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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
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Additional Information about Class 12 Physics: CBSE Marking Scheme (2023-24) for Class 12 Preparation
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The content of Class 12 Physics: CBSE Marking Scheme (2023-24) is prepared as per the latest Class 12 syllabus.
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