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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
(ii) Nuclear fusion-Z 0.5M
Reason: As Z has binding energy per nucleon more then Y and X and nucleus is smaller
in size. 0.5M
Ans. 21 - (I) ?????????? ???????????????? ?
?? ???????????????????? ????????
1 M
(II) Alternating current changes direction every half cycle. 0.5 M
So average drift velocity is zero 0.5 M
[SECTION – C]
(3 Marks)
Ans.22 -(I) X = Full wave rectifier½
Y = Filter½
(Output Waveform for X) ½
(Output Waveform for Y) ½
(ii) 1
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The "Class 12 Physics: CBSE Marking Scheme (2024-25) Class 12 Questions" guide is a valuable resource for all aspiring students preparing for the
Class 12 exam. It focuses on providing a wide range of practice questions to help students gauge
their understanding of the exam topics. These questions cover the entire syllabus, ensuring comprehensive preparation.
The guide includes previous years' question papers for students to familiarize themselves with the exam's format and difficulty level.
Additionally, it offers subject-specific question banks, allowing students to focus on weak areas and improve their performance.
Study Class 12 Physics: CBSE Marking Scheme (2024-25) on the App
Students of Class 12 can study Class 12 Physics: CBSE Marking Scheme (2024-25) alongwith tests & analysis from the EduRev app,
which will help them while preparing for their exam. Apart from the Class 12 Physics: CBSE Marking Scheme (2024-25),
students can also utilize the EduRev App for other study materials such as previous year question papers, syllabus, important questions, etc.
The EduRev App will make your learning easier as you can access it from anywhere you want.
The content of Class 12 Physics: CBSE Marking Scheme (2024-25) is prepared as per the latest Class 12 syllabus.
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