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Class 12 Physics: CBSE Marking Scheme (2025-26)
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Page 1
PHYSICS – Code No. 042
MARKING SCHEME
CLASS – XII (2025 – 26)
SECTION A
Q.No Questions
Marks
1. Answer: (A)
Both are having equal charges
For two bodies to be in equilibrium, both should have same potential(V).
As V=
?? ??
Where C of sphere is 4?? ?? ?? ?? . Which is independent of all the factors
mentioned in options.
1
2. Answer: (A)
Diameter of copper wire d,
Diameter of cylindrical iron is D
No.of turns N,(D>>d)
Length=N x Circumference of cylinder
L= NpD
R=
????
?? =
???????? ?? 2
?? 4
R=
4?????? ?? 2
1
3. Answer: (A)
When the frequency of the AC source is increased than the impedance of
the device decreases.
As in phasor diagram current leads the voltage, so given appliance is
capacitor.
1
4. Answer: (D)
The energy of radio waves is lesser than that of the gamma rays.
Since the frequency of radio waves is less than gamma waves.
E = h?
Hence, energy of radio waves is less than gamma waves
1
Page 2
PHYSICS – Code No. 042
MARKING SCHEME
CLASS – XII (2025 – 26)
SECTION A
Q.No Questions
Marks
1. Answer: (A)
Both are having equal charges
For two bodies to be in equilibrium, both should have same potential(V).
As V=
?? ??
Where C of sphere is 4?? ?? ?? ?? . Which is independent of all the factors
mentioned in options.
1
2. Answer: (A)
Diameter of copper wire d,
Diameter of cylindrical iron is D
No.of turns N,(D>>d)
Length=N x Circumference of cylinder
L= NpD
R=
????
?? =
???????? ?? 2
?? 4
R=
4?????? ?? 2
1
3. Answer: (A)
When the frequency of the AC source is increased than the impedance of
the device decreases.
As in phasor diagram current leads the voltage, so given appliance is
capacitor.
1
4. Answer: (D)
The energy of radio waves is lesser than that of the gamma rays.
Since the frequency of radio waves is less than gamma waves.
E = h?
Hence, energy of radio waves is less than gamma waves
1
5. Answer: (A)
Total Internal reflection
For VI- Students
Answer: (D)
?? 1
?? =
?????? O
?? ?????? 90
cSinO
1
6. Answer: (D)
Slit width increases hence amplitude will increase, so intensity will also
increase.
For VI- Students
Answer: (B)
Interference
1
1
7. Answer: (C)
IV
Transition III, V , VI corresponds to absorption of energy.
Maximum emitted wavelength corresponds minimum energy difference.
??? ??
> ??? ????
> ??? ????
Therefore, maximum emitted wavelength corresponds to transition IV .
For VI- Students
Transition III, V , VI corresponds to absorption of energy.
Maximum emitted wavelength corresponds minimum energy difference.
??? ????
> ??? ??
> ??? ????
Therefore, maximum emitted wavelength corresponds to transition IV .
1
8. Answer: (D)
The charged particle will move with constant velocity.
As charge particle is moving parallel to magnetic field, there will be no
acceleration.
1
Page 3
PHYSICS – Code No. 042
MARKING SCHEME
CLASS – XII (2025 – 26)
SECTION A
Q.No Questions
Marks
1. Answer: (A)
Both are having equal charges
For two bodies to be in equilibrium, both should have same potential(V).
As V=
?? ??
Where C of sphere is 4?? ?? ?? ?? . Which is independent of all the factors
mentioned in options.
1
2. Answer: (A)
Diameter of copper wire d,
Diameter of cylindrical iron is D
No.of turns N,(D>>d)
Length=N x Circumference of cylinder
L= NpD
R=
????
?? =
???????? ?? 2
?? 4
R=
4?????? ?? 2
1
3. Answer: (A)
When the frequency of the AC source is increased than the impedance of
the device decreases.
As in phasor diagram current leads the voltage, so given appliance is
capacitor.
1
4. Answer: (D)
The energy of radio waves is lesser than that of the gamma rays.
Since the frequency of radio waves is less than gamma waves.
E = h?
Hence, energy of radio waves is less than gamma waves
1
5. Answer: (A)
Total Internal reflection
For VI- Students
Answer: (D)
?? 1
?? =
?????? O
?? ?????? 90
cSinO
1
6. Answer: (D)
Slit width increases hence amplitude will increase, so intensity will also
increase.
For VI- Students
Answer: (B)
Interference
1
1
7. Answer: (C)
IV
Transition III, V , VI corresponds to absorption of energy.
Maximum emitted wavelength corresponds minimum energy difference.
??? ??
> ??? ????
> ??? ????
Therefore, maximum emitted wavelength corresponds to transition IV .
For VI- Students
Transition III, V , VI corresponds to absorption of energy.
Maximum emitted wavelength corresponds minimum energy difference.
??? ????
> ??? ??
> ??? ????
Therefore, maximum emitted wavelength corresponds to transition IV .
1
8. Answer: (D)
The charged particle will move with constant velocity.
As charge particle is moving parallel to magnetic field, there will be no
acceleration.
1
9. Answer: (C)
more for the magnet falling through the solenoid.
Emf will be induced in solenoid due to motion of magnet through it. As
per Lenz’s law induced emf will oppose the motion of magnet.
1
10. Answer: (C)
V=2V
o
sin 2?t
As V= NBA?? sin ????
1
11. Answer: (D)
1:1
Nuclear density does not depend on mass number.
1
12. Answer: (B)
The deflection of the magnetic needle at P and Q will be in the opposite
directions.
As magnetic field at equator is antiparallel to magnetic field at pole.
1
13. Answer: (B)
both Assertion and Reason are true but Reason is not the correct
explanation of Assertion.
1
14. Answer: (C)
Assertion is true but Reason is false.
1
15. Answer: (D)
both Assertion and Reason are false
1
16. Answer: (B)
both Assertion and Reason are true but Reason is not the correct
explanation of Assertion.
If three point charges are in equilibrium then forces acting on each
charges should be linearly opposite.
1
Page 4
PHYSICS – Code No. 042
MARKING SCHEME
CLASS – XII (2025 – 26)
SECTION A
Q.No Questions
Marks
1. Answer: (A)
Both are having equal charges
For two bodies to be in equilibrium, both should have same potential(V).
As V=
?? ??
Where C of sphere is 4?? ?? ?? ?? . Which is independent of all the factors
mentioned in options.
1
2. Answer: (A)
Diameter of copper wire d,
Diameter of cylindrical iron is D
No.of turns N,(D>>d)
Length=N x Circumference of cylinder
L= NpD
R=
????
?? =
???????? ?? 2
?? 4
R=
4?????? ?? 2
1
3. Answer: (A)
When the frequency of the AC source is increased than the impedance of
the device decreases.
As in phasor diagram current leads the voltage, so given appliance is
capacitor.
1
4. Answer: (D)
The energy of radio waves is lesser than that of the gamma rays.
Since the frequency of radio waves is less than gamma waves.
E = h?
Hence, energy of radio waves is less than gamma waves
1
5. Answer: (A)
Total Internal reflection
For VI- Students
Answer: (D)
?? 1
?? =
?????? O
?? ?????? 90
cSinO
1
6. Answer: (D)
Slit width increases hence amplitude will increase, so intensity will also
increase.
For VI- Students
Answer: (B)
Interference
1
1
7. Answer: (C)
IV
Transition III, V , VI corresponds to absorption of energy.
Maximum emitted wavelength corresponds minimum energy difference.
??? ??
> ??? ????
> ??? ????
Therefore, maximum emitted wavelength corresponds to transition IV .
For VI- Students
Transition III, V , VI corresponds to absorption of energy.
Maximum emitted wavelength corresponds minimum energy difference.
??? ????
> ??? ??
> ??? ????
Therefore, maximum emitted wavelength corresponds to transition IV .
1
8. Answer: (D)
The charged particle will move with constant velocity.
As charge particle is moving parallel to magnetic field, there will be no
acceleration.
1
9. Answer: (C)
more for the magnet falling through the solenoid.
Emf will be induced in solenoid due to motion of magnet through it. As
per Lenz’s law induced emf will oppose the motion of magnet.
1
10. Answer: (C)
V=2V
o
sin 2?t
As V= NBA?? sin ????
1
11. Answer: (D)
1:1
Nuclear density does not depend on mass number.
1
12. Answer: (B)
The deflection of the magnetic needle at P and Q will be in the opposite
directions.
As magnetic field at equator is antiparallel to magnetic field at pole.
1
13. Answer: (B)
both Assertion and Reason are true but Reason is not the correct
explanation of Assertion.
1
14. Answer: (C)
Assertion is true but Reason is false.
1
15. Answer: (D)
both Assertion and Reason are false
1
16. Answer: (B)
both Assertion and Reason are true but Reason is not the correct
explanation of Assertion.
If three point charges are in equilibrium then forces acting on each
charges should be linearly opposite.
1
SECTION B
17. Given, B
o
= 510 nT = 510 x 10
-9
T
? = 60 x 10
6
rad/sec
E
o
= cB
o
= 153 N/C
k = ?/c = 20 x 10
-2
rad/m
E = E
o
sin (?t – kz)
E = 153 sin (60 x 10
6
t – 20 x10
-2
x) N/C
1
1
18. (I) E.m.f of the cell is 6V, As when load current is zero potential
difference becomes equal to emf of the cell.
(II) Explanation: The internal resistance of a cell can be determined as the
negative slope of its voltage–current graph.
First, we can determine the slope by choosing two points on the line:
Slope =
?? -?? ???? -?? = - 0.5
This means that the internal resistance must be 0.50 ohm (O).
For VI-Candidates
E = V + v = IR + Ir
(where V is potential drop in the external circuit and v is potential drop in
the cell)
Or, E = I (R + r)
Or, I = E / (R + r)
This is the relation.
1
1
1
1
19. From Gauss’s theorem
Ø=
?? ?? ?? ?? ?? [Where ?? ?? is relative permittivity of medium inside Gaussian
surface]
For sphere,
Ø
sphere
=
?? ?? ?????????? ?? ?? ……………..(i)
For cube
Ø
cube
=
2?? ?? ?? …………………(ii)
Dividing (i) by (ii)
Øsphere
Øcube
=
1
2?? ?????????? =
1
160
½
½
1
Page 5
PHYSICS – Code No. 042
MARKING SCHEME
CLASS – XII (2025 – 26)
SECTION A
Q.No Questions
Marks
1. Answer: (A)
Both are having equal charges
For two bodies to be in equilibrium, both should have same potential(V).
As V=
?? ??
Where C of sphere is 4?? ?? ?? ?? . Which is independent of all the factors
mentioned in options.
1
2. Answer: (A)
Diameter of copper wire d,
Diameter of cylindrical iron is D
No.of turns N,(D>>d)
Length=N x Circumference of cylinder
L= NpD
R=
????
?? =
???????? ?? 2
?? 4
R=
4?????? ?? 2
1
3. Answer: (A)
When the frequency of the AC source is increased than the impedance of
the device decreases.
As in phasor diagram current leads the voltage, so given appliance is
capacitor.
1
4. Answer: (D)
The energy of radio waves is lesser than that of the gamma rays.
Since the frequency of radio waves is less than gamma waves.
E = h?
Hence, energy of radio waves is less than gamma waves
1
5. Answer: (A)
Total Internal reflection
For VI- Students
Answer: (D)
?? 1
?? =
?????? O
?? ?????? 90
cSinO
1
6. Answer: (D)
Slit width increases hence amplitude will increase, so intensity will also
increase.
For VI- Students
Answer: (B)
Interference
1
1
7. Answer: (C)
IV
Transition III, V , VI corresponds to absorption of energy.
Maximum emitted wavelength corresponds minimum energy difference.
??? ??
> ??? ????
> ??? ????
Therefore, maximum emitted wavelength corresponds to transition IV .
For VI- Students
Transition III, V , VI corresponds to absorption of energy.
Maximum emitted wavelength corresponds minimum energy difference.
??? ????
> ??? ??
> ??? ????
Therefore, maximum emitted wavelength corresponds to transition IV .
1
8. Answer: (D)
The charged particle will move with constant velocity.
As charge particle is moving parallel to magnetic field, there will be no
acceleration.
1
9. Answer: (C)
more for the magnet falling through the solenoid.
Emf will be induced in solenoid due to motion of magnet through it. As
per Lenz’s law induced emf will oppose the motion of magnet.
1
10. Answer: (C)
V=2V
o
sin 2?t
As V= NBA?? sin ????
1
11. Answer: (D)
1:1
Nuclear density does not depend on mass number.
1
12. Answer: (B)
The deflection of the magnetic needle at P and Q will be in the opposite
directions.
As magnetic field at equator is antiparallel to magnetic field at pole.
1
13. Answer: (B)
both Assertion and Reason are true but Reason is not the correct
explanation of Assertion.
1
14. Answer: (C)
Assertion is true but Reason is false.
1
15. Answer: (D)
both Assertion and Reason are false
1
16. Answer: (B)
both Assertion and Reason are true but Reason is not the correct
explanation of Assertion.
If three point charges are in equilibrium then forces acting on each
charges should be linearly opposite.
1
SECTION B
17. Given, B
o
= 510 nT = 510 x 10
-9
T
? = 60 x 10
6
rad/sec
E
o
= cB
o
= 153 N/C
k = ?/c = 20 x 10
-2
rad/m
E = E
o
sin (?t – kz)
E = 153 sin (60 x 10
6
t – 20 x10
-2
x) N/C
1
1
18. (I) E.m.f of the cell is 6V, As when load current is zero potential
difference becomes equal to emf of the cell.
(II) Explanation: The internal resistance of a cell can be determined as the
negative slope of its voltage–current graph.
First, we can determine the slope by choosing two points on the line:
Slope =
?? -?? ???? -?? = - 0.5
This means that the internal resistance must be 0.50 ohm (O).
For VI-Candidates
E = V + v = IR + Ir
(where V is potential drop in the external circuit and v is potential drop in
the cell)
Or, E = I (R + r)
Or, I = E / (R + r)
This is the relation.
1
1
1
1
19. From Gauss’s theorem
Ø=
?? ?? ?? ?? ?? [Where ?? ?? is relative permittivity of medium inside Gaussian
surface]
For sphere,
Ø
sphere
=
?? ?? ?????????? ?? ?? ……………..(i)
For cube
Ø
cube
=
2?? ?? ?? …………………(ii)
Dividing (i) by (ii)
Øsphere
Øcube
=
1
2?? ?????????? =
1
160
½
½
1
20.
(I)
?? ?? =
µ
0
?? 1
?? 2
2????
(I
1
is the current in first wire and I
2
is the current in second wire)
Thus we define ampere as the current flowing in each conductor separated
by a
unit distance so that one conductor applies a force of 2 x 10
-7
N on a unit
length of another parallel conductor.
Or
1
1
20
(II)
For VI-Candidates
Gauss’s law for magnetism is: The net magnetic flux through any closed
surface is zero.
Hence magnetic flux linked to given sphere will also be zero.
1
1
1
1
21A. Smaller is the impact parameter, larger is the angle at which a – particles
scatters.
Larger is the impact parameter, a – particles scatter less keeping its
original trajectory.
For head on collision, the value of impact parameter is zero.
OR
1
1
Read MoreFAQs on Class 12 Physics: CBSE Marking Scheme (2025-26)
| 1. What is the CBSE marking scheme for Class 12 Physics? |  |
Ans. The CBSE marking scheme for Class 12 Physics typically includes a distribution of marks across various sections of the examination. The question paper is generally divided into sections that include very short answer questions, short answer questions, and long answer questions. Each section carries a specific weightage, contributing to the total marks of the exam. The scheme aims to assess a student's understanding of concepts, problem-solving skills, and application of Physics in real-life situations.
| 2. How can students prepare effectively for the Class 12 Physics exam? | |
Ans. Students can prepare effectively for the Class 12 Physics exam by following a structured study plan. This includes understanding the syllabus thoroughly, focusing on concepts rather than rote learning, and practising numerical problems regularly. Making use of NCERT textbooks, previous years' question papers, and sample papers can also enhance preparation. Group studies and discussions can further clarify doubts and reinforce knowledge. Regular revision and staying updated with the latest exam patterns are also crucial.
| 3. What are the key topics that students should focus on for the Class 12 Physics exam? | |
Ans. Key topics for the Class 12 Physics exam typically include Electrostatics, Current Electricity, Magnetic Effects of Current, Electromagnetic Induction, Optics, Dual Nature of Radiation and Matter, Atoms and Nuclei, and Electronic Devices. Understanding these concepts is essential as they form the basis for many applications in Physics. Students should also pay attention to practical experiments and their underlying principles, as they are often included in the examination.
| 4. Is there any practical component in the Class 12 Physics exam? | |
Ans. Yes, there is a practical component in the Class 12 Physics exam. Students are required to perform experiments and demonstrate their understanding of practical applications of Physics. The practical examination assesses students' skills in conducting experiments, recording observations, and analysing results. It is important for students to be familiar with the laboratory equipment and procedures, as well as the principles behind the experiments they perform.
| 5. How does the marking scheme for NEET differ from that of the Class 12 Physics exam? | |
Ans. The marking scheme for NEET differs significantly from that of the Class 12 Physics exam. NEET is a competitive entrance examination for medical courses and primarily focuses on Biology, Chemistry, and Physics. Each subject has a specific number of questions, and the marking scheme typically includes positive marking for correct answers and negative marking for incorrect responses. In contrast, the Class 12 Physics exam assesses overall understanding of the subject with a greater emphasis on concepts and practical applications, without the competitive aspect of negative marking.
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