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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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