NEET Exam > NEET Notes > Physics Class 12 > Class 12 Physics: CBSE Marking Scheme (2024-25)
Class 12 Physics: CBSE Marking Scheme (2024-25) | Physics Class 12 - NEET PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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
Read More
|
97 videos|369 docs|104 tests
|
About this Document
|
4.79/5 Rating |
|
Nov 22, 2024 Last updated |
Document Description: Class 12 Physics: CBSE Marking Scheme (2024-25) for NEET 2024 is part of Physics Class 12 preparation.
The notes and questions for Class 12 Physics: CBSE Marking Scheme (2024-25) have been prepared according to the NEET exam syllabus. Information about Class 12 Physics: CBSE Marking Scheme (2024-25) covers topics
like and Class 12 Physics: CBSE Marking Scheme (2024-25) Example, for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Class 12 Physics: CBSE Marking Scheme (2024-25).
Introduction of Class 12 Physics: CBSE Marking Scheme (2024-25) in English is available as part of our Physics Class 12
for NEET & Class 12 Physics: CBSE Marking Scheme (2024-25) in Hindi for Physics Class 12 course.
Download more important topics related with notes, lectures and mock test series for NEET
Exam by signing up for free. NEET: Class 12 Physics: CBSE Marking Scheme (2024-25) | Physics Class 12 - NEET
Description
Full syllabus notes, lecture & questions for Class 12 Physics: CBSE Marking Scheme (2024-25) | Physics Class 12 - NEET - NEET | Plus excerises question with solution to help you revise complete syllabus for Physics Class 12 | Best notes, free PDF download
Information about Class 12 Physics: CBSE Marking Scheme (2024-25)
In this doc you can find the meaning of Class 12 Physics: CBSE Marking Scheme (2024-25) defined & explained in the simplest way possible. Besides explaining types of
Class 12 Physics: CBSE Marking Scheme (2024-25) theory, EduRev gives you an ample number of questions to practice Class 12 Physics: CBSE Marking Scheme (2024-25) tests, examples and also practice NEET
tests
|
Explore Courses for NEET exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
Viva Questions
,past year papers
,study material
,Class 12 Physics: CBSE Marking Scheme (2024-25) | Physics Class 12 - NEET
,ppt
,Sample Paper
,Exam
,Objective type Questions
,Class 12 Physics: CBSE Marking Scheme (2024-25) | Physics Class 12 - NEET
,video lectures
,Summary
,Free
,Extra Questions
,Important questions
,Previous Year Questions with Solutions
,Semester Notes
,shortcuts and tricks
,mock tests for examination
,MCQs
,practice quizzes
,Class 12 Physics: CBSE Marking Scheme (2024-25) | Physics Class 12 - NEET
;
Additional Information about Class 12 Physics: CBSE Marking Scheme (2024-25) for NEET Preparation
Class 12 Physics: CBSE Marking Scheme (2024-25) Free PDF Download
The Class 12 Physics: CBSE Marking Scheme (2024-25) is an invaluable resource that delves deep into the core of the NEET exam.
These study notes are curated by experts and cover all the essential topics and concepts, making your preparation more efficient and effective.
With the help of these notes, you can grasp complex subjects quickly, revise important points easily,
and reinforce your understanding of key concepts. The study notes are presented in a concise and easy-to-understand manner,
allowing you to optimize your learning process. Whether you're looking for best-recommended books, sample papers, study material,
or toppers' notes, this PDF has got you covered. Download the Class 12 Physics: CBSE Marking Scheme (2024-25) now and kickstart your journey towards success in the NEET exam.
Importance of Class 12 Physics: CBSE Marking Scheme (2024-25)
The importance of Class 12 Physics: CBSE Marking Scheme (2024-25) cannot be overstated, especially for NEET aspirants.
This document holds the key to success in the NEET exam.
It offers a detailed understanding of the concept, providing invaluable insights into the topic.
By knowing the concepts well in advance, students can plan their preparation effectively.
Utilize this indispensable guide for a well-rounded preparation and achieve your desired results.
Class 12 Physics: CBSE Marking Scheme (2024-25) Notes
Class 12 Physics: CBSE Marking Scheme (2024-25) Notes offer in-depth insights into the specific topic to help you master it with ease.
This comprehensive document covers all aspects related to Class 12 Physics: CBSE Marking Scheme (2024-25).
It includes detailed information about the exam syllabus, recommended books, and study materials for a well-rounded preparation.
Practice papers and question papers enable you to assess your progress effectively.
Additionally, the paper analysis provides valuable tips for tackling the exam strategically.
Access to Toppers' notes gives you an edge in understanding complex concepts.
Whether you're a beginner or aiming for advanced proficiency, Class 12 Physics: CBSE Marking Scheme (2024-25) Notes on EduRev are your ultimate resource for success.
Class 12 Physics: CBSE Marking Scheme (2024-25) NEET Questions
The "Class 12 Physics: CBSE Marking Scheme (2024-25) NEET Questions" guide is a valuable resource for all aspiring students preparing for the
NEET 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 NEET 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 NEET syllabus.
|
© EduRev
|
Education Revolution
|
Follow Us
|
Signup to see your scores
go up within 7 days!
Access 1000+ FREE Docs, Videos and Tests
Takes less than 10 seconds to signup