of size 3 cm is placed at 30 cm from this combination of
lenses. Calculate the position and size of the image formed.
26. In an experiment of photoelectric effect, Neet plotted graphs for different
observation between photoelectric current and collector plate potential but her
friend Megha has to help her in plotting the correct graph. Neeta thanked
Megha for timely help.
(a)
What value was displaed by Megha and Neeta.
(b)
Draw the correct graph between I and V.
27. What is induced emft? Write Faraday’s law of electromagnetic induction.
Express it mathematically. A conducting rod of length I, with one pivoted, is
rotated with a uniform angular speed  in a vertical plane, normal to a
uniform magnetic field B. Deduce an expression for the emf induced in this
rod.
In India, domestic power supply is at 220V 50 Hz, while in USA it is 110 V,
50 Hz. Give one advantage and one disadvantage of 220 V supply over 110V
supply.
Or
Explain the phenomenon of self induction. Define coefficient of self
inductance. What are its units Calculate self inductance of a long solenoid.
28. (a)
(c)
With the help of a circuit diagram explain the working of transistor as
oscillator.
Draw a circuit diagram for a two inputs oR gate and explain its
working with the help of input, output waveforms.
Or
Define the terms potential barrier and depletion region for a p-n
junction. Explain with the help of a circuit diagram, the use of a p-n
diode as a full wave rectifier. Draw the input and output wave form
29.
Define magnifying power of an optical telescope. Draw ray diagram for an
astronomical refracting telescope in normal adjustment showing the paths
through the instrumental of three rays from a distant object. Derive an
expression for its magnifying power. Write the significance of diameter of
the objective lens on the optical performance of a telescope.
Or
State Huygens principle and prove laws of reflection and refraction on the
basis of Huygens principle.
SOLUTIONS (Paper-II)
Max. Marks : 70
Sub : Physics
Class : XII
1.
Alloys are used for making standard resistance coils because they have low
value of temperature coefficient (less temperature sensitivity) of resistance
and high resistivity
2.
According to lens maker’s formula,
 1
1
1 

   1 

f
 R1 R 2 
As  b   r  f b  f r
3.
4.
5.
6.
7.
8.
9.
10.
i.e. focal length of lens decreases.
Paschen series, Brackett series and Pfund series.
No, signals of frequency greater than 30 MHz will not be reflected by the
ionosphere, but will pentetrate through the ionosphere.
Given, resistance = 3.5k 5% = 35 x 102  5%
Colour code of given resistor is orange, green, red and gold.
Magnetic field induction has the unit Wb m-2. It is a vector quantity.
The current will decrease. As the iron core is inserted in the solenoid, the
magnetic field increase and the flux increase. Lenz’s Law implies that
induced emf should resist this increase, which can be achieved by a decrease
in current.
In ground wave propagation, the loss of energy due to interaction with
matter increase with the increase in frequency of wave. Therefore, the waves
of frequency above 1500 kHz get heavily damaged in ground wave
progagation. Hence the ground wave propagation is restricted to a frequency
of 1500 kHz.
Work done in moving a charge along the closed rectangular path would be
zero, because field in the entire space is uniform and electrostatic forces are
conservative forces.
A p-n junction diode which offers zero resistance when forward biased and
infinite resistance when reverse biased is called an ideal diode.
The output waveform across R is as shown in the figure below.
11.
In Young’s double slit experiment, the resultant intensity at any point on the
screen is
I  I1  I 2  2 I1I 2 cos 
Where  is the phase difference between the waves at that point.
Here, we consider l1 = l2 = l0
Therefore,
When  = 00, the intensity at point P is

I p  I 0  I 0  2 I 0 I 0 cos 0 0  4I 0

the resultant intensity at point Q is
2

I Q  I 0  I 0  2 I 0 I 0 cos  2I 0
2
Ip 2


IQ 1
When  
12.
Resistance of conductor A,
l
RA 
(0.5x10 3 ) 2
Resistance of conductor B,
l
RB 
2
 1x10 3   (0.5x10 3 ) 2


R A (1x10 3 ) 2  0.5x10 3  0.75 3




RB
0.25 1
0.5x10 3 2
2
13.
Given I s  I s 
20
120
Is 
I s , R '  2R
100
100
Then, initial voltage sensitivity, Vs 
New Voltage sensitivity,
I'
3
 120  1
Vs'  s'  
I s x
 Vs
R
 100  2R 5
 % decrease in voltage sensitivity
3
Vs  Vs
V V
5 x100  40%
 s
x100 
Vs
Vs
'
s
ls
R
14.
Here, R = 1,  = 50 Hz,
The inductive reactance is
L= 0.01 H
X L  L  2vL  2.x 3.14 x50x 0.01  3.14
The impedance of the circuit is
Z  R 2  X 2L 
15.
12  3.142
 10.86  3.3
The phase difference between current and voltage is
X
tan   L  3.14
R
72 x
  tan 1 3.14   72 0 
rad
180

72 x
1
Time lag , t  

s
 180 x 2x50 250
Consider that -decays of nucleus
A
z
A  AZ24 Y  42 He
238
92
234
U  90
Th  42 He
238  92
 1.58
92
234  90
 1.60
After decay, neutron to proton ratio, n/p 
90
Before decay, neutron to proton ratio, n/p 
Thus ratio increases.
16.
(i)
a
(ii)
17.
When a battery is connected in series to two long parallel wires, the
currents in the two wires will be in opposite directions. Due to which
force of repulsion will be acting between them and they move further
apart.
When a battery is connected in parallel to two long parallel wires, the
currents in the two wires will be same direction. Due to which a force
of attraction will be acting between them and they come closer to each
other.
Here , C 2 and C 3 are in series

1
1
1
2
1




Cs 200 200 200 100
Cs  100 pF
Cs and C1 are in parallel

C p  Cs  C1  100  20 pF
Again C p and C 4 are in series , therefore the equivalent capaci tan ce of network is

1
1
1
1
1
3





C C p C 4 200 100 200

C

Vp  V4  300
200
pF  66.7x10 12 F
3
As C p and C 4 are in series
200
x10 12 x 300
3
 2x10 8 C
Ch arg e on C 4 is q 4  CV 
Potential difference across C4 is
V4 
q4
2x10 8

 200V
C 4 100x10 12
From (i), Vp  300  V4  300  200  100V
Potential difference across C1 is V1  Vp  100V
Charge on C1 is q1  C1V1  100x1012 x100  108 C
Potential difference across C2 and C3 in series  100V
Charge on C2 is q 2  C2 V2  200x1012 x50  108 C
Charge on C3 is q 3  C3V3  200x10 12 x50  10 8 C
18.
(i) Phenomenon of polarisaiton is shown by transverse waves only. X-rays
are transverse in nature, and hence they can be polarized.
According to Malus law
I  I 0 cos 2 
Where
I = intensity of light coming from analyser
I0 = initial intensity.
When   450 , then I  I 0 cos 2 450 
I0
2
When   900 , then I  I 0 cos 2 900  0
When   1800 , then I  I 0 cos 2 1800  l 0
19.
A brief description of the various element is as given below:(i)
A microphone converts sound waves into electrical waves i.e. audio
signal.
(ii) An oscillator generates carrier waves.
(iii) There is a mixing of carrier waves and audio signal in modulator.
(iv) The modulated waves are fed to transmitter, these waves are then
radiated through transmitting antenna.
(v) The receiving antenna receives the transmitting signal.
(vi) The detector demodulates (separate out) the audio signal from the
modulated waves.
(vii) The loudspeaker converts the audio signal back into sound waves.
20.
Applying Kirchhoff’s junction rule at junction B, we get
I1= I + I2
(i)
Applying Kirchhoff’s loop rule for closed loop ABCDEFA give
10 = IR + 10I1
Applying Kirchhoff’s loop rule for closed loop BCDEB gives
2 = 5I2 – IR = 5(I2-I)-IR
(Using (i))
Or
4 = 1011-10I – 2IR
(iii)
Equation (i) – (iii) gives
6 = 3IR + 10I
Or
10 

2  1 R  
3

Or
2  R  R eff I
Comparing with Veff  R  R eff I, we get
Veff  2V and R eff 
21.
10

3
Initially, the source has 90% of
32
15
P nuclides and 10%
say after 1 days, the source has 10% of
nuclides.

Initial no.
32
15
P of nuclides = 9x;
Initial no.
33
15
P of nuclides = x
Final no.
32
15
P of nuclides = y;
Final no.
33
15
P of nuclides = 9y
n
1 1
As
   
N
2
2
0  
N
t
or N  N ( 2)
0
t
32
15
33
15
P nuclides. Finally,
P nuclides and 90% of
33
15
P
T
1/ 2  2  t T
1/ 2
T
1/ 2
For 1533P isotope
N 0 x; N  9 y; T  25.3 days
1

21.
2
9 y  x ( 2)
t
25.3
Here, distance between parallel plates d = 4mm = 0.004m, K =, thickness, t3mm = 0.003m
Let the new distance between the plate bed d1.
Dividing Eq. (ii ) by Eq. (i), we get
11t
1 t  14.3  25.3 
9  ( 2)
or 841  2 14.3x 25.3
9
11t

log10 81 
log10 2
14.3x 25.3
11tx 0.3010
or
1.9085 
14.3x 25.3
1.9085 x14.3x 25.3

t
 208 .5 days
11x 0.3010
A
0 A

C  0 and C1 
1
d

d 1  t 1  
 K
2
Since C1  C
3
0 A
2 0 A


1 3 d

d 1  t 1  
 K
1
2

1  3d

d 1  t 1  
 K
1
1

 1  3x 0.004
d1  0.0031  
 3
1
1

2 0.006
d1  0.003x
3
1
1

d1  0.002 0.006
1
1
d1  0.002  0.006
d1  0.006  0.002  0.008 m  8mm
Or
(Given )
Electric potential at a point is the amount of work done to bring a unit
positive charge from infinity to that point against the electrostatic forces.
AC  BD 
 2   2
2
2
 2m
 AO  OC  BO  OD  1 m
Potential at the Centre of the square Ois
V
qC
1  q
qB
qD 



4 o  AO BO OC OD 
100 x10 6 50 x10 6 20 x10 6 60 x10 6 
V  9x10 



1
1
1
1 

 9x10 9 x10 x10 6  9x10 4 V
9
23.
Here I D  2.0A, C  1F  10 6 F
dE
d
dE
  0 ( EA )   0 A
dt
dt
dt
d V
A dV
dV
0 A    0
C
dt  d 
d dt
dt
0 A 
V

 E  and C 

d
d 

dV I D 2.0
or

 6  2x10 6 V s 1
dt
C 10
I D  0
Thus a displacement current of 2.0 A can be set up by changing the potential
difference across the parallel plates of capacitor at the rate of 2 x 106 Vs-1
24.
Force between wires PQ and CD
F1 
 0 I1I 2 I
2 r
2x10 7 x15x 25x 0.25

 93.75x10 5
2
2x10
 9.375 x10 4 N (Re pulsive )
Force between wires PQ and AB
2x10 7 x15x 25x 0.25
 1.56 x10 4 N Attractive 
0.12
Net force on the rec tan gular loop
F2 
F  F1  F2  (9.375  1.56) x10 4
 7.815x10 4 N (Re pulsive i.e., towards left ).
25.
(a) Here, P1 = 15D, P2 = -5D
Power of the combination, P = P1 + P2
= 15 D – 5 D = 10 D
Focal length, f 
(b)

or
1 1
 x100  10 cm
P 10
1 1 1
 
f v u
1 1 1 1
1
1 3 1 2
  

or 

v f u 10  30
v
30
30
v  15cm
m
v h1

u h0
h1 
v
15

x 3  1.5cm
u  30
26.
(a) The values displayed by them is sharing and caring.
(b)
27.
Def – ½ Law ½ penvation -3
Advantage : At 220 V supply power loss due to heating effect is lesser.
Disadvantage : At 220 V peak value of current is more. Thus, it is more
dangerous.
(1)
Or
Explanation -1 Define -1 1 units & calculation
(3)
28.
Calculate
29.
Power – ½ rays dia -2 derivation – 2 significance – ½
Or
Statement -1
Diagrams -1 proof 1½, 1½
(a)
(b)
Working
OR Gate
(3)
(2)