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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.
The first loop ADBA gives
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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FAQs on Physics: Sample Solution Paper- 5, Class 12 - Sample Papers for Class 12 Medical and Non-Medical
| 1. What is the concept of potential energy in physics? |  |
Ans. Potential energy is the energy possessed by an object due to its position or state. It is stored energy that can be converted into other forms of energy, such as kinetic energy. In physics, potential energy is commonly expressed as gravitational potential energy or elastic potential energy, depending on the system.
| 2. What is the difference between speed and velocity? | |
Ans. Speed and velocity are both measures of how fast an object is moving, but they have different meanings. Speed is a scalar quantity that refers to the rate at which an object covers distance. Velocity, on the other hand, is a vector quantity that includes both the speed and the direction of motion. It specifies how fast an object is moving in a particular direction.
| 3. How does Newton's first law of motion relate to inertia? | |
Ans. Newton's first law of motion states that an object at rest will stay at rest, and an object in motion will stay in motion with the same velocity unless acted upon by an external force. This property of objects to resist changes in their state of motion is called inertia. Inertia is directly related to Newton's first law, as it explains why objects tend to maintain their current state of motion.
| 4. What are the different types of forces in physics? | |
Ans. In physics, there are several types of forces. Some of the commonly encountered forces include gravitational force, electromagnetic force, frictional force, normal force, tension force, and applied force. Each type of force has its own characteristics and effects on objects.
| 5. How does the conservation of energy apply to a closed system? | |
Ans. The principle of conservation of energy states that energy cannot be created or destroyed but can only be transferred or transformed from one form to another. This principle applies to closed systems, which are isolated from external influences. In a closed system, the total amount of energy remains constant, meaning the sum of all forms of energy (such as potential energy and kinetic energy) remains unchanged over time. This principle allows for the analysis and understanding of energy transfers and transformations within closed systems.
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Dec 25, 2024 Last updated |
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