CBSE Class 12 Marking Scheme 2022 for Physics

Physics
Marking Scheme
For SQP – 45
XII – I Term
Q.1 Which of the_____________________
Ans. 1 (iii)
As all other statements are correct. In uniform electric field equipotential surfaces are never
concentric spheres but are planes  to Electric field lines.
Q.2 Two Point charges_____________________
Ans. 2 (iii)
Let P is the observation point at a distance r from -2q
and at (L+r) from +8q.
Given Now, Net EFI at P = 0

 E1 = EFI (Electric Field Intensity) at P due to +8q

E 2 = EFI (Electric Field Intensity) at P due to -2q

E1 = E 2

k(8q) k(2q)
  2
(L  r) 2
r
4 1
  2
(L  r) 2
(r)

4r2 = (L+r)2

2r = L+r

r=L
 P is at x = L + L = 2L from origin
 Correct Option is (iii) 2L
Q3. An electric______________________
Ans. 2 (ii)
W = pE (cos1 – cos2)
1 = 0

2 = 90
W = pE (cos 0 - cos 90)
= pE (1 – 0) = pE
Q4. Three Capacitors______________
Ans.4. (ii)

1 1 1 1
  
Cseries C1 C2 C3
1 1 1 1
  
Cseries 2 3 6
3+2+1 6

6 6
Cseries  1F

Q5. Two Point Charges___________________
Ans.5. (i)
Q2 1 Q1 Q2
r K=5 F
4ok r 2
Q2 Force in the charges in the air is

1 Q1 Q2
F 
4o r 2
=K F

=5F

Q6. Which statement is true________________
Ans.6. (iv)
All other statements except (iv) are in correct
The electric field over the Gaussian surface remains continuous and uniform at every point.
Q7. A capacitor plates
Ans.7. (iii)
+ C
–
+ –

+ –
Battery is disconnected`ʹ

+ kC
–
+ –

+ – Q = Charge remains context
Cʹ = K C
Qʹ = Cʹ Vʹ
Q = Cʹ Vʹ
Q = K C Vʹ
𝑄 𝑉
Vʹ= 𝐾 𝐶 = 𝐾

Q.8. The best instrument for_________________
Ans.8. (i)
Potentiometer
Q9. An electric current___________________
Ans.9. (iii) 8:27
R1
l1 : l2 = 3 : 2
R2 r1 : r2 = 2 : 3
+ -
I1 : I2 = ?

l1
R1  
r12
l2
R2   2
r 2

R1 l1 r 2 2 l1 r 22
 = 
R 2 l2 r12 l2 r12
2
3 3 (3)3 27
   = =
2 2 (2) 3
8
I1 V / R1 R2
 = =  8 / 27
I2 V / R2 R1
Q.10. By increasing the temperature___________________
Ans. 10. (iii) Specific resistance of a conductor increases and for a semiconductor decreases with
increase in temperature because for a conductor, a temperature.

coefficient of resistivity  = + ve
and for a semiconductor,  = -ve

Q.11. We use alloys___________________
Ans. 11 (i) Alloys have low temperature coefficient of resistivity and high specific resistance. If
 = low , the value of ‘R’ with temperature will not change much and specific resistance is high
then required length of the wire will be less.
Q.12. A constant Voltage___________________

Ans. 12. (iii)
l 2l
R  R  
A (2r) 2
l 2l
R = 2 R  
r 4r 2
V2 V2
H 
t & H  1t
R R
V = constant
H 2
V R t

H R V2 t
R l 2r 2
 =  2
R r l
H 2
=
H 1

H  2H
Correct option is (iii)

Q.13. If the potential diff____________________
Ans.13. We know
eE
Vd  
ml
V
=e 
ml

If temperature is kept constant, relaxation time  - will remain constant, and e, m are also constants.

Vd  V
Vd  2V
Correct option is (ii)

Q.14. The equivalent resistance _________________
Ans. 14. (iii)
Redrawing the circuit, we get

3 8
6 B

30
3 & 6 are in parallel.
3  6 18
 R1    2
3 6 9
Now R1 and 8 in series
 R2 = R1 + 8 = 2 + 8 = 10
Now R2 and 30 in parallel
R 2  30 10  30
Rep = 
R 2  30 10  30

300 30 15
=  
40 4 2
= 7.5 (iii) correct option

Q.15. The SI unit of magnetic field intensity is ______________
Ans.15. We know

F
B
Il sin
N
SI Unit of B =  NA 1m 1
Am
Correct option is (ii)

Q16. The coil of _____________________
Ans. (iv) Correct Option
The coil of a moving coil galvanometer is wound over metallic frame to provide electromagnetic
damping so it becomes dead beat galvanometer.

Q.17. Two wires of__________________
Ans.17. Correct option (iii)

A B
l l

Square a r Circle
a
Area of a Square Area of a Circle
 a2  r 2
Also here l = 4a Also here, 2r  l
l l
a= r=
4 2
2
l2  l 
 Area = Now Area =   
16  2 
2
l l 2
A1 = A2 =
16 4

Now Magnetic moment = I A
M1 = IA, & M2 = I A 2
Since I (current) is same in both

M1 A1 l 2 4 
 =   2 
M2 A2 16 l 4
M1 M2 =  : 4
Correct option is (iii)

Q.18. The horizontal comp________________
Ans.18. Correct option (i)
Target law Bv = BH tan 

BV
tan =
BH
Given BH = 3 Bv
Bv 1
tan = 
3 Bv 3

 = 300 or radians.
6
Q.19. The small___________________
Ans. 19. Correct option is Magnetic declination or Angle of declination. It is the small angle
between geographic axis & magnetic axis.
Q.20. Two coils___________________
Ans.20. Correct option is (ii)
Mutual inductance of a pair of two coils depends on the relative position and orientation of two
coils, other statements are incorrect.
Q.21. A conducting__________________
Ans. 21. Correct option is (iv)
lel
Current induced is I =
R
d
Now lel =
dt

But there is no change of flux with time, as B, A &  all remain constant with time.

 No current is induced
Q22. The magnetic flux___________
Ans.22.
  5t 2  3t  16
d
e
dt
d
 5t 2  3t  16 
dt 
 10t + 3
e t=4 = 10(4)+3 = 43V
e = -43Volts
Correct option is (ii)

Q23 Which of the following_______________
Ans.23. Correct option is (iii)
V
I in Pure Capacitor
XC
V
=  V 2fc
1
2fc
 If f
Straight line paragraph
other parameters kept cosntant

Q24. A 20 Volt AC________________
Ans.24. Correct option is (i)

VR VL

VR = Effective Voltage across R
 VR  Ieff R
VL = Effective Voltage across L
VL  Ieff  L
Net V = VR 2  VL2
= Ieff 2 R 2  Ieff 2  L2
20 = (12)2  VL2
(20)2  (12)2  VL2
400 = 144 + VL2
VL = 400  144 = 256  16 Volts

Q25. The instantaneous_________________
Ans. 25.
E = E0 sin t


I = I0 sin t + 
3 
Correct option is (iv)
as I can lead the Voltage in RC and LCR circuit, so it can be RC or LCR circuit.
(iv) is correct option.

Section - B
Q26.
r
Correct option is (i)
Since –ve electric flux
l = + ve flux electric flux enclosed with a cylinder
here
 Total Electric
Flux = 0.

Q27. Two Parallel______________________
Ans. 27. (iv) Correct option.

E

+ +

C
Surface Charge density, = 26.4 10 12
m2
 
E= +
20 20
2 
= =
20 0
26.4 1012 N
=
8.85 1012 C
N
= 3
C
Correct option is (iv)
Q28. Consider__________________
+Q
-Q
+

Ans.28. Equal and Opposite charges appear on the nearby conductor due to
induction, but still net charge on the conductor is zero. Correct option (iv)

Q29. Three Charges_________________
Ans.29.

Net E F I at G  O
Net Potential at G,
K2Q KQ
V= 
r r r
E2 G r E2 KQ
-
r
r
=0
-q E1 -q
Correct option is (iii)
Q30. Two parallel______________
Ans.30.

k=2 A = Same
d = Same

Q = Same in Series
12V

0 A 20 A
Cx = Cy =
d d
2
Q Q2
Ux = Uy =
2Cx 2Cy
Ux Cy 2Cx 2
 =  =
Uy Cx Cx 1
Correct Option is (iii)
Q31. Which among_________________
Ans.31. Correct statement is option (iv) as Primary coil made of Thick Coper wire has very less
R. Therefore negligible power loss. Rest all options are reasons for power losses in a transformer.
Q32. An alternating Voltage_______________
Ans.32.
B C

V


1 1
Xc    i.e. Xc 
2fc c
I   Brightness of the bulb will  .
Correct option is (ii)

Q.33. A solid Sphere______________
Ans.33. Correct option is (4)
As all other statements seem incorrect in context with the given figure.
Q. 34. A battery is connected …..

Ans. Correct option is (iv).

Rest all quantities change with area of cross-section of a conductor.

Q. 35. Three resistors……
R1 R2 R3

+ –
Ans.

Given

I = 2 A, R2 = 3 Ω, P3 = 6 W

Power across R3 = V3I

6 W = I2 R3

6 3
= R3 = = 1.5 Ω
4 2
V3 = I R3 = 2 (1.5) = 3 V

Correct option is (iii).

Q. 36. A straight line……

Ans. I = O, V = E,  E = 5.6 V

E 5.6
r =   2.8 
I 2.0
Correct option is (i).

Q. 37. A 10 m long potentiometer …..
Ans. Let PQ is a potentiometer wore of length 10m,
E 5 5
I =  
R  R 480  20 500

1
=  0.01 A
100
VPQ = I RPQ = 0.01 × 20

= 0.2 V

If 10 m potentiometer wire balances  0.2 V

0.2
Then 1 m potentiometer wire balances  V
10
0.2
Then 6 m potentiometer wire balances  6V
10
1.2
=  0.12 V
10
Correct option is (iv).

Q. 38. The current sensitivity……

20
Ans. Given, I g = I g  Ig
100
120
= I g  1.2 I g
100
25 125
R = R  R R
100 100
= 1.25 R

Vg = ?

I g 1.2 I g
Vg = 
R 1.25 R
120 25
= Vg  Vg
125 25

Vg  Vg
% change =  100
Vg

 24 
 Vg  Vg 
=
 25   100
Vg

(24  25)
=  100
25
1
=  100 = 4%
25
Decrease by 4%. Correct option is (iv).

Q. 39. Three infinitely long parallel …..
F1
r r

F2 F2
2I I I

Ans. A B C

Let F1 is force per unit, length between A & C

0 2 I  I
 i.e. F1 =
4 2r
And F2 is force per unit, length between B & C

0 I  I
 F2 =
4 r
Now net force on ‘C’ is per unit length

 I2
F1  F2 = (1  1)
4 r

20 I 2
=  F (given)
4 r

B C
I I

1
F1
2I
1
F1
F21 F21
Now Fig. r r b

F1 = Repulsive force between A & C

0 2 I 2
=
4 2r
F2 = F2 = A reactive force between B & C

 Net force on ‘C’ F1  F2 = 0

 2I 2
 F1 = F2 =
4 2r
 Net Force on ‘C’ is zero.

Correct option is (i).

Q. 40. In a H-atom …….

Ans. R = 0.5 A°

 = 10 rps = 10 × 2 rad/s

 = 10 Hz

M = I A = e   r2

= 1.6 × 10–19 × 10 × 3.14 × 0.5 × 0.5 × 10–10 × 10–10

= 1.256 × 10–38 Am2

Ans. (ii).

Q. 41. An air-cored solenoid …..

Ans. Magnetic field inside a solenoid

N
B = 0 I'
l
Flux linked with ‘N’ turns

N
Initial flux 1 = N B A = N 0 IA
l

N2
= 0 IA
l

4  107  800  800  2.5  2.5  104
=
0.30
= 16.74 × 10– 3 Wb

Final flux 2 = 0

d  16.74  103  0
Average back emf |e| = =
dt 103
= 16.74 V

Correct option is (ii).

Q. 42. Vo = 283 V, f = 50 Hz

R = 3 Ω, L = 25.48 mH

C = 796 F

P |at resonance = ?

Power dissipated P = I2 R

I0 1  283 
I =   
2 2 3 
= 66.7 A

P = I2 R

= (66.7)2 3

= 13.35 kW

Correct option is (iii).

Q. 43. A circular loop …..

Ans. Let flux linked with smaller loop is 1 and with bigger loop is 2.

2 1 R1
I1
Fig.

Given R2 = 0.2 m

R1 = 0.003 m

x = 15 cm = 0.15 m

Now 1 = B2 A1

 0  2 R22 I 2 
  R1
2
= 
4  ( R22  x 2 )3 / 2 

1 0 2 R22  R12
M = 
I 2 4 ( R22  x 2 )3/ 2

Now 2 = M I1

0 2 R22  R12
= . I1
4 ( R22  x 2 )3/ 2

= 9.1 × 10–11 Weber

Correct answer is (iv).

Q. 44. If both the no. of turns…..

N2
Ans. L = 0 A
l

(2 N )2
L =  0 A
2l

N2
= 20 A  2L
l
Correct answer is (ii). Doubted.

Q.45. Given below___________________
To increase the range

Ans. 45. Correct option is (iv) as both statements are false. To increase the range of an ammeter, suitable
low R (or shunt) should be connected in parallel to it. The ammeter with increased range has low
resistance.

Q.46. An electron____________________

Ans.46. Correct option is (iii)

Statements correct but reason is wrong because electrons move from a region of low potential to high
potential.

Q. 47. A magnetic needle ……..

Ans. The given statement is correct and reason is the correct explanation of the above statement. At
poles, magnetic needle orients itself vertically because horizontal components of earth’s field is zero
there. (correct option is (i))

Q. 48. A proton and an electron, …….
𝑚𝑣 2
Ans. we know 𝑟
= 𝐵𝑞𝑣𝑠𝑖𝑛 𝜃 = 𝐵𝑞𝑣 𝑆𝑖𝑛 𝜃

Centripetal force = magnetic Lorentz force

⃗ &𝐵
sin 𝜃 = sin 90𝜃 = 1 (∠ between 𝑉 ⃗ = 90°)
𝑚𝑣 2
= 𝐵𝑞𝑣
𝑟
𝑚𝑣
𝑟
= 𝐵𝑞
𝑚𝑣 𝑝 𝑙𝑖𝑛𝑒𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚
𝑟 = 𝐵𝑞 = 𝐵𝑞 = 𝐵𝑞

𝑝
Since 𝑟 =
𝐵𝑞

Given p, B are same

Also q for proton & electron is same except its sign

 Radius is same. So statement is correct but

reason is not the correct explanation of the given assertion.

correct option is (ii)

Q. 49. On increasing…………..

Ans. 49. When we increase current sensitivity by increasing no. of turns, then resistance of coil also
increases. So increasing current sensitivity does not necessarily imply that voltage sensitivity will
𝐼𝑔
increase because 𝑉𝑔 = 𝑅

∴ if 𝐼𝑔 ↑ & 𝑅 ↑ by different amounts, then 𝑉𝑔 may increase or decrease.

Correct option is (i).

Q.50. A small object………

Ans. 50. Ans is (ii)

F𝑒 = 𝑚𝑔 𝑡𝑎𝑛𝜃
𝑞𝐸 = 𝑚𝑔 𝑡𝑎𝑛𝜃
𝑚𝑔
𝑞 = ( 𝐸 ) tan 𝜃

𝑒 𝐹
tan 𝜃 = 𝑚𝑔

E

q
T qE

Correct ans. is (ii)

Q. 51. A free electron…………

Ans. 51. Correct ans. (ii) i.e. II only

∵ 𝐹𝑝 = 𝐹𝑒 ∵ 𝐹 = 𝑞𝐸

𝐸 = 𝑠𝑎𝑚𝑒
‘𝑞’ = 𝑠𝑎𝑚𝑒
Now, 𝑃𝜀 = 𝑞 𝑉(𝑟)

(𝑃. 𝜀)𝑝 > (𝑃. 𝜀)𝑒

Q. 52. Correct ans is (iv) i.e. step down transformer decreases the ac voltage.

Q.53. correct ans is (i)
𝑁𝑠 𝐸
i.e. = 𝑠
𝑁𝑝 𝐸𝑝

i.e. if no. of turns in secondary coil are more than no. of turns in primary, then voltage is increased or
stepped up in secondary, so called step up transformer.

Q.54 Correct ans. is (i).

i.e. current is reduced if voltage is stepped – up so corresponding 𝐼 2 𝑅 losses are cut down.

Q. 55. Correct ans is (iii)

Given 𝐸𝑖 = 2300𝑉

𝐸𝑜 = 230𝑉
𝑁𝑝 = 4000

𝑁𝑠 = ?
𝐸𝑖 𝑁
𝐸𝑜
= 𝑁𝑝
𝑠

2300 4000
230
= 𝑥

𝑥 = 400 = 𝑁𝑠 = No of turns in secondary coil