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Class 12 Physics: CBSE Marking Scheme (2024-25)
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Page 1
MARKING SCHEME
PHYSICS
Subject Code – 042
CLASS – XII
Academic Session 2024 – 25
Maximum Marks:70 Time Allowed: 3hours
[SECTION – A]
Ans.1- (B) (1 mark)
VA> VB [VA = VC]
In the direction of electric field, the electric potential decreases.
Ans.2- (B) In the state of equilibrium, (1 mark)
The potential on the surface of bigger sphere = the potential at the surface of the smaller sphere
12
12
=
kq kq
rr
?
11
22
=
qr
qr
?
22
1 1 2 1 2 2
22
2 2 2 1
11
E
E
= = ? =
q r r r r
q r r
rr
Ans.3 - (C) (1 mark)
AtP2, B2 =
00
II
3 3
2
2
??
=
? ??
?
??
??
a a
AtP1, B1 =
( )
00
(I 4) I
2 2 4
??
=
?? aa
?
0
2
0 1
I
B 3
I B
4
? ??
??
?
??
=
? ??
??
?
??
a
a
?
2
1
B 4
B3
=
Ans.4 - (D)Sound waves as well as light waves (1 mark)
Ans.5 -(A) (1 mark)
Ans.6 - (C)When all the given components are connected(1 mark)
Page 2
MARKING SCHEME
PHYSICS
Subject Code – 042
CLASS – XII
Academic Session 2024 – 25
Maximum Marks:70 Time Allowed: 3hours
[SECTION – A]
Ans.1- (B) (1 mark)
VA> VB [VA = VC]
In the direction of electric field, the electric potential decreases.
Ans.2- (B) In the state of equilibrium, (1 mark)
The potential on the surface of bigger sphere = the potential at the surface of the smaller sphere
12
12
=
kq kq
rr
?
11
22
=
qr
qr
?
22
1 1 2 1 2 2
22
2 2 2 1
11
E
E
= = ? =
q r r r r
q r r
rr
Ans.3 - (C) (1 mark)
AtP2, B2 =
00
II
3 3
2
2
??
=
? ??
?
??
??
a a
AtP1, B1 =
( )
00
(I 4) I
2 2 4
??
=
?? aa
?
0
2
0 1
I
B 3
I B
4
? ??
??
?
??
=
? ??
??
?
??
a
a
?
2
1
B 4
B3
=
Ans.4 - (D)Sound waves as well as light waves (1 mark)
Ans.5 -(A) (1 mark)
Ans.6 - (C)When all the given components are connected(1 mark)
IR = IXC = IXL = 10V
XC = XL = R
Z =
22
CL
R (X X ) +-
Z =
22
R (R R) +-
Z = R
VS = IZ = IR = 10 V
So, the source voltage is also 10 V
When the capacitor is short circuited then
Z =
22
L
R (X ) +
= v?? 2
+ ?? 2
= ?? v2
VL = I ? XL =
10
R 5 2
2R
?= V
Ans.7 - (B)(1 mark)
Ans.8 - (B) The distance of closest approach(1 mark)
2
1
const
V
= d ...(1)
2
2
const
2
V
=
d
...(2)
From equations (1) and (2),
2
2
2
1
V
2
V
= ?V2 = 2 V1
? V2 = 2 V Given, (V1 = V)
Ans.9 - (C)(1 mark)
?? 2
?? -
?? 1
?? =
?? 2
- ?? 1
??
1 3 [1 3 2]
2[ 6] 6 v
-
-=
--
Page 3
MARKING SCHEME
PHYSICS
Subject Code – 042
CLASS – XII
Academic Session 2024 – 25
Maximum Marks:70 Time Allowed: 3hours
[SECTION – A]
Ans.1- (B) (1 mark)
VA> VB [VA = VC]
In the direction of electric field, the electric potential decreases.
Ans.2- (B) In the state of equilibrium, (1 mark)
The potential on the surface of bigger sphere = the potential at the surface of the smaller sphere
12
12
=
kq kq
rr
?
11
22
=
qr
qr
?
22
1 1 2 1 2 2
22
2 2 2 1
11
E
E
= = ? =
q r r r r
q r r
rr
Ans.3 - (C) (1 mark)
AtP2, B2 =
00
II
3 3
2
2
??
=
? ??
?
??
??
a a
AtP1, B1 =
( )
00
(I 4) I
2 2 4
??
=
?? aa
?
0
2
0 1
I
B 3
I B
4
? ??
??
?
??
=
? ??
??
?
??
a
a
?
2
1
B 4
B3
=
Ans.4 - (D)Sound waves as well as light waves (1 mark)
Ans.5 -(A) (1 mark)
Ans.6 - (C)When all the given components are connected(1 mark)
IR = IXC = IXL = 10V
XC = XL = R
Z =
22
CL
R (X X ) +-
Z =
22
R (R R) +-
Z = R
VS = IZ = IR = 10 V
So, the source voltage is also 10 V
When the capacitor is short circuited then
Z =
22
L
R (X ) +
= v?? 2
+ ?? 2
= ?? v2
VL = I ? XL =
10
R 5 2
2R
?= V
Ans.7 - (B)(1 mark)
Ans.8 - (B) The distance of closest approach(1 mark)
2
1
const
V
= d ...(1)
2
2
const
2
V
=
d
...(2)
From equations (1) and (2),
2
2
2
1
V
2
V
= ?V2 = 2 V1
? V2 = 2 V Given, (V1 = V)
Ans.9 - (C)(1 mark)
?? 2
?? -
?? 1
?? =
?? 2
- ?? 1
??
1 3 [1 3 2]
2[ 6] 6 v
-
-=
--
1 3 1 2 1
12 12 12 6 v
- - -
= + = =
?? = –6 cm
Ans.10 - (B)Diffraction (1 mark)
Ans.11- (A)doping level (1 mark)
Ans.12- (C)+0.4% (1 mark)
Ans.13- (A) (1 mark)
Ans.14- (A)(1 mark)
Ans.15- (D)(1 mark)
Ans.16- (A)(1 mark)
[SECTION – B]
Ans.17–
Given Ø
?? = ?? .???????? = ?? .???? × ?? .?? × ????
-????
??
?? = ?? .?? × ????
????
????
?? .?? .= ???? - Ø
?? =
????
?? ½
?? =
????
???? -Ø
?? ½
=
?? .???? × ????
-????
×?? ×????
?? ?? .???? × ????
-????
×?? .?? ×????
????
- ?? .???? ×?? .?? ×????
-????
½
=
???? .???? × ????
-????
?? .?? × ????
-????
(?? .???? - ?? .???? )
=
???? .???? × ????
-????
?? .?? ×????
????
= ???? .?? × ????
-?? ?? ½
Ans.18 - ?? 1
= 4 × 10
-7
?? ?? 2
= 6× 10
-7
??
Distance at which dark fringe is observed ?? = (?? +
1
2
)
????
?? ½
First Dark fringe for ?? 1
?? 1
=
1
2
4×10
-7
10
-2
?? = 2 × 10
-5
?? ½
Page 4
MARKING SCHEME
PHYSICS
Subject Code – 042
CLASS – XII
Academic Session 2024 – 25
Maximum Marks:70 Time Allowed: 3hours
[SECTION – A]
Ans.1- (B) (1 mark)
VA> VB [VA = VC]
In the direction of electric field, the electric potential decreases.
Ans.2- (B) In the state of equilibrium, (1 mark)
The potential on the surface of bigger sphere = the potential at the surface of the smaller sphere
12
12
=
kq kq
rr
?
11
22
=
qr
qr
?
22
1 1 2 1 2 2
22
2 2 2 1
11
E
E
= = ? =
q r r r r
q r r
rr
Ans.3 - (C) (1 mark)
AtP2, B2 =
00
II
3 3
2
2
??
=
? ??
?
??
??
a a
AtP1, B1 =
( )
00
(I 4) I
2 2 4
??
=
?? aa
?
0
2
0 1
I
B 3
I B
4
? ??
??
?
??
=
? ??
??
?
??
a
a
?
2
1
B 4
B3
=
Ans.4 - (D)Sound waves as well as light waves (1 mark)
Ans.5 -(A) (1 mark)
Ans.6 - (C)When all the given components are connected(1 mark)
IR = IXC = IXL = 10V
XC = XL = R
Z =
22
CL
R (X X ) +-
Z =
22
R (R R) +-
Z = R
VS = IZ = IR = 10 V
So, the source voltage is also 10 V
When the capacitor is short circuited then
Z =
22
L
R (X ) +
= v?? 2
+ ?? 2
= ?? v2
VL = I ? XL =
10
R 5 2
2R
?= V
Ans.7 - (B)(1 mark)
Ans.8 - (B) The distance of closest approach(1 mark)
2
1
const
V
= d ...(1)
2
2
const
2
V
=
d
...(2)
From equations (1) and (2),
2
2
2
1
V
2
V
= ?V2 = 2 V1
? V2 = 2 V Given, (V1 = V)
Ans.9 - (C)(1 mark)
?? 2
?? -
?? 1
?? =
?? 2
- ?? 1
??
1 3 [1 3 2]
2[ 6] 6 v
-
-=
--
1 3 1 2 1
12 12 12 6 v
- - -
= + = =
?? = –6 cm
Ans.10 - (B)Diffraction (1 mark)
Ans.11- (A)doping level (1 mark)
Ans.12- (C)+0.4% (1 mark)
Ans.13- (A) (1 mark)
Ans.14- (A)(1 mark)
Ans.15- (D)(1 mark)
Ans.16- (A)(1 mark)
[SECTION – B]
Ans.17–
Given Ø
?? = ?? .???????? = ?? .???? × ?? .?? × ????
-????
??
?? = ?? .?? × ????
????
????
?? .?? .= ???? - Ø
?? =
????
?? ½
?? =
????
???? -Ø
?? ½
=
?? .???? × ????
-????
×?? ×????
?? ?? .???? × ????
-????
×?? .?? ×????
????
- ?? .???? ×?? .?? ×????
-????
½
=
???? .???? × ????
-????
?? .?? × ????
-????
(?? .???? - ?? .???? )
=
???? .???? × ????
-????
?? .?? ×????
????
= ???? .?? × ????
-?? ?? ½
Ans.18 - ?? 1
= 4 × 10
-7
?? ?? 2
= 6× 10
-7
??
Distance at which dark fringe is observed ?? = (?? +
1
2
)
????
?? ½
First Dark fringe for ?? 1
?? 1
=
1
2
4×10
-7
10
-2
?? = 2 × 10
-5
?? ½
First Dark fringe for ?? 2
?? 2
=
1
2
6×10
-7
10
-2
?? = 3 × 10
-5
??
First dark fringe will be the distance where both dark fringes will coincide i.e LCM of ?? 1
&?? 1
½
i.e. 2 × 10
-5
?? × 3 × 10
-5
??
= 6 × 10
-5
?? ½
OR
(II) ?????? ?? = ?? 1
+ ?? 2
+ 2v?? 1
v?? 2
cos? 0.5 M
Since, ?? 1
= ?? 2
= ??
Net I = I + I + 2 I cos?
= 2I (1 + cos?)
= 2I (2 cos
2
Ø
2
) 0.5 M
For path difference ?/4 , phase difference is p/2 0.5 M
Net I = 4 I cos
2
?? 4
Net I = 2 I 0.5 M
(2 Marks)
Ans.19 - (I)The direction of the magnetic field is perpendicular and inward into the plane of thepaper 0.5M
(II) For a head-on collision to take place, the radius of the path of each ion should be equal to 0.5
m.
0.5
mv
r
qB
== m 0.5M
26 5
19
4 10 2.4 10
B
4.8 10 0.5
mv
qr
-
-
? ? ?
==
??
0.5M
B = 0.04 T 0.5M
For VI Candidate
(a) As Pitch (p)=
2????? ?????? O
????
0.5M
Or, p=
2 ?? 3.14 ?? 1.7?? 10
-27
?? 2 ?? 10
5
?????? 30
0
1.6 ?? 10
-19
?? 1.5
m
Or, P=7.7X10
-3
m 0.5M
(b)As, done by magnetic field is always zero K.E=1/2mv
2
0.5M
KE=3.4 X 10
-17
J 0.5M
Ans.20 –(i) Nuclear fission –W
0.5M
Reason: As W has binding energy per nucleon less then Y and X and nucleus is larger
in size.
0.5M
Page 5
MARKING SCHEME
PHYSICS
Subject Code – 042
CLASS – XII
Academic Session 2024 – 25
Maximum Marks:70 Time Allowed: 3hours
[SECTION – A]
Ans.1- (B) (1 mark)
VA> VB [VA = VC]
In the direction of electric field, the electric potential decreases.
Ans.2- (B) In the state of equilibrium, (1 mark)
The potential on the surface of bigger sphere = the potential at the surface of the smaller sphere
12
12
=
kq kq
rr
?
11
22
=
qr
qr
?
22
1 1 2 1 2 2
22
2 2 2 1
11
E
E
= = ? =
q r r r r
q r r
rr
Ans.3 - (C) (1 mark)
AtP2, B2 =
00
II
3 3
2
2
??
=
? ??
?
??
??
a a
AtP1, B1 =
( )
00
(I 4) I
2 2 4
??
=
?? aa
?
0
2
0 1
I
B 3
I B
4
? ??
??
?
??
=
? ??
??
?
??
a
a
?
2
1
B 4
B3
=
Ans.4 - (D)Sound waves as well as light waves (1 mark)
Ans.5 -(A) (1 mark)
Ans.6 - (C)When all the given components are connected(1 mark)
IR = IXC = IXL = 10V
XC = XL = R
Z =
22
CL
R (X X ) +-
Z =
22
R (R R) +-
Z = R
VS = IZ = IR = 10 V
So, the source voltage is also 10 V
When the capacitor is short circuited then
Z =
22
L
R (X ) +
= v?? 2
+ ?? 2
= ?? v2
VL = I ? XL =
10
R 5 2
2R
?= V
Ans.7 - (B)(1 mark)
Ans.8 - (B) The distance of closest approach(1 mark)
2
1
const
V
= d ...(1)
2
2
const
2
V
=
d
...(2)
From equations (1) and (2),
2
2
2
1
V
2
V
= ?V2 = 2 V1
? V2 = 2 V Given, (V1 = V)
Ans.9 - (C)(1 mark)
?? 2
?? -
?? 1
?? =
?? 2
- ?? 1
??
1 3 [1 3 2]
2[ 6] 6 v
-
-=
--
1 3 1 2 1
12 12 12 6 v
- - -
= + = =
?? = –6 cm
Ans.10 - (B)Diffraction (1 mark)
Ans.11- (A)doping level (1 mark)
Ans.12- (C)+0.4% (1 mark)
Ans.13- (A) (1 mark)
Ans.14- (A)(1 mark)
Ans.15- (D)(1 mark)
Ans.16- (A)(1 mark)
[SECTION – B]
Ans.17–
Given Ø
?? = ?? .???????? = ?? .???? × ?? .?? × ????
-????
??
?? = ?? .?? × ????
????
????
?? .?? .= ???? - Ø
?? =
????
?? ½
?? =
????
???? -Ø
?? ½
=
?? .???? × ????
-????
×?? ×????
?? ?? .???? × ????
-????
×?? .?? ×????
????
- ?? .???? ×?? .?? ×????
-????
½
=
???? .???? × ????
-????
?? .?? × ????
-????
(?? .???? - ?? .???? )
=
???? .???? × ????
-????
?? .?? ×????
????
= ???? .?? × ????
-?? ?? ½
Ans.18 - ?? 1
= 4 × 10
-7
?? ?? 2
= 6× 10
-7
??
Distance at which dark fringe is observed ?? = (?? +
1
2
)
????
?? ½
First Dark fringe for ?? 1
?? 1
=
1
2
4×10
-7
10
-2
?? = 2 × 10
-5
?? ½
First Dark fringe for ?? 2
?? 2
=
1
2
6×10
-7
10
-2
?? = 3 × 10
-5
??
First dark fringe will be the distance where both dark fringes will coincide i.e LCM of ?? 1
&?? 1
½
i.e. 2 × 10
-5
?? × 3 × 10
-5
??
= 6 × 10
-5
?? ½
OR
(II) ?????? ?? = ?? 1
+ ?? 2
+ 2v?? 1
v?? 2
cos? 0.5 M
Since, ?? 1
= ?? 2
= ??
Net I = I + I + 2 I cos?
= 2I (1 + cos?)
= 2I (2 cos
2
Ø
2
) 0.5 M
For path difference ?/4 , phase difference is p/2 0.5 M
Net I = 4 I cos
2
?? 4
Net I = 2 I 0.5 M
(2 Marks)
Ans.19 - (I)The direction of the magnetic field is perpendicular and inward into the plane of thepaper 0.5M
(II) For a head-on collision to take place, the radius of the path of each ion should be equal to 0.5
m.
0.5
mv
r
qB
== m 0.5M
26 5
19
4 10 2.4 10
B
4.8 10 0.5
mv
qr
-
-
? ? ?
==
??
0.5M
B = 0.04 T 0.5M
For VI Candidate
(a) As Pitch (p)=
2????? ?????? O
????
0.5M
Or, p=
2 ?? 3.14 ?? 1.7?? 10
-27
?? 2 ?? 10
5
?????? 30
0
1.6 ?? 10
-19
?? 1.5
m
Or, P=7.7X10
-3
m 0.5M
(b)As, done by magnetic field is always zero K.E=1/2mv
2
0.5M
KE=3.4 X 10
-17
J 0.5M
Ans.20 –(i) Nuclear fission –W
0.5M
Reason: As W has binding energy per nucleon less then Y and X and nucleus is larger
in size.
0.5M
Read MoreFAQs on Class 12 Physics: CBSE Marking Scheme (2024-25)
| 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 paper, including theoretical questions, numerical problems, and practicals. It usually assigns higher marks to conceptual understanding and application of theories, while also considering the clarity of presentation and accuracy in calculations.
| 2. How important is the NEET exam for Class 12 students aspiring for medical studies? | |
Ans. The NEET exam is crucial for Class 12 students who wish to pursue undergraduate courses in medicine and dentistry in India. It serves as a single entrance examination for admissions into various medical colleges, thus determining the eligibility and selection of candidates based on their performance in Physics, Chemistry, and Biology.
| 3. What are the key topics covered in Class 12 Physics that are important for NEET? | |
Ans. Key topics in Class 12 Physics that are important for NEET include electrostatics, current electricity, magnetic effects of current, optics, dual nature of radiation and matter, and atomic structure. Mastery of these topics is essential as they form the basis for numerous questions in the NEET exam.
| 4. How can students effectively prepare for Physics in the NEET exam? | |
Ans. Students can effectively prepare for Physics in the NEET exam by understanding core concepts, solving previous years' question papers, and taking mock tests to enhance problem-solving speed and accuracy. Regular revision and clarification of doubts are also essential for grasping complex topics.
| 5. What practical skills are assessed in the Class 12 Physics practical examination? | |
Ans. The practical examination in Class 12 Physics assesses skills such as conducting experiments, recording observations accurately, analysing data, and drawing conclusions based on experimental findings. Students are also evaluated on their ability to handle apparatus safely and effectively, as well as their report writing skills.
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