Class 12 Exam > Class 12 Notes > Sample Papers for Class 12 Medical and Non-Medical > Class 12 Physics: CBSE Marking Scheme (2023-24)
Class 12 Physics: CBSE Marking Scheme (2023-24) | Sample Papers for Class 12 Medical and Non-Medical PDF Download
| Download, print and study this document offline |
Please wait while the PDF view is loading
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 More
|
159 docs|4 tests
|
About this Document
|
4.70/5 Rating |
|
Nov 09, 2024 Last updated |
Document Description: Class 12 Physics: CBSE Marking Scheme (2023-24) for Class 12 2024 is part of Sample Papers for Class 12 Medical and Non-Medical preparation.
The notes and questions for Class 12 Physics: CBSE Marking Scheme (2023-24) have been prepared according to the Class 12 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 Class 12 2024 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 Sample Papers for Class 12 Medical and Non-Medical
for Class 12 & Class 12 Physics: CBSE Marking Scheme (2023-24) in Hindi for Sample Papers for Class 12 Medical and Non-Medical course.
Download more important topics related with notes, lectures and mock test series for Class 12
Exam by signing up for free. Class 12: Class 12 Physics: CBSE Marking Scheme (2023-24) | Sample Papers for Class 12 Medical and Non-Medical
Description
Full syllabus notes, lecture & questions for Class 12 Physics: CBSE Marking Scheme (2023-24) | Sample Papers for Class 12 Medical and Non-Medical - Class 12 | Plus excerises question with solution to help you revise complete syllabus for Sample Papers for Class 12 Medical and Non-Medical | Best notes, free PDF download
Information about Class 12 Physics: CBSE Marking Scheme (2023-24)
In this doc you can find the meaning of Class 12 Physics: CBSE Marking Scheme (2023-24) defined & explained in the simplest way possible. Besides explaining types of
Class 12 Physics: CBSE Marking Scheme (2023-24) theory, EduRev gives you an ample number of questions to practice Class 12 Physics: CBSE Marking Scheme (2023-24) tests, examples and also practice Class 12
tests
|
159 docs|4 tests
|
Download as PDF
|
Explore Courses for Class 12 exam
|
|
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
Related Searches
mock tests for examination
,Class 12 Physics: CBSE Marking Scheme (2023-24) | Sample Papers for Class 12 Medical and Non-Medical
,MCQs
,video lectures
,Free
,Important questions
,Sample Paper
,practice quizzes
,ppt
,Extra Questions
,Class 12 Physics: CBSE Marking Scheme (2023-24) | Sample Papers for Class 12 Medical and Non-Medical
,Previous Year Questions with Solutions
,Exam
,Viva Questions
,Class 12 Physics: CBSE Marking Scheme (2023-24) | Sample Papers for Class 12 Medical and Non-Medical
,Summary
,Objective type Questions
,shortcuts and tricks
,study material
,Semester Notes
,past year papers
;
Additional Information about Class 12 Physics: CBSE Marking Scheme (2023-24) for Class 12 Preparation
Class 12 Physics: CBSE Marking Scheme (2023-24) Free PDF Download
The Class 12 Physics: CBSE Marking Scheme (2023-24) is an invaluable resource that delves deep into the core of the Class 12 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 (2023-24) now and kickstart your journey towards success in the Class 12 exam.
Importance of Class 12 Physics: CBSE Marking Scheme (2023-24)
The importance of Class 12 Physics: CBSE Marking Scheme (2023-24) cannot be overstated, especially for Class 12 aspirants.
This document holds the key to success in the Class 12 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 (2023-24) Notes
Class 12 Physics: CBSE Marking Scheme (2023-24) 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 (2023-24).
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 (2023-24) Notes on EduRev are your ultimate resource for success.
Class 12 Physics: CBSE Marking Scheme (2023-24) Class 12 Questions
The "Class 12 Physics: CBSE Marking Scheme (2023-24) 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 (2023-24) on the App
Students of Class 12 can study Class 12 Physics: CBSE Marking Scheme (2023-24) 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 (2023-24),
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 (2023-24) is prepared as per the latest Class 12 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