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# Sample Solution Paper 2 - Physics, Class 12 NEET Notes | EduRev

## NEET : Sample Solution Paper 2 - Physics, Class 12 NEET Notes | EduRev

``` Page 1

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

CBSE Board
Class XII – Physics
Sample Paper – 2 Solution

1. Since the current does not obey triangular law of vector addition.

2. The current induced in the coil will oppose the approach of magnet, so nearer face of
coil will act as north pole. Therefore on viewing from the magnet side the current in the
coil will be anticlockwise.

3. In forward bias the current is represented by

0
B
eV
i i exp( 1)
2k T
??
Where
0
i is called reverse saturation current, V is voltage across the diode,
B
k is
Boltzmann constant.

4. All electromagnetic waves travel in vacuum with same speed.

infrared
ultraviolet
Ratio = =1
c
c

5. The size of the object will be magnified by a factor of

? ?
o
e
f 60
M 12
f5
? ? ? ? ? ?

6. Nuclear density is independent of mass number, so ratio of nuclear densities is 1 : 1.

7. As the voltage of a. c. power transmitted from one station to another is very large, the
magnitude of the current is very low. Therefore, the power loss is reduced in
transmission of power as the power loss is given by
2
P i R ? .

8. Because at absolute zero temperature semiconductor behaves as insulator. Therefore
semiconductor devices do not work at absolute zero temperature.

9. Torque pEsin ? ? ?

9 4 1 o
C p 4 10 m, E 5 10 NC , 30
??
? ? ? ? ? ?

9 4o
sin 4 10 5 10 30
?
? ? ? ? ? ? ?

4
10 Nm
?
??

Page 2

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

CBSE Board
Class XII – Physics
Sample Paper – 2 Solution

1. Since the current does not obey triangular law of vector addition.

2. The current induced in the coil will oppose the approach of magnet, so nearer face of
coil will act as north pole. Therefore on viewing from the magnet side the current in the
coil will be anticlockwise.

3. In forward bias the current is represented by

0
B
eV
i i exp( 1)
2k T
??
Where
0
i is called reverse saturation current, V is voltage across the diode,
B
k is
Boltzmann constant.

4. All electromagnetic waves travel in vacuum with same speed.

infrared
ultraviolet
Ratio = =1
c
c

5. The size of the object will be magnified by a factor of

? ?
o
e
f 60
M 12
f5
? ? ? ? ? ?

6. Nuclear density is independent of mass number, so ratio of nuclear densities is 1 : 1.

7. As the voltage of a. c. power transmitted from one station to another is very large, the
magnitude of the current is very low. Therefore, the power loss is reduced in
transmission of power as the power loss is given by
2
P i R ? .

8. Because at absolute zero temperature semiconductor behaves as insulator. Therefore
semiconductor devices do not work at absolute zero temperature.

9. Torque pEsin ? ? ?

9 4 1 o
C p 4 10 m, E 5 10 NC , 30
??
? ? ? ? ? ?

9 4o
sin 4 10 5 10 30
?
? ? ? ? ? ? ?

4
10 Nm
?
??

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

10. Resistance
2
R
Ar
??
??
?

New radius r ' r 2 ?
New length '4 ?
?New resistance
? ?
? ?
1
1
' 2 2 2
/4
RR
rr
r / 2
?
??
? ? ? ?
??
?

Resistance of the wire will remain unchanged.

11. The fringe width in Young’s double slit experiment comes out to be
D
d
?
?? , where D is
the distance between slits and screen, d is slit width and ? is wavelength of light used.

If ? is small and d is large, then the fringe width of interference pattern becomes so
small that interference pattern becomes non-observable.
Therefore Young’s double slit experiment the slit size is taken of the order of the
wavelength of light used.

12.  The magnetization of ferromagnetic materials depends both on the applied magnetic
field and also on the history of magnetization (i. e., how many cycles of magnetization it
has gone through etc.)
In other words, the value of magnetization is a record or memory of its cycles of
magnetization. Since information bits can be made to correspond to these cycles.
Therefore ferromagnetic materials which show hysteresis loop are used in memory
devices.         1

13. Wavelength
8
5
12
c 3 10
5 10 m
6 10
?
?
?
? ? ? ? ?
?

This wavelength corresponds to infrared waves.
Applications of infrared waves:
(i) They are used in green houses to warm the plants.
(ii) They are used in taking photographs during fogs.

14. Given that
Refractive index of lens material with respect to air
l
n = 1.5
Focal length of lens in air
air
f = 18 cm
Refractive index of water
w
n = 4 / 3
Since,

12
12
1 1 1
( n 1)( )
f R R
? ? ?
1
2

For air medium
Page 3

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

CBSE Board
Class XII – Physics
Sample Paper – 2 Solution

1. Since the current does not obey triangular law of vector addition.

2. The current induced in the coil will oppose the approach of magnet, so nearer face of
coil will act as north pole. Therefore on viewing from the magnet side the current in the
coil will be anticlockwise.

3. In forward bias the current is represented by

0
B
eV
i i exp( 1)
2k T
??
Where
0
i is called reverse saturation current, V is voltage across the diode,
B
k is
Boltzmann constant.

4. All electromagnetic waves travel in vacuum with same speed.

infrared
ultraviolet
Ratio = =1
c
c

5. The size of the object will be magnified by a factor of

? ?
o
e
f 60
M 12
f5
? ? ? ? ? ?

6. Nuclear density is independent of mass number, so ratio of nuclear densities is 1 : 1.

7. As the voltage of a. c. power transmitted from one station to another is very large, the
magnitude of the current is very low. Therefore, the power loss is reduced in
transmission of power as the power loss is given by
2
P i R ? .

8. Because at absolute zero temperature semiconductor behaves as insulator. Therefore
semiconductor devices do not work at absolute zero temperature.

9. Torque pEsin ? ? ?

9 4 1 o
C p 4 10 m, E 5 10 NC , 30
??
? ? ? ? ? ?

9 4o
sin 4 10 5 10 30
?
? ? ? ? ? ? ?

4
10 Nm
?
??

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

10. Resistance
2
R
Ar
??
??
?

New radius r ' r 2 ?
New length '4 ?
?New resistance
? ?
? ?
1
1
' 2 2 2
/4
RR
rr
r / 2
?
??
? ? ? ?
??
?

Resistance of the wire will remain unchanged.

11. The fringe width in Young’s double slit experiment comes out to be
D
d
?
?? , where D is
the distance between slits and screen, d is slit width and ? is wavelength of light used.

If ? is small and d is large, then the fringe width of interference pattern becomes so
small that interference pattern becomes non-observable.
Therefore Young’s double slit experiment the slit size is taken of the order of the
wavelength of light used.

12.  The magnetization of ferromagnetic materials depends both on the applied magnetic
field and also on the history of magnetization (i. e., how many cycles of magnetization it
has gone through etc.)
In other words, the value of magnetization is a record or memory of its cycles of
magnetization. Since information bits can be made to correspond to these cycles.
Therefore ferromagnetic materials which show hysteresis loop are used in memory
devices.         1

13. Wavelength
8
5
12
c 3 10
5 10 m
6 10
?
?
?
? ? ? ? ?
?

This wavelength corresponds to infrared waves.
Applications of infrared waves:
(i) They are used in green houses to warm the plants.
(ii) They are used in taking photographs during fogs.

14. Given that
Refractive index of lens material with respect to air
l
n = 1.5
Focal length of lens in air
air
f = 18 cm
Refractive index of water
w
n = 4 / 3
Since,

12
12
1 1 1
( n 1)( )
f R R
? ? ?
1
2

For air medium

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

12
1 1 1
(1.5 1)( )
18 RR
? ? ?  …… (1)
For water medium

m 1 2
1 1.5 1 1
( 1)( )
f 4 / 3 R R
? ? ?  …… (2)
Dividing (1) by (2) and simplifying, we get

m
f = + 72 cm
Therefore change in focal length = 72 - 18 = 54 cm.

15. Capacitance of parallel plate capacitor
o
A
CA
d
?
?
?
Plate area of
2
C is double to that of
1 2 1
C ,thus, C 2C . ?
Slope of q – V graph ??
q
C
V

As slope of A is greater than slope of B, so A corresponds to larger capacitance, i.e.,
2
and CB to smaller capacitance, i.e.,
1
C .

16. According to Einstein, when a photon of incident light strikes on a bound electron of
metal, its energy is used in two ways.
(i) In overcoming work function of metal to free metallic electron.
(ii) In imparting kinetic energy to this freed electron.
i.e.,
k
h w E ? ? ?
When
k0
E 0, ? ? ? ? (threshold frequency), then

0
0
h w 0
wh
? ? ?
??

Therefore,
0k
h h E ? ? ? ?
? ?
0 k
E h v v ??
So, as the frequency of incident radiation ? increases the max. KE of photoelectrons also
increases.      OR

From Einstein’s photoelectric equation

k0
E h h ? ? ? ?

0
eV h h ? ? ? ? (V = stopping potential)
0
hh
V
ee
? ? ? ?

Page 4

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

CBSE Board
Class XII – Physics
Sample Paper – 2 Solution

1. Since the current does not obey triangular law of vector addition.

2. The current induced in the coil will oppose the approach of magnet, so nearer face of
coil will act as north pole. Therefore on viewing from the magnet side the current in the
coil will be anticlockwise.

3. In forward bias the current is represented by

0
B
eV
i i exp( 1)
2k T
??
Where
0
i is called reverse saturation current, V is voltage across the diode,
B
k is
Boltzmann constant.

4. All electromagnetic waves travel in vacuum with same speed.

infrared
ultraviolet
Ratio = =1
c
c

5. The size of the object will be magnified by a factor of

? ?
o
e
f 60
M 12
f5
? ? ? ? ? ?

6. Nuclear density is independent of mass number, so ratio of nuclear densities is 1 : 1.

7. As the voltage of a. c. power transmitted from one station to another is very large, the
magnitude of the current is very low. Therefore, the power loss is reduced in
transmission of power as the power loss is given by
2
P i R ? .

8. Because at absolute zero temperature semiconductor behaves as insulator. Therefore
semiconductor devices do not work at absolute zero temperature.

9. Torque pEsin ? ? ?

9 4 1 o
C p 4 10 m, E 5 10 NC , 30
??
? ? ? ? ? ?

9 4o
sin 4 10 5 10 30
?
? ? ? ? ? ? ?

4
10 Nm
?
??

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

10. Resistance
2
R
Ar
??
??
?

New radius r ' r 2 ?
New length '4 ?
?New resistance
? ?
? ?
1
1
' 2 2 2
/4
RR
rr
r / 2
?
??
? ? ? ?
??
?

Resistance of the wire will remain unchanged.

11. The fringe width in Young’s double slit experiment comes out to be
D
d
?
?? , where D is
the distance between slits and screen, d is slit width and ? is wavelength of light used.

If ? is small and d is large, then the fringe width of interference pattern becomes so
small that interference pattern becomes non-observable.
Therefore Young’s double slit experiment the slit size is taken of the order of the
wavelength of light used.

12.  The magnetization of ferromagnetic materials depends both on the applied magnetic
field and also on the history of magnetization (i. e., how many cycles of magnetization it
has gone through etc.)
In other words, the value of magnetization is a record or memory of its cycles of
magnetization. Since information bits can be made to correspond to these cycles.
Therefore ferromagnetic materials which show hysteresis loop are used in memory
devices.         1

13. Wavelength
8
5
12
c 3 10
5 10 m
6 10
?
?
?
? ? ? ? ?
?

This wavelength corresponds to infrared waves.
Applications of infrared waves:
(i) They are used in green houses to warm the plants.
(ii) They are used in taking photographs during fogs.

14. Given that
Refractive index of lens material with respect to air
l
n = 1.5
Focal length of lens in air
air
f = 18 cm
Refractive index of water
w
n = 4 / 3
Since,

12
12
1 1 1
( n 1)( )
f R R
? ? ?
1
2

For air medium

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

12
1 1 1
(1.5 1)( )
18 RR
? ? ?  …… (1)
For water medium

m 1 2
1 1.5 1 1
( 1)( )
f 4 / 3 R R
? ? ?  …… (2)
Dividing (1) by (2) and simplifying, we get

m
f = + 72 cm
Therefore change in focal length = 72 - 18 = 54 cm.

15. Capacitance of parallel plate capacitor
o
A
CA
d
?
?
?
Plate area of
2
C is double to that of
1 2 1
C ,thus, C 2C . ?
Slope of q – V graph ??
q
C
V

As slope of A is greater than slope of B, so A corresponds to larger capacitance, i.e.,
2
and CB to smaller capacitance, i.e.,
1
C .

16. According to Einstein, when a photon of incident light strikes on a bound electron of
metal, its energy is used in two ways.
(i) In overcoming work function of metal to free metallic electron.
(ii) In imparting kinetic energy to this freed electron.
i.e.,
k
h w E ? ? ?
When
k0
E 0, ? ? ? ? (threshold frequency), then

0
0
h w 0
wh
? ? ?
??

Therefore,
0k
h h E ? ? ? ?
? ?
0 k
E h v v ??
So, as the frequency of incident radiation ? increases the max. KE of photoelectrons also
increases.      OR

From Einstein’s photoelectric equation

k0
E h h ? ? ? ?

0
eV h h ? ? ? ? (V = stopping potential)
0
hh
V
ee
? ? ? ?

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

Thus,  vs V ? graph is a straight line of form y mx c ?? and the slope of graph is
h
m
e
? .

17. (i)According to Gauss’s law electric flux through S 1
and

1
0
2
00
1
2
1
Q
11
(Q 2Q) .3Q
1

3
??
?
? ? ? ?
??
?
?
?

(ii) When a medium of dielectric constant K = 5 is introduced in the space inside S 1 in place
of air, flux through S 1 will be modified to
11
1
0
11
' Q Q
k k 5
??
? ? ? ? ?
??

Thus, the flux will be reduced to (1/5) of its previous value.

18. The described arrangement is equivalent to two capacitors joined in parallel where
area of plates of either capacitor is
2
A
.
Thus,
10
10
1
A
k ( )
kA
2
C
d 2d
?
?
??
and
20
20
2
A
k ( )
kA
2
C
d 2d
?
?
??
Therefore net capacitance of the capacitor

12
C C C ?? =
1 0 2 0
k A k A
2d 2d
??
?
=
0 12
Akk
()
d2
? ?

Page 5

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

CBSE Board
Class XII – Physics
Sample Paper – 2 Solution

1. Since the current does not obey triangular law of vector addition.

2. The current induced in the coil will oppose the approach of magnet, so nearer face of
coil will act as north pole. Therefore on viewing from the magnet side the current in the
coil will be anticlockwise.

3. In forward bias the current is represented by

0
B
eV
i i exp( 1)
2k T
??
Where
0
i is called reverse saturation current, V is voltage across the diode,
B
k is
Boltzmann constant.

4. All electromagnetic waves travel in vacuum with same speed.

infrared
ultraviolet
Ratio = =1
c
c

5. The size of the object will be magnified by a factor of

? ?
o
e
f 60
M 12
f5
? ? ? ? ? ?

6. Nuclear density is independent of mass number, so ratio of nuclear densities is 1 : 1.

7. As the voltage of a. c. power transmitted from one station to another is very large, the
magnitude of the current is very low. Therefore, the power loss is reduced in
transmission of power as the power loss is given by
2
P i R ? .

8. Because at absolute zero temperature semiconductor behaves as insulator. Therefore
semiconductor devices do not work at absolute zero temperature.

9. Torque pEsin ? ? ?

9 4 1 o
C p 4 10 m, E 5 10 NC , 30
??
? ? ? ? ? ?

9 4o
sin 4 10 5 10 30
?
? ? ? ? ? ? ?

4
10 Nm
?
??

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

10. Resistance
2
R
Ar
??
??
?

New radius r ' r 2 ?
New length '4 ?
?New resistance
? ?
? ?
1
1
' 2 2 2
/4
RR
rr
r / 2
?
??
? ? ? ?
??
?

Resistance of the wire will remain unchanged.

11. The fringe width in Young’s double slit experiment comes out to be
D
d
?
?? , where D is
the distance between slits and screen, d is slit width and ? is wavelength of light used.

If ? is small and d is large, then the fringe width of interference pattern becomes so
small that interference pattern becomes non-observable.
Therefore Young’s double slit experiment the slit size is taken of the order of the
wavelength of light used.

12.  The magnetization of ferromagnetic materials depends both on the applied magnetic
field and also on the history of magnetization (i. e., how many cycles of magnetization it
has gone through etc.)
In other words, the value of magnetization is a record or memory of its cycles of
magnetization. Since information bits can be made to correspond to these cycles.
Therefore ferromagnetic materials which show hysteresis loop are used in memory
devices.         1

13. Wavelength
8
5
12
c 3 10
5 10 m
6 10
?
?
?
? ? ? ? ?
?

This wavelength corresponds to infrared waves.
Applications of infrared waves:
(i) They are used in green houses to warm the plants.
(ii) They are used in taking photographs during fogs.

14. Given that
Refractive index of lens material with respect to air
l
n = 1.5
Focal length of lens in air
air
f = 18 cm
Refractive index of water
w
n = 4 / 3
Since,

12
12
1 1 1
( n 1)( )
f R R
? ? ?
1
2

For air medium

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

12
1 1 1
(1.5 1)( )
18 RR
? ? ?  …… (1)
For water medium

m 1 2
1 1.5 1 1
( 1)( )
f 4 / 3 R R
? ? ?  …… (2)
Dividing (1) by (2) and simplifying, we get

m
f = + 72 cm
Therefore change in focal length = 72 - 18 = 54 cm.

15. Capacitance of parallel plate capacitor
o
A
CA
d
?
?
?
Plate area of
2
C is double to that of
1 2 1
C ,thus, C 2C . ?
Slope of q – V graph ??
q
C
V

As slope of A is greater than slope of B, so A corresponds to larger capacitance, i.e.,
2
and CB to smaller capacitance, i.e.,
1
C .

16. According to Einstein, when a photon of incident light strikes on a bound electron of
metal, its energy is used in two ways.
(i) In overcoming work function of metal to free metallic electron.
(ii) In imparting kinetic energy to this freed electron.
i.e.,
k
h w E ? ? ?
When
k0
E 0, ? ? ? ? (threshold frequency), then

0
0
h w 0
wh
? ? ?
??

Therefore,
0k
h h E ? ? ? ?
? ?
0 k
E h v v ??
So, as the frequency of incident radiation ? increases the max. KE of photoelectrons also
increases.      OR

From Einstein’s photoelectric equation

k0
E h h ? ? ? ?

0
eV h h ? ? ? ? (V = stopping potential)
0
hh
V
ee
? ? ? ?

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

Thus,  vs V ? graph is a straight line of form y mx c ?? and the slope of graph is
h
m
e
? .

17. (i)According to Gauss’s law electric flux through S 1
and

1
0
2
00
1
2
1
Q
11
(Q 2Q) .3Q
1

3
??
?
? ? ? ?
??
?
?
?

(ii) When a medium of dielectric constant K = 5 is introduced in the space inside S 1 in place
of air, flux through S 1 will be modified to
11
1
0
11
' Q Q
k k 5
??
? ? ? ? ?
??

Thus, the flux will be reduced to (1/5) of its previous value.

18. The described arrangement is equivalent to two capacitors joined in parallel where
area of plates of either capacitor is
2
A
.
Thus,
10
10
1
A
k ( )
kA
2
C
d 2d
?
?
??
and
20
20
2
A
k ( )
kA
2
C
d 2d
?
?
??
Therefore net capacitance of the capacitor

12
C C C ?? =
1 0 2 0
k A k A
2d 2d
??
?
=
0 12
Akk
()
d2
? ?

CBSE XII  |  PHYSICS
Sample Paper – 2 Solution

19. The given square ABCD of side 10 cm is one face of a cube of side 10 cm.  At the
centre of this cube a charge + q = +10 C ? is placed.

According to Gauss’s theorem, total electric through the six faces of cube
0
q
?
?

?Total electric flux through square face ABCD of the cube
0
1
6
q
?
?
6
5 2 1
12
1 10 10
1.88 10
6 8.85 10
?
?
?
?
? ? ? ?
?
Nm C .
OR

Since the air is replaced by another dielectric medium of dielectric constant 10 without
disconnecting the capacitor from d. c. source. Hence, potential difference between the
plates of capacitor remains unchanged. Consequently, electric field between the plates (
V
E
d
? ) remains unchanged.

Now capacitance of capacitor with dielectric
m
C kC 10C ??
Therefore, the new energy stored in capacitor is
2 2 2
mm
1 1 1
U C V (10C)V 10( CV ) 10U
2 2 2
? ? ? ?
Thus, energy stored becomes 10 times the value with air as dielectric.

```
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## Physics Class 12

204 videos|288 docs|125 tests

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