# Physics: Sample Solution Paper- 5, Class 12 Notes | Study Physics Class 12 - NEET

## NEET: Physics: Sample Solution Paper- 5, Class 12 Notes | Study Physics Class 12 - NEET

The document Physics: Sample Solution Paper- 5, Class 12 Notes | Study Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
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``` Page 1

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

CBSE
Class XII – Physics
Sample Paper – 5 Solution

Section A
1. The net work done in moving a charge q along the closed rectangular path ABCDA is
zero.

2.
(i) Pole strength does not change.
(ii) Magnetic moment reduces to half.
OR
There will be an induced emf in the second position when the loop is exiting the field.

3. Microwaves possess greater energy and least angular speed, so they are used in RADAR.

OR
X-rays, infra red waves, visible rays, microwaves

4.

Image will be less bright.

5. Frequency of the incident radiation was kept constant.

Page 2

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

CBSE
Class XII – Physics
Sample Paper – 5 Solution

Section A
1. The net work done in moving a charge q along the closed rectangular path ABCDA is
zero.

2.
(i) Pole strength does not change.
(ii) Magnetic moment reduces to half.
OR
There will be an induced emf in the second position when the loop is exiting the field.

3. Microwaves possess greater energy and least angular speed, so they are used in RADAR.

OR
X-rays, infra red waves, visible rays, microwaves

4.

Image will be less bright.

5. Frequency of the incident radiation was kept constant.

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

Section B

6. The length of the edge of the cube isa , so

2
a
OA OB OC OD ? ? ? ?

o
12
and and 90 E E E AOB ? ? ? ?

The net electric field at O in the figure (i) is

22
o 1 2
2 ( ) E E E E i ? ? ?

o
E is directed at
o
45 from OD.
In figure (ii), the magnitude of E remains the same but the direction changes.

For potential in (i)
Potential at
? ? ? ? ? ? ? ?
o
1
4
Q Q Q Q
O
OA OB OC OD ??
? ? ? ? ? ?
? ? ? ?
??
??
= 0

and in case (ii)
Potential at
? ? ? ? ? ? ? ?
o
1
4
Q Q Q Q
O
OA OB OC OD ??
? ? ? ? ? ?
? ? ? ?
??
??
= 0
Hence, the potential remains the same.

Page 3

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

CBSE
Class XII – Physics
Sample Paper – 5 Solution

Section A
1. The net work done in moving a charge q along the closed rectangular path ABCDA is
zero.

2.
(i) Pole strength does not change.
(ii) Magnetic moment reduces to half.
OR
There will be an induced emf in the second position when the loop is exiting the field.

3. Microwaves possess greater energy and least angular speed, so they are used in RADAR.

OR
X-rays, infra red waves, visible rays, microwaves

4.

Image will be less bright.

5. Frequency of the incident radiation was kept constant.

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

Section B

6. The length of the edge of the cube isa , so

2
a
OA OB OC OD ? ? ? ?

o
12
and and 90 E E E AOB ? ? ? ?

The net electric field at O in the figure (i) is

22
o 1 2
2 ( ) E E E E i ? ? ?

o
E is directed at
o
45 from OD.
In figure (ii), the magnitude of E remains the same but the direction changes.

For potential in (i)
Potential at
? ? ? ? ? ? ? ?
o
1
4
Q Q Q Q
O
OA OB OC OD ??
? ? ? ? ? ?
? ? ? ?
??
??
= 0

and in case (ii)
Potential at
? ? ? ? ? ? ? ?
o
1
4
Q Q Q Q
O
OA OB OC OD ??
? ? ? ? ? ?
? ? ? ?
??
??
= 0
Hence, the potential remains the same.

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

OR

(i) 10 volts ? V x = 1 cm = 0.01 m

3
10
10 volt/metre
0.01
? ? ?
V
E
x

Direction of E is from surface A towards surface B.

(ii)
6
20 20 10 q C C ?
?
? ? ?
Work done
64
20 10 10 2 10 J
??
? ? ? ? ? ? W qV

7.
At higher values of voltage, the resistor shows non-ohmic character, so the resistor does
not obey Ohm’s law. This is the reason for the deviation of the I–V graph from the
straight line.

8. For the same glow of a bulb, the current in the resistance R (i.e. bulb) should remain the
same, i.e. the value of impedance
? ?
2
2
LC
Z R X X ? ? ? will remain the same, i.e. the value of ? ?
LC
XX ? should
remain the same.
We have
2
L
XL ?? ?
So, on doubling the value of ? , L should be halved for the same value of .
L
X
Also, we have

1
2
C
X
c ??
?

So, on doubling the value of ? ,C should be halved for the same value of
C
X .

OR
On inserting the magnetic core (rod) inside the solenoid, the value of
2
o r
NA
L
l
?? ??
?
??
??

will increase, and hence, the induced emf also increases which opposes the flow of
current in the circuit. Thus, the bulb will glow with less brightness.
Page 4

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

CBSE
Class XII – Physics
Sample Paper – 5 Solution

Section A
1. The net work done in moving a charge q along the closed rectangular path ABCDA is
zero.

2.
(i) Pole strength does not change.
(ii) Magnetic moment reduces to half.
OR
There will be an induced emf in the second position when the loop is exiting the field.

3. Microwaves possess greater energy and least angular speed, so they are used in RADAR.

OR
X-rays, infra red waves, visible rays, microwaves

4.

Image will be less bright.

5. Frequency of the incident radiation was kept constant.

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

Section B

6. The length of the edge of the cube isa , so

2
a
OA OB OC OD ? ? ? ?

o
12
and and 90 E E E AOB ? ? ? ?

The net electric field at O in the figure (i) is

22
o 1 2
2 ( ) E E E E i ? ? ?

o
E is directed at
o
45 from OD.
In figure (ii), the magnitude of E remains the same but the direction changes.

For potential in (i)
Potential at
? ? ? ? ? ? ? ?
o
1
4
Q Q Q Q
O
OA OB OC OD ??
? ? ? ? ? ?
? ? ? ?
??
??
= 0

and in case (ii)
Potential at
? ? ? ? ? ? ? ?
o
1
4
Q Q Q Q
O
OA OB OC OD ??
? ? ? ? ? ?
? ? ? ?
??
??
= 0
Hence, the potential remains the same.

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

OR

(i) 10 volts ? V x = 1 cm = 0.01 m

3
10
10 volt/metre
0.01
? ? ?
V
E
x

Direction of E is from surface A towards surface B.

(ii)
6
20 20 10 q C C ?
?
? ? ?
Work done
64
20 10 10 2 10 J
??
? ? ? ? ? ? W qV

7.
At higher values of voltage, the resistor shows non-ohmic character, so the resistor does
not obey Ohm’s law. This is the reason for the deviation of the I–V graph from the
straight line.

8. For the same glow of a bulb, the current in the resistance R (i.e. bulb) should remain the
same, i.e. the value of impedance
? ?
2
2
LC
Z R X X ? ? ? will remain the same, i.e. the value of ? ?
LC
XX ? should
remain the same.
We have
2
L
XL ?? ?
So, on doubling the value of ? , L should be halved for the same value of .
L
X
Also, we have

1
2
C
X
c ??
?

So, on doubling the value of ? ,C should be halved for the same value of
C
X .

OR
On inserting the magnetic core (rod) inside the solenoid, the value of
2
o r
NA
L
l
?? ??
?
??
??

will increase, and hence, the induced emf also increases which opposes the flow of
current in the circuit. Thus, the bulb will glow with less brightness.

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

9. As X-rays travel with the speed of light in vacuum, it shows that these are
electromagnetic waves.
Phenomenon of diffraction supports the wave-like characteristics of X-rays.
Polarisation depicts that X-rays are transverse in nature.

10.  When a strong current is passed through a semiconductor, it heats up. Thus, a large
number of covalent bonds break up in a semiconductor, resulting in a large number of
charge carriers. As a result, the material starts behaving as a conductor. At this stage,
the semiconductor loses the property of low conduction; hence, it is damaged.

11.
The side bands are at 1000 + 10 = 1010 kHz
And 1000 10 = 990 kHz.

12. Leti be the angle of incidence,
e be the angle of emergence,
A be the angle of the prism,
? be the angle of deviation.
The required expression is
? ? i e A ? ? ? ?
The refractive index ? of the material of prism is related with A ? and
m
? as

? ?
sin
2
sin 2
m
A
A
?
?
? ??
??
??
?

Section C

13. In the first case,

1
1
(1)
100
Rl
Sl
?
?

In the second case,

2
2
(2)
100
Rl
X S l
?
??

Dividing equation (2) by (1), we get

? ?
? ?
21
12
100
100
ll S
X S l l
?
?
??

Page 5

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

CBSE
Class XII – Physics
Sample Paper – 5 Solution

Section A
1. The net work done in moving a charge q along the closed rectangular path ABCDA is
zero.

2.
(i) Pole strength does not change.
(ii) Magnetic moment reduces to half.
OR
There will be an induced emf in the second position when the loop is exiting the field.

3. Microwaves possess greater energy and least angular speed, so they are used in RADAR.

OR
X-rays, infra red waves, visible rays, microwaves

4.

Image will be less bright.

5. Frequency of the incident radiation was kept constant.

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

Section B

6. The length of the edge of the cube isa , so

2
a
OA OB OC OD ? ? ? ?

o
12
and and 90 E E E AOB ? ? ? ?

The net electric field at O in the figure (i) is

22
o 1 2
2 ( ) E E E E i ? ? ?

o
E is directed at
o
45 from OD.
In figure (ii), the magnitude of E remains the same but the direction changes.

For potential in (i)
Potential at
? ? ? ? ? ? ? ?
o
1
4
Q Q Q Q
O
OA OB OC OD ??
? ? ? ? ? ?
? ? ? ?
??
??
= 0

and in case (ii)
Potential at
? ? ? ? ? ? ? ?
o
1
4
Q Q Q Q
O
OA OB OC OD ??
? ? ? ? ? ?
? ? ? ?
??
??
= 0
Hence, the potential remains the same.

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

OR

(i) 10 volts ? V x = 1 cm = 0.01 m

3
10
10 volt/metre
0.01
? ? ?
V
E
x

Direction of E is from surface A towards surface B.

(ii)
6
20 20 10 q C C ?
?
? ? ?
Work done
64
20 10 10 2 10 J
??
? ? ? ? ? ? W qV

7.
At higher values of voltage, the resistor shows non-ohmic character, so the resistor does
not obey Ohm’s law. This is the reason for the deviation of the I–V graph from the
straight line.

8. For the same glow of a bulb, the current in the resistance R (i.e. bulb) should remain the
same, i.e. the value of impedance
? ?
2
2
LC
Z R X X ? ? ? will remain the same, i.e. the value of ? ?
LC
XX ? should
remain the same.
We have
2
L
XL ?? ?
So, on doubling the value of ? , L should be halved for the same value of .
L
X
Also, we have

1
2
C
X
c ??
?

So, on doubling the value of ? ,C should be halved for the same value of
C
X .

OR
On inserting the magnetic core (rod) inside the solenoid, the value of
2
o r
NA
L
l
?? ??
?
??
??

will increase, and hence, the induced emf also increases which opposes the flow of
current in the circuit. Thus, the bulb will glow with less brightness.

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

9. As X-rays travel with the speed of light in vacuum, it shows that these are
electromagnetic waves.
Phenomenon of diffraction supports the wave-like characteristics of X-rays.
Polarisation depicts that X-rays are transverse in nature.

10.  When a strong current is passed through a semiconductor, it heats up. Thus, a large
number of covalent bonds break up in a semiconductor, resulting in a large number of
charge carriers. As a result, the material starts behaving as a conductor. At this stage,
the semiconductor loses the property of low conduction; hence, it is damaged.

11.
The side bands are at 1000 + 10 = 1010 kHz
And 1000 10 = 990 kHz.

12. Leti be the angle of incidence,
e be the angle of emergence,
A be the angle of the prism,
? be the angle of deviation.
The required expression is
? ? i e A ? ? ? ?
The refractive index ? of the material of prism is related with A ? and
m
? as

? ?
sin
2
sin 2
m
A
A
?
?
? ??
??
??
?

Section C

13. In the first case,

1
1
(1)
100
Rl
Sl
?
?

In the second case,

2
2
(2)
100
Rl
X S l
?
??

Dividing equation (2) by (1), we get

? ?
? ?
21
12
100
100
ll S
X S l l
?
?
??

CBSE XII  |  PHYSICS
Sample Paper – 5 Solution

? ?
? ?
12
21
100
100
ll XS
S l l
? ?
?
?

? ?
? ?
12
21
100
1
100
ll X
S l l
?
??
?

? ?
? ?
12
21
100
1
100
ll
XS
ll
?? ?
??
??
??
?
??

OR
A Wheatstone bridge is an arrangement of four resistances which can be used to
measure
one of them in terms of the rest.

I1 = I3 and I2 = I4.
Next, we apply Kirchhoff’s loop rule to closed loops ADBA and CBDC.
I1 R1 + 0 - I2 R2 = 0 (Ig = 0)
and the second loop CBDC gives, upon using I 3 = I 1, I 4 = I 2
-I4 R4 + 0 + I3 R3 = 0
Upon using I3 = I1, I4 = I2 and taking the ratio of the above two equations, we obtain

12
34
RR
RR
?
This last equation relating to the four resistors is called the balance condition for the
galvanometer to give zero or null deflection.

```
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