Physics

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
kq1 kq2 q r
=  1 = 1
r1 r2 q2 r2

E1 q1 r22 r1 r22 r2
 = =  =
E 2 q2 r12 r2 r12 r1

Ans.3 - (C) (1 mark)




0 I  I
AtP2, B2 = = 0
  3a
3 a
2  
 2 
0 (I 4) 0 I
AtP1, B1 = =
2 ( a 2 ) 4a

 0 I 
B2  3a  B 4
 =  2 =
B1  0 I  B1 3
 4a 
 

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= R 2 + (X C − X L )2

Z= R 2 + (R − R)2
Z=R
VS = IZ = IR = 10 V
So, the source voltage is also 10 V
When the capacitor is short circuited then

Z= R 2 + (X L )2

= √𝑅 2 + 𝑅 2 = 𝑅√2
10
VL = I XL = R =5 2 V
2R

Ans.7 - (B)(1 mark)

Ans.8 - (B) The distance of closest approach(1 mark)
const
d= ...(1)
V12
d const
= ...(2)
2 V22
From equations (1) and (2),
V22
2= V2 = 2 V1
V12

 V2 = 2V Given, (V1 = V)

Ans.9 - (C)(1 mark)




𝑛2 𝑛1 𝑛2 − 𝑛1
− =
𝑣 𝑢 𝑅
1 3 [1 − 3 2]
− =
v 2[−6] −6

, 1 −3 1 −2 −1
= + = =
v 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 ∅𝟎 = 𝟓. 𝟔𝟑𝒆𝑽 = 𝟓. 𝟔𝟑 × 𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 𝑱
𝝂 = 𝟏. 𝟔 × 𝟏𝟎𝟏𝟓 𝑯𝒛

𝒉𝒄
𝑲. 𝑬. = 𝒉𝝂 − ∅𝟎 = 𝝀
½


𝒉𝒄
𝝀= ½
𝒉𝝂−∅𝟎




𝟔.𝟔𝟑 × 𝟏𝟎−𝟑𝟒 ×𝟑 ×𝟏𝟎𝟖
= 𝟔.𝟔𝟑 × 𝟏𝟎−𝟑𝟒 ×𝟏.𝟔×𝟏𝟎𝟏𝟓 − 𝟓.𝟔𝟑 ×𝟏.𝟔×𝟏𝟎−𝟏𝟗
½


𝟏𝟗. 𝟖𝟗 × 𝟏𝟎−𝟐𝟔
=
𝟏. 𝟔 × 𝟏𝟎−𝟏𝟗 (𝟔. 𝟔𝟑 − 𝟓. 𝟔𝟑)

𝟏𝟗.𝟖𝟗 × 𝟏𝟎−𝟐𝟔
= 𝟏.𝟔×𝟏𝟎𝟏𝟓
= 𝟏𝟐. 𝟒 × 𝟏𝟎−𝟕 𝒎 ½




Ans.18 - 𝜆1 = 4 × 10−7 𝑚𝜆2 = 6 × 10−7 𝑚
1 𝜆𝐷
Distance at which dark fringe is observed 𝑥 = (𝑛 + 2) 𝑑
½
1 4×10−7
First Dark fringe for 𝜆1 𝑑1 = 2 10−2
𝑚 = 2 × 10−5 𝑚 ½