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Class 12 Physics: CBSE Marking Scheme (2023-24)
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Page 1
Class: XII Session 2023-24
SUBJECT: PHYSICS(THEORY)
MARKING SCHEME
SECTION A
A1: c 1M
A2: c q = ?/[(2a) E sin ?] =
4
2×10
-2
× 2 × 10
5
sin 30°
1M
= 2 × 10
–3
C = 2 mC
A3: d Higher the frequency, greater is the stopping potential 1M
A4: c 1M
A5: b 1M
A6: d 1M
A7: b 1M
9 x S = 1 × 0.81
S =
0.81
9
= 0.09 ?
A8: a 1M
A9: d 1M
A10: a 1M
A11: d e =
1
,I
t R t
?? ??
=
??
1M
It
R
??
?= = Area under I – t graph, R = 100 ohm
?
1
100 10 0.5 250
2
?? = ? ? ? = Wb.
A12: b 1M
A13: a 1M
A14: a 1M
A15: c 1M
Q16: c 1M
SECTION B
A17: (a) Rectifier 1M
(b) Circuit diagram of full wave rectifier 1M
Page 2
Class: XII Session 2023-24
SUBJECT: PHYSICS(THEORY)
MARKING SCHEME
SECTION A
A1: c 1M
A2: c q = ?/[(2a) E sin ?] =
4
2×10
-2
× 2 × 10
5
sin 30°
1M
= 2 × 10
–3
C = 2 mC
A3: d Higher the frequency, greater is the stopping potential 1M
A4: c 1M
A5: b 1M
A6: d 1M
A7: b 1M
9 x S = 1 × 0.81
S =
0.81
9
= 0.09 ?
A8: a 1M
A9: d 1M
A10: a 1M
A11: d e =
1
,I
t R t
?? ??
=
??
1M
It
R
??
?= = Area under I – t graph, R = 100 ohm
?
1
100 10 0.5 250
2
?? = ? ? ? = Wb.
A12: b 1M
A13: a 1M
A14: a 1M
A15: c 1M
Q16: c 1M
SECTION B
A17: (a) Rectifier 1M
(b) Circuit diagram of full wave rectifier 1M
A18: As ? = h / mv , v= h /m? -----------------(i) 1/2M
Energy of photon E = hc /? 1/2M
& Kinetic energy of electron K =1/2 mv
2
= ½ mh
2
/ m
2
?
2
--------------(ii) 1/2M
Simplifying equation i & ii we get E / K = 2?mc /h 1/2M
A19: Here angle of prism A = 60°, angle of incidence i = angle of emergence e and
under this condition angle of deviation is minimum
? i = e =
3
4
A =
3
4
× 60° = 45° and i + e = A + D,
hence D m = 2i – A = 2 × 45° – 60° = 30° 1M
? Refractive index of glass prism
n =
60 30
sin sin
sin 45 1 2 2 2
2.
60 sin30 1 2
sin sin
22
m
AD
A
+ ? + ? ?? ??
? ? ? ?
?
? ? ? ?
= = = =
? ? ? ? ? ?
? ? ? ?
? ? ? ?
1M
A20:Given: V=230 V, I 0= 3.2A, I=2.8A, ?? 0
=27
0
C, a=1.70 × 10
–4
°C
–1
.
Using equation R =R 0 (1+ a ?T) ½ M
i.e V/ I = {V/ I 0 } [1 + a ?T] ½ M
and solving ?T= 840 , i.e. T = 840 + 27 = 867
0
C 1M
A21: Let d be the least distance between object and image for a real image formation.
½ M
1 1 1
f v u
=- ,
1
?? =
1
?? +
1
?? -?? =
?? ?? (?? -?? )
½ M
fd = xd – x
2
, x
2
– dx + fd = 0 , x =
?? ±v?? 2
-4????
2
½ M
For real roots of x, d
2
– 4fd ? 0 ½ M
d ? 4f.
OR
Let f o and f e be the focal length of the objective and eyepiece respectively.
For normal adjustment the distance from objective to eyepiece is f o + f e.
Taking the line on the objective as object and eyepiece as lens
u = –(f o + f e ) and f = f e
1
?? -
1
[ -{ ?? o + ?? ?? } ]
=
1
?? ?? ? 1M
Page 3
Class: XII Session 2023-24
SUBJECT: PHYSICS(THEORY)
MARKING SCHEME
SECTION A
A1: c 1M
A2: c q = ?/[(2a) E sin ?] =
4
2×10
-2
× 2 × 10
5
sin 30°
1M
= 2 × 10
–3
C = 2 mC
A3: d Higher the frequency, greater is the stopping potential 1M
A4: c 1M
A5: b 1M
A6: d 1M
A7: b 1M
9 x S = 1 × 0.81
S =
0.81
9
= 0.09 ?
A8: a 1M
A9: d 1M
A10: a 1M
A11: d e =
1
,I
t R t
?? ??
=
??
1M
It
R
??
?= = Area under I – t graph, R = 100 ohm
?
1
100 10 0.5 250
2
?? = ? ? ? = Wb.
A12: b 1M
A13: a 1M
A14: a 1M
A15: c 1M
Q16: c 1M
SECTION B
A17: (a) Rectifier 1M
(b) Circuit diagram of full wave rectifier 1M
A18: As ? = h / mv , v= h /m? -----------------(i) 1/2M
Energy of photon E = hc /? 1/2M
& Kinetic energy of electron K =1/2 mv
2
= ½ mh
2
/ m
2
?
2
--------------(ii) 1/2M
Simplifying equation i & ii we get E / K = 2?mc /h 1/2M
A19: Here angle of prism A = 60°, angle of incidence i = angle of emergence e and
under this condition angle of deviation is minimum
? i = e =
3
4
A =
3
4
× 60° = 45° and i + e = A + D,
hence D m = 2i – A = 2 × 45° – 60° = 30° 1M
? Refractive index of glass prism
n =
60 30
sin sin
sin 45 1 2 2 2
2.
60 sin30 1 2
sin sin
22
m
AD
A
+ ? + ? ?? ??
? ? ? ?
?
? ? ? ?
= = = =
? ? ? ? ? ?
? ? ? ?
? ? ? ?
1M
A20:Given: V=230 V, I 0= 3.2A, I=2.8A, ?? 0
=27
0
C, a=1.70 × 10
–4
°C
–1
.
Using equation R =R 0 (1+ a ?T) ½ M
i.e V/ I = {V/ I 0 } [1 + a ?T] ½ M
and solving ?T= 840 , i.e. T = 840 + 27 = 867
0
C 1M
A21: Let d be the least distance between object and image for a real image formation.
½ M
1 1 1
f v u
=- ,
1
?? =
1
?? +
1
?? -?? =
?? ?? (?? -?? )
½ M
fd = xd – x
2
, x
2
– dx + fd = 0 , x =
?? ±v?? 2
-4????
2
½ M
For real roots of x, d
2
– 4fd ? 0 ½ M
d ? 4f.
OR
Let f o and f e be the focal length of the objective and eyepiece respectively.
For normal adjustment the distance from objective to eyepiece is f o + f e.
Taking the line on the objective as object and eyepiece as lens
u = –(f o + f e ) and f = f e
1
?? -
1
[ -{ ?? o + ?? ?? } ]
=
1
?? ?? ? 1M
Linear magnification (eyepiece) =
?? ?? =
?????????? ???????? ???????????? ???????? =
?? ?? ?? ?? =
?? ?? ½ M
? Angular magnification of telescope
M =
?? 0
?? ?? =
?? ?? ½ M
SECTION C
A22: Number of atoms in 3 gram of Cu coin = (6.023 x 10
23
X 3 ) / 63 = 2.86 X 10
22
½ M
Each atom has 29 Protons & 34 Neutrons
Thus Mass defect ?m= 29X 1.00783 + 34X 1.00867 – 62.92960 u =0.59225u 1M
Nuclear energy required for one atom =0.59225 X 931.5 MeV ½ M
Nuclear energy required for 3 gram of Cu =0.59225 X 931.5 X 2.86X 10
22
MeV
= 1.58 X 10
25
MeV 1M
A23:
V C = 0, 1M
V D =
00
1
4 3L L 6 L
q q q - ??
-=
??
?? ??
??
1M
W = Q [V D – V C] =
0
6L
Qq -
??
1M
A24 : formula K=-E , U = -2K 1M
(a) K = 3.4 eV & (b) U= -6.8 eV 1M
(c) The kinetic energy of the electron will not change. The value of potential energy and
consequently, the value of total energy of the electron will change. 1M
A25:
1.5M
As the points B and P are at the same potential,
1
1
=
(1+?? )
(2+?? )
(1-?? )
? ?? = (v2 - 1) ?? h?? 1.5M
Page 4
Class: XII Session 2023-24
SUBJECT: PHYSICS(THEORY)
MARKING SCHEME
SECTION A
A1: c 1M
A2: c q = ?/[(2a) E sin ?] =
4
2×10
-2
× 2 × 10
5
sin 30°
1M
= 2 × 10
–3
C = 2 mC
A3: d Higher the frequency, greater is the stopping potential 1M
A4: c 1M
A5: b 1M
A6: d 1M
A7: b 1M
9 x S = 1 × 0.81
S =
0.81
9
= 0.09 ?
A8: a 1M
A9: d 1M
A10: a 1M
A11: d e =
1
,I
t R t
?? ??
=
??
1M
It
R
??
?= = Area under I – t graph, R = 100 ohm
?
1
100 10 0.5 250
2
?? = ? ? ? = Wb.
A12: b 1M
A13: a 1M
A14: a 1M
A15: c 1M
Q16: c 1M
SECTION B
A17: (a) Rectifier 1M
(b) Circuit diagram of full wave rectifier 1M
A18: As ? = h / mv , v= h /m? -----------------(i) 1/2M
Energy of photon E = hc /? 1/2M
& Kinetic energy of electron K =1/2 mv
2
= ½ mh
2
/ m
2
?
2
--------------(ii) 1/2M
Simplifying equation i & ii we get E / K = 2?mc /h 1/2M
A19: Here angle of prism A = 60°, angle of incidence i = angle of emergence e and
under this condition angle of deviation is minimum
? i = e =
3
4
A =
3
4
× 60° = 45° and i + e = A + D,
hence D m = 2i – A = 2 × 45° – 60° = 30° 1M
? Refractive index of glass prism
n =
60 30
sin sin
sin 45 1 2 2 2
2.
60 sin30 1 2
sin sin
22
m
AD
A
+ ? + ? ?? ??
? ? ? ?
?
? ? ? ?
= = = =
? ? ? ? ? ?
? ? ? ?
? ? ? ?
1M
A20:Given: V=230 V, I 0= 3.2A, I=2.8A, ?? 0
=27
0
C, a=1.70 × 10
–4
°C
–1
.
Using equation R =R 0 (1+ a ?T) ½ M
i.e V/ I = {V/ I 0 } [1 + a ?T] ½ M
and solving ?T= 840 , i.e. T = 840 + 27 = 867
0
C 1M
A21: Let d be the least distance between object and image for a real image formation.
½ M
1 1 1
f v u
=- ,
1
?? =
1
?? +
1
?? -?? =
?? ?? (?? -?? )
½ M
fd = xd – x
2
, x
2
– dx + fd = 0 , x =
?? ±v?? 2
-4????
2
½ M
For real roots of x, d
2
– 4fd ? 0 ½ M
d ? 4f.
OR
Let f o and f e be the focal length of the objective and eyepiece respectively.
For normal adjustment the distance from objective to eyepiece is f o + f e.
Taking the line on the objective as object and eyepiece as lens
u = –(f o + f e ) and f = f e
1
?? -
1
[ -{ ?? o + ?? ?? } ]
=
1
?? ?? ? 1M
Linear magnification (eyepiece) =
?? ?? =
?????????? ???????? ???????????? ???????? =
?? ?? ?? ?? =
?? ?? ½ M
? Angular magnification of telescope
M =
?? 0
?? ?? =
?? ?? ½ M
SECTION C
A22: Number of atoms in 3 gram of Cu coin = (6.023 x 10
23
X 3 ) / 63 = 2.86 X 10
22
½ M
Each atom has 29 Protons & 34 Neutrons
Thus Mass defect ?m= 29X 1.00783 + 34X 1.00867 – 62.92960 u =0.59225u 1M
Nuclear energy required for one atom =0.59225 X 931.5 MeV ½ M
Nuclear energy required for 3 gram of Cu =0.59225 X 931.5 X 2.86X 10
22
MeV
= 1.58 X 10
25
MeV 1M
A23:
V C = 0, 1M
V D =
00
1
4 3L L 6 L
q q q - ??
-=
??
?? ??
??
1M
W = Q [V D – V C] =
0
6L
Qq -
??
1M
A24 : formula K=-E , U = -2K 1M
(a) K = 3.4 eV & (b) U= -6.8 eV 1M
(c) The kinetic energy of the electron will not change. The value of potential energy and
consequently, the value of total energy of the electron will change. 1M
A25:
1.5M
As the points B and P are at the same potential,
1
1
=
(1+?? )
(2+?? )
(1-?? )
? ?? = (v2 - 1) ?? h?? 1.5M
A26:
(a) Consider the case r > a. The Amperian loop, labelled 2, is a circle concentric with the cross-section.
For this loop, L = 2 p r
Using Ampere circuital Law, we can write,
B (2p r) = µ 0 I , B =
0
2
I
r
?
?
, B ?
1
r
(r > a) 1.5 M
(b)Consider the case r < a. The Amperian loop is a circle labelled 1. For this loop, taking the radius of
the circle to be r, L = 2 p r
Now the current enclosed I e is not I, but is less than this value. Since the current distribution is uniform,
the current enclosed is,
I e = I
2
2
r
a
??
?
??
??
?
??
=
2
2
Ir
a
Using Ampere’s law, B (2?r) = µ 0
2
2
Ir
a
B =
0
2
2
I
a
? ??
??
? ??
r B ? r (r < a) 1.5M
A27: (a) Infrared (b) Ultraviolet (c) X rays ½ + ½ + ½ M
Any one method of the production of each one ½ + ½ + ½ M
A28 (a ) : Definition and S.I. Unit. ½ + ½ M
(b)
Let a current I P flow through the circular loop of radius R. The magnetic induction at the centre of the
loop is
½ M
As, r << R, the magnetic induction B P may be considered to be constant over the entire cross sectional
area of inner loop of radius r. Hence magnetic flux linked with the smaller loop will be
½ M
Also, ½ M
Page 5
Class: XII Session 2023-24
SUBJECT: PHYSICS(THEORY)
MARKING SCHEME
SECTION A
A1: c 1M
A2: c q = ?/[(2a) E sin ?] =
4
2×10
-2
× 2 × 10
5
sin 30°
1M
= 2 × 10
–3
C = 2 mC
A3: d Higher the frequency, greater is the stopping potential 1M
A4: c 1M
A5: b 1M
A6: d 1M
A7: b 1M
9 x S = 1 × 0.81
S =
0.81
9
= 0.09 ?
A8: a 1M
A9: d 1M
A10: a 1M
A11: d e =
1
,I
t R t
?? ??
=
??
1M
It
R
??
?= = Area under I – t graph, R = 100 ohm
?
1
100 10 0.5 250
2
?? = ? ? ? = Wb.
A12: b 1M
A13: a 1M
A14: a 1M
A15: c 1M
Q16: c 1M
SECTION B
A17: (a) Rectifier 1M
(b) Circuit diagram of full wave rectifier 1M
A18: As ? = h / mv , v= h /m? -----------------(i) 1/2M
Energy of photon E = hc /? 1/2M
& Kinetic energy of electron K =1/2 mv
2
= ½ mh
2
/ m
2
?
2
--------------(ii) 1/2M
Simplifying equation i & ii we get E / K = 2?mc /h 1/2M
A19: Here angle of prism A = 60°, angle of incidence i = angle of emergence e and
under this condition angle of deviation is minimum
? i = e =
3
4
A =
3
4
× 60° = 45° and i + e = A + D,
hence D m = 2i – A = 2 × 45° – 60° = 30° 1M
? Refractive index of glass prism
n =
60 30
sin sin
sin 45 1 2 2 2
2.
60 sin30 1 2
sin sin
22
m
AD
A
+ ? + ? ?? ??
? ? ? ?
?
? ? ? ?
= = = =
? ? ? ? ? ?
? ? ? ?
? ? ? ?
1M
A20:Given: V=230 V, I 0= 3.2A, I=2.8A, ?? 0
=27
0
C, a=1.70 × 10
–4
°C
–1
.
Using equation R =R 0 (1+ a ?T) ½ M
i.e V/ I = {V/ I 0 } [1 + a ?T] ½ M
and solving ?T= 840 , i.e. T = 840 + 27 = 867
0
C 1M
A21: Let d be the least distance between object and image for a real image formation.
½ M
1 1 1
f v u
=- ,
1
?? =
1
?? +
1
?? -?? =
?? ?? (?? -?? )
½ M
fd = xd – x
2
, x
2
– dx + fd = 0 , x =
?? ±v?? 2
-4????
2
½ M
For real roots of x, d
2
– 4fd ? 0 ½ M
d ? 4f.
OR
Let f o and f e be the focal length of the objective and eyepiece respectively.
For normal adjustment the distance from objective to eyepiece is f o + f e.
Taking the line on the objective as object and eyepiece as lens
u = –(f o + f e ) and f = f e
1
?? -
1
[ -{ ?? o + ?? ?? } ]
=
1
?? ?? ? 1M
Linear magnification (eyepiece) =
?? ?? =
?????????? ???????? ???????????? ???????? =
?? ?? ?? ?? =
?? ?? ½ M
? Angular magnification of telescope
M =
?? 0
?? ?? =
?? ?? ½ M
SECTION C
A22: Number of atoms in 3 gram of Cu coin = (6.023 x 10
23
X 3 ) / 63 = 2.86 X 10
22
½ M
Each atom has 29 Protons & 34 Neutrons
Thus Mass defect ?m= 29X 1.00783 + 34X 1.00867 – 62.92960 u =0.59225u 1M
Nuclear energy required for one atom =0.59225 X 931.5 MeV ½ M
Nuclear energy required for 3 gram of Cu =0.59225 X 931.5 X 2.86X 10
22
MeV
= 1.58 X 10
25
MeV 1M
A23:
V C = 0, 1M
V D =
00
1
4 3L L 6 L
q q q - ??
-=
??
?? ??
??
1M
W = Q [V D – V C] =
0
6L
Qq -
??
1M
A24 : formula K=-E , U = -2K 1M
(a) K = 3.4 eV & (b) U= -6.8 eV 1M
(c) The kinetic energy of the electron will not change. The value of potential energy and
consequently, the value of total energy of the electron will change. 1M
A25:
1.5M
As the points B and P are at the same potential,
1
1
=
(1+?? )
(2+?? )
(1-?? )
? ?? = (v2 - 1) ?? h?? 1.5M
A26:
(a) Consider the case r > a. The Amperian loop, labelled 2, is a circle concentric with the cross-section.
For this loop, L = 2 p r
Using Ampere circuital Law, we can write,
B (2p r) = µ 0 I , B =
0
2
I
r
?
?
, B ?
1
r
(r > a) 1.5 M
(b)Consider the case r < a. The Amperian loop is a circle labelled 1. For this loop, taking the radius of
the circle to be r, L = 2 p r
Now the current enclosed I e is not I, but is less than this value. Since the current distribution is uniform,
the current enclosed is,
I e = I
2
2
r
a
??
?
??
??
?
??
=
2
2
Ir
a
Using Ampere’s law, B (2?r) = µ 0
2
2
Ir
a
B =
0
2
2
I
a
? ??
??
? ??
r B ? r (r < a) 1.5M
A27: (a) Infrared (b) Ultraviolet (c) X rays ½ + ½ + ½ M
Any one method of the production of each one ½ + ½ + ½ M
A28 (a ) : Definition and S.I. Unit. ½ + ½ M
(b)
Let a current I P flow through the circular loop of radius R. The magnetic induction at the centre of the
loop is
½ M
As, r << R, the magnetic induction B P may be considered to be constant over the entire cross sectional
area of inner loop of radius r. Hence magnetic flux linked with the smaller loop will be
½ M
Also, ½ M
? ½ M
OR
The magnetic induction B 1 set up by the current I 1 flowing in first conductor at a point somewhere in the
middle of second conductor is
B 1 =
01
I
2 a
?
?
...(1) ½ M
The magnetic force acting on the portion P 2Q 2 of length
2
of second conductor is
F 2 = I 2
2
B 1 sin 90° ...(2)
From equation (1) and (2),
F 2 =
0 1 2 2
II
2 a
?
?
, towards first conductor ½ M
0 1 2 2
2
II F
2 a
?
=
?
...(3)
The magnetic induction B 2 set up by the current I 2 flowing in second conductor at a point somewhere in
the middle of first conductor is
B 2 =
02
I
2 a
?
?
...(4) ½ M
The magnetic force acting on the portion P 1Q 1 of length
1
of first conductor is
F 1 =
1 1 2
IB sin 90° ...(5)
From equation (3) and (5)
F 1 =
0 1 2 1
II
2 a
?
?
, towards second conductor ½ M
0 1 2 1
1
2
II F
a
?
=
?
...(6)
The standard definition of 1A
If I 1 = I 2 = 1A
12
= = 1m
a = 1m in V/A then
7 0 12
12
11
2 10
21
FF
-
? ? ?
= = = ?
??
N/m
? One ampere is that electric current which when flows in each one of the two infinitely long
straight parallel conductors placed 1m apart in vacuum causes each one of them to experience a force
of 2 × 10
–7
N/m. 1M
Read MoreFAQs on Class 12 Physics: CBSE Marking Scheme (2023-24)
| 1. What are the key topics covered in the Class 12 Physics syllabus for CBSE? |  |
Ans. The Class 12 Physics syllabus for CBSE includes important topics such as Electrostatics, Current Electricity, Magnetic Effects of Current and Magnetism, Electromagnetic Induction and Alternating Currents, Optics, Dual Nature of Matter and Radiation, Atoms and Nuclei, and Electronic Devices. Each topic encompasses essential concepts, laws, and applications relevant to the understanding of physics principles.
| 2. How is the marking scheme structured for the Class 12 Physics exam? | |
Ans. The marking scheme for the Class 12 Physics exam typically consists of different sections, including objective type questions, short answer questions, and long answer questions. The weightage for each section is defined, with objective questions carrying less weight compared to descriptive questions, which assess deeper understanding and analytical skills. Specific marking criteria such as accuracy, clarity, and presentation are also considered in evaluation.
| 3. What types of questions can be expected in the NEET exam related to Physics? | |
Ans. In the NEET exam, candidates can expect a variety of questions related to Physics that test their conceptual understanding and application skills. These include multiple-choice questions (MCQs) on topics such as Mechanics, Thermodynamics, Waves, Optics, and Modern Physics. Questions may require problem-solving skills and the ability to apply formulas to real-life scenarios, along with conceptual questions that assess understanding of fundamental principles.
| 4. How important is numerical problem-solving in the Class 12 Physics exam? | |
Ans. Numerical problem-solving is a crucial aspect of the Class 12 Physics exam as it tests students' ability to apply theoretical concepts to practical situations. A significant portion of the exam consists of numerical questions that require students to understand and use relevant formulas, perform calculations, and interpret results. Mastery of numerical problems not only helps in scoring well in exams but also builds a solid foundation for future studies in science and engineering.
| 5. What strategies can students use to prepare effectively for the Class 12 Physics exam? | |
Ans. To prepare effectively for the Class 12 Physics exam, students should start by thoroughly understanding the syllabus and marking scheme. Creating a study schedule that allocates time for each topic is essential. Regular practice of numerical problems, revision of concepts, and solving previous years' question papers can enhance problem-solving skills and familiarity with the exam format. Additionally, group study sessions and seeking clarification on difficult topics can further aid in effective preparation.
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Document Description: Class 12 Physics: CBSE Marking Scheme (2023-24) for NEET 2026 is part of Physics Class 12 preparation. The notes and questions for Class 12 Physics: CBSE Marking Scheme (2023-24) have been prepared according to the NEET exam syllabus. Information about Class 12 Physics: CBSE Marking Scheme (2023-24) covers topics like and Class 12 Physics: CBSE Marking Scheme (2023-24) Example, for NEET 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for Class 12 Physics: CBSE Marking Scheme (2023-24).
Introduction of Class 12 Physics: CBSE Marking Scheme (2023-24) in English is available as part of our Physics Class 12 for NEET & Class 12 Physics: CBSE Marking Scheme (2023-24) 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 (2023-24)
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