CBSE Class 12 Physics Question Paper 2020 Set 55-4-3 Solutions

Strictly Confidential: (For Internal and Restricted use only)
Senior School Certificate Examination-2020
Marking Scheme – PHYSICS THEORY (042)
(55/4/3)
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Page 1 of 15

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Page 2 of 15

MARKING SCHEME: PHYSICS
QUESTION PAPER CODE: 55/4/3
Q.No. Value Points/Expected Answer Marks Total
Marks

SECTION-A

1. 𝐼 1 1
(b)
4
2. (c) helical with its axis parallel to x-axis 1 1
3. (b) Zero 1 1
4. (b) 1 1

5. (b) 2 × 10-5 T, acting towards north 1 1
6. (b) R 1 1
7. (a) Only on impact parameter 1 1
8. (a) Infra red region 1 1
9. (b) 1:1 1 1
10. (c) π 1 1
11. Middle/mid point /center 1 1
OR
Decrease
𝜋
12. 900 or 1 1
2
13. 7
/ 73𝑋 1 1
3𝐿𝑖
14. Decreasing/Lower 1 1
15. Zero 1 1
16. The displacement current will decrease. 1 1
𝑉 𝑉
𝐻𝑖𝑛𝑡 ∶ − (𝐼𝐶 = = 1 = ωCV) / the rate of change of electric
𝑋𝐶 (ωc)
flux/electric field will decrease
17. No ½ 1
As there will be discontinuity for the flow of charge carriers / no
contact at atomic level. ½
(Any One Justification)
OR
The forward current is large due to majority charge carriers which
are very large in number. Hence resistance in forward bias is low. 1
Alternatively: Depletion region decreases or barrier potential
decreases.
18. Reflecting type telescope ½ 1
Reason/Justification :-
Mirror have large aperture/high resolving power/ free from ½
chromatic aberration /free from spherical aberration. ( Any one)
19. It increases form 0° to 90° 1 1

20. Increases 1 1

Page 3 of 15

SECTION - B
21.
Explanation of set up of potential barrier 2

Diffusion current is set up across the junction due to the
concentration difference of the majority charge carriers on the two 1
sides of the junction.
This diffusion develops an electric field from n- side to p- side ½
across the junction which creates a drift current in the opposite
direction.
When diffusion and drift current become equal in magnitude the 2
potential difference across the junction is the barrier potential. ½
22.
(a) Comparison of frequencies 1
(b) Justification 1
(a) Let υ0A , υ0B and υ0C be their threshold frequencies for
the surfaces A,B and C
Therefore 𝜐0𝐴 > 𝜐0𝐵 > 𝜐0𝐶 1

(b) Justification :-
If the frequency of incident light/photon is υ
½
ℎ𝜐 = ℎ𝜐0 + 𝐸𝑘
Therefore 𝜐0𝐴 > 𝜐 , 𝜐0𝐵 = 𝜐 𝑎𝑛𝑑 𝜐0𝐶 < 𝜐 ½ 2

OR
(i) Effect on the energy of the photo electrons 1
(ii) Effect on photoelectric current 1

(i) The energy of the emitted photoelectrons increases ½
As 𝐸𝑘 = ℎ𝜐 − 𝜙0
As υ increases, Ek also increases ½

(ii) Photo current will not be affected ½

As, increase of 𝜐, Ek will increase but not the number of
photoelectrons ½ 2
[Alternatively photocurrent depends upon intensity of light and not
on frequency]
23.
a) Plotting graph of R vs l 1
b) Finding final resistance 1

a)

1

Page 4 of 15

(b) Resistance becomes 4 R0
𝜌𝑙 𝜌𝑙 2 𝜌𝑙 2
( Hint :- As 𝑅 = = = ) ( V= Volume)
𝐴 𝐴𝑙 𝑉
2
𝑅∝𝑙 1 2

24.
a) Formation of combination 1
b) Maximum and minimum value of equivalent
capacitance 1

a)

1

𝑖𝑓 𝑡ℎ𝑒 𝑠𝑡𝑢𝑑𝑒𝑛𝑡 𝑒𝑥𝑝𝑙𝑎𝑖𝑛𝑠 𝑡ℎ𝑒 𝑓𝑜𝑟𝑚𝑎𝑡𝑖𝑜𝑛 𝑜𝑓 𝑡ℎ𝑖𝑠
[ ] ½
𝑐𝑜𝑚𝑏𝑖𝑛𝑎𝑡𝑖𝑜𝑛 𝑤𝑖𝑡ℎ𝑜𝑢𝑡 𝑑𝑖𝑎𝑔𝑟𝑎𝑚 𝑎𝑤𝑎𝑟𝑑 ℎ𝑖𝑚 1 𝑚𝑎𝑟𝑘
b) C max = 2 +3 + 4 = 9 µF
1 1 1 1 12 ½ 2
= + + = µF
𝐶𝑚𝑖𝑛 2 3 4 13
25
(a) Energy bands in solids 1
(b) Drawing energy band diagram
(i) Metal ; (ii) Semiconductor ½+½
(a) Note: Out of syllabus; marks are distributed in part(b)

(b) [A student may draw both or any one]

(i)

1

(ii)

1 2

Page 5 of 15

26.
(i) Finding average power dissipated 1
(ii) Finding instantaneous current 1

(i) Average power dissipated
𝑃1 = 𝐼𝑒𝑓𝑓 × 𝑉𝑒𝑓𝑓 × cos 00 ½
2 𝑉0 2
= 𝐼𝑒𝑓𝑓 × 𝐼𝑒𝑓𝑓 × 𝑅 × 1 = 𝐼𝑒𝑓𝑓 𝑅= ½
𝑅
(ii) Instantaneous Current
𝑉 𝑉 ½+ ½ 2
𝐼= = 0 sin 𝜔𝑡 = 𝐼0 sin 𝜔𝑡
𝑅 𝑅
27
(a) Photo diode in reverse biasing 1
(b) V-I Characteristics of photodiode 1

(a) Because the fractional change in the minority carriers
dominated very weak reverse current is more easily 1
measurable than fractional change in forward biased large
current

(b)

1 2

I1>I2
OR

(a) Level of doping and biasing in LED 1
(b) Any two advantages of LED 1

(a) It is a heavily doped p-n junction. ½
It operates in forward biasing ½
(b) Advantages
Low operational voltage/less power /fast action / nearly ½+½ 2
monochromatic / long life ( Any two)
SECTION C
28
(a) Phase difference between the waves 1
(b) Resultant intensity at the point 1
(c) Resultant intensity in terms of intensity at
maximum 1

Page 6 of 15

2𝜋 𝜆 𝜋 ½+ ½
(a) Phase difference ∅ = × =
𝜆 6 3
(b) 𝐼1 = 𝐼2 + 𝐼3 + 2√𝐼3 𝐼2 𝑐𝑜𝑠 ∅
1 ½+ ½
=𝐼+𝐼+2𝐼 × =3𝐼
2
= 15 × 10−2 𝑊𝑚−2
(c) 𝐼𝑚𝑎𝑥 = 4𝐼
3𝐼 3 ½+ ½ 3
𝐼1 = × 4𝐼 = 𝐼𝑚𝑎𝑥
4𝐼 4

29.
(a) Production of electromagnetic waves 1
(b) Depiction of electromagnetic waves 1
(c) Two characteristics of electromagnetic waves 1

a) Electromagnetic waves are produced by accelerated 1
(oscillating) charges

1

b) i) An e.m.wave consists of oscillating ⃗⃗⃗𝐸 and 𝐵
⃗ mutually
perpendicular and also perpendicular to the direction of
propagation of the wave. 1 3
ii) They carry energy and momentum.
(or any other two characteristics)
30.
Finding the focal length of the mirror 3

𝑓
m= ½
𝑓−𝑢
In first case: u =-9 cm
𝑓
m1 = ½
𝑓+9
In second case: u = -15 cm
𝑓
m2 = ½
𝑓+15
but
m1 = 4m2
𝑓 𝑓
=4 ½
𝑓+9 𝑓+15
f+15 = 4f + 36 ½
f= -7cm ½ 3

31.
(a) Difference between self-inductance and mutual
inductance 1
(b) Finding
(i) Change in magnetic flux 1
(ii) EMF induced 1

Page 7 of 15

(a) Self inductance is the response of the coil/ solenoid to the
charge in current in the coil/ solenoid itself (or definition of
self inductance) ½
Mutual inductance is the response of a coil to the charge of
current in a neighbouring coil (or definition of mutual ½
inductance)
Alternatively
Self-inductance is the property of given coil/solenoid
Mutual inductance is the property of given pair of coils
/solenoids
(b)
(i) Δ𝜙 = 𝑀∆𝐼 = 2 × 0.5 = 1𝑊𝑏 ½+½

𝑑𝜙 1 ½+ ½ 3
(ii) 𝑒=− = = 10𝑉
𝑑𝑡 100×10−3

32.

a) Labelled diagram of A.C. generator 1
b) Working 1
c) Obtaining expression for EMF 1
1
(a) Labelled diagram

1

(b) Working :-
When a coil is rotated in a magnetic field; the magnetic flux
linked with it changes; hence an EMF is induced which lasts
so long as the coil rotates. 1
(c) Let initially the plane of coil is perpendicular to the direction
of magnetic field ( ⃗⃗⃗⃗
𝐵).After time t , the normal to the plane
of the coil makes an angle 𝜃,when the coil is rotated with a
constant angular speed ω
𝜃 = 𝜔𝑡
−𝑑𝜙 ⃗⃗⃗ 𝑑𝑆)
𝑑(𝐵. ½
𝐸= =−
𝑑𝑡 𝑑𝑡
𝑑(𝐵𝐴𝑁𝑐𝑜𝑠𝜔𝑡)
=−
𝑑𝑡

= 𝐵𝐴𝑁𝜔 sin 𝜔𝑡
𝐸 = 𝐸0 sin 𝜔𝑡 ½ 3
33
(a) Giving the value of surface charge density of
(i) Inner surface (ii) Outer Surface ½+½
(b) Deriving expression for electric field 2

Page 8 of 15

𝑞
(a) Surface charge density on the inner surface=
4𝜋𝑟12 ½
𝑄−𝑞
On the outer surface=
4𝜋𝑟22 ½
(b) For a spherical Gaussian surface 𝑥 > 𝑟2

𝑄−𝑞 1
⃗⃗⃗⃗ =
∮ 𝐸⃗ 𝑑𝑠
𝜖0

𝑄−𝑞
𝐸 × 4𝜋𝑥 2 = ½
𝜖0
1 𝑄−𝑞 3
𝐸= ½ .
4𝜋𝜖0 𝑥 2

OR

(a) Derivation for electric field due to a uniformly
charged straight wire 2
(b) Graph showing variation of electric field E vs
distance x 1
(a)

½

𝑞
𝐸. ⃗⃗⃗⃗
∮ ⃗⃗⃗ 𝑑𝑆 =
𝜖0
𝜆𝑙
∫ 𝐸⃗ . ⃗⃗⃗⃗⃗⃗
𝑑𝑆1 + ∫ 𝐸⃗ . ⃗⃗⃗⃗⃗⃗
𝑑𝑆2 + ∫ 𝐸⃗ . ⃗⃗⃗⃗⃗⃗
𝑑𝑆3 =
𝜖0
𝜆𝑙
𝐸 𝑑𝑆1 cos 90 + 𝐸𝑑𝑆2 cos 90 + 𝐸𝑑𝑆3 cos 00 =
0 0
½
𝜖0
𝜆𝑙
0 + 0 + 𝐸 × 2𝜋𝑥𝑙 = ½
𝜖0
𝜆
𝐸=
2𝜋𝜖0 𝑥 ½
(b)

3
1

34
(a) Principle of working of potentiometer 1
(b) Finding emf of two cells 1+1

Page 9 of 15

(a) For a steady current flowing through a uniform wire , the
potential difference between any two points is directly
proportional to the length of the wire between the two 1
points
5 ½
(b) Potential gradient = 𝑉𝑐𝑚−1
1000
5 ½
𝐸1 + 𝐸2 = 700 × = 3.5 𝑉 (i) ½
1000
5
𝐸1 − 𝐸2 = 100 × = 0.5 𝑉 (ii)
1000
3
Solving these two equations, we get
½
E1=2V and E2 = 1.5 V
SECTION D
35
(a) (i) Ray diagram showing refraction in a prism 1
( 𝐴+ 𝛿𝑚 )⁄
sin 2
(ii) Derivation 𝜇= 𝐴 2
sin 2

(b) (i)Tracing the path of the ray 1
(ii) Effect on path of the ray 1

(a)
(i)

1

(ii) Derivation

From the figure
∠ 𝐴 + ∠ 𝑄𝑁𝑅 = 1800 (i)
0 ½
In ∆ 𝑄𝑁𝑅 𝑟1 + 𝑟2 + ∠ 𝑄𝑁𝑅 = 180 (ii)
Comparing equation (i) and (ii) we get
𝑟1 + 𝑟2 = 𝐴 (iii)
Total deviation produced 𝛿 = (𝑖 − 𝑟1 ) + ( 𝑒 − 𝑟2 )
𝛿 = 𝑖 + 𝑒 − ( 𝑟1 + 𝑟2 ) = 𝑖 + 𝑒 − 𝐴 (iv) ½

From the graph 𝛿 𝑣𝑠 𝑖 we find that when 𝛿 becomes
minimum i.e. 𝛿𝑚
i = e and r1 = r2 ½

Page 10 of 15

( 𝐴+ 𝛿𝑚 )
From (iv) 𝑖=
2
𝐴
and from (iii) 𝑟 =
2
( 𝐴+ 𝛿𝑚 )⁄ ½
sin 𝑖 sin 2
𝜇= = 𝐴
sin 𝑟 sin 2

(a) (i)

1

(ii) If 𝜇 = 1.4 Total Internal Reflection will not occur as
shown in the figure

(Note: Award this last one mark if student does not draw the 1 5
diagram and conclude correctly.)

OR
(a) Expression for focal length of combination with
labelled diagram 3
(b) Finding refractive index of the liquid 2

(a)

½

For lens A
1 1 1
− = (i)
𝑣1 𝑢 𝑓1

½
𝐹𝑜𝑟 𝑙𝑒𝑛𝑠 𝐵 ∶
𝑣𝑖𝑟𝑡𝑢𝑎𝑙 𝑖𝑚𝑎𝑔𝑒 𝐼1 𝑓𝑜𝑟𝑚𝑒𝑑 𝑏𝑦 𝐴 𝑎𝑐𝑡𝑖𝑛𝑔 𝑎𝑠 𝑜𝑏𝑗𝑒𝑐𝑡
1 1 1
− = (ii)
𝑣 𝑣1 𝑓2
½
Adding equations (i) and (ii)
1 1 1 1 ½
+ = −
𝑓1 𝑓2 𝑣 𝑢

Page 11 of 15

1 1 1 ½
Hence − =
𝑣 𝑢 𝑓
1 1 1
Therefore + = ½
𝑓1 𝑓2 𝐹

(b)
1 1 1 ½
In air 𝑃1 = = ( 𝑎𝜇𝑔 − 1)( − ) (i)
𝑓1 𝑅1 𝑅2
For Liquid ½
1 𝜇 1 1
𝑃2 = = (𝑙 𝑔 − 1)( − ) (ii)
𝑓2 𝑅1 𝑅2
From (i) and (ii)

𝑃1 ( 𝑎𝜇𝑔 −1) ½
=
𝑃2 ( 𝑙 𝜇𝑔 −1)

10 ( 1.5−1 )
= 1.5
−2 ( −1)
𝜇𝑙
5 ½
𝜇𝑙 =
3 5
36
(a) Derivation for decay law 2½
(b) Calculation of mean life 1½
(c) Calculation of fraction of initial mass 1
(a) Let N0 be the initial ( t =0 ) number of radioactive substance
and N be the number of radioactive substance at interval t =
t
𝑑𝑁
Hence rate of radioactive decay = − ∝𝑁 ½
𝑑𝑡
𝑑𝑁
= −𝜆𝑁
𝑑𝑡
𝑁 𝑡
𝑑𝑁
∫ = − ∫ 𝜆 𝑑𝑡 ½
𝑁0 𝑁 0

ln 𝑁 − ln 𝑁0 = −𝜆 𝑡 ½
𝑁
= 𝑒 −𝜆𝑡
𝑁0 ½
𝑁 = 𝑁0 𝑒 −𝜆𝑡 ½

(𝑏 )
𝑇1
2 ½
𝜏=
log 2
4.5 × 109 ½
=
0.693 ½
𝜏 = 6.493 × 109 𝑦𝑒𝑎𝑟𝑠

𝑁 1
(c) = 𝑛
𝑁0 2
𝑁 1
= ( )5 ½
𝑁0 2
𝑁 1 1 31 5
= therefore fraction decaying= (1- )=
𝑁0 32 32 32 ½
OR

Page 12 of 15

(a) Bohr’s Postulate and Derivation of expression 3
(b) Finding ratio of wavelengths 2
(a) Bohr’s Postulates:-

1) An electron in an atom could revolve in certain stable
orbits without the emission of radiant energy ½
2) The electron revolves around the nucleus only in those
orbits for which the angular momentum is some integral
multiple of h/2π where h is the Planck’s constant ½
3) The frequency of the emitted photon when an electron
makes a transition from higher orbit to lower energy orbit
is given by
h ν = E2 – E1 ½
𝑛ℎ
Ln = mvnrn = (i)
2𝜋
𝑚𝑣𝑛 2 1 𝑒2 ½
=
𝑟𝑛 4𝜋𝜖0 𝑟𝑛2
𝑒
𝑣𝑛 =
√4𝜋𝜖0 𝑚𝑟𝑛 ½
Combining wit h equation (i)
1 𝑒2 1
𝑣𝑛 =
𝑛 4𝜋𝜖0 (ℎ⁄𝜋)
𝑛2 ℎ 2 4𝜋𝜖0
𝑟𝑛 = ( ) ( ) ( 2 ) ½
𝑚 2𝜋 𝑒
(b) For shortest wave length
1 1 1
= 𝑅( 2 − )
𝜆𝑆 2 ∞

1 𝑅
= (i) ½
𝜆𝑆 4
For longest wave length
1 1 1
= 𝑅( 2 − 2 )
𝜆𝐿 2 3
1 1
= 𝑅( − )
4 9
5
= 𝑅( ) (ii) ½
36
Dividing equation (i) by equation (ii) we get
(1⁄𝜆 ) (𝑅⁄4)
𝑆 ½
=
(1⁄𝜆 ) (5𝑅⁄36)
𝐿
𝜆𝐿 9
= OR 𝜆𝐿 : 𝜆𝑆 = 9: 5 ½ 5
𝜆𝑆 5

37
(a) Solenoid as a small bar magnet 1
Expression for magnitude of magnetic field 2
(b) Magnitude of magnetic dipole moment 1½
Direction ½

(a) A solenoid may be regarded as a combination of large
number of identical circular current loops in which each
behaves like a magnetic dipole. Hence, the current carrying
solenoid will behave like a small bar magnet. ½
Page 13 of 15

Expression for magnetic field :-

½

Figure shows a solenoid consisting of n turns per unit length
Consider a circular element of thickness dx at a distance x
from the centre of the solenoid
Therefore magnetic field at point P due to this circular
element

𝜇0 𝑛𝑑𝑥𝐼𝑎2
𝑑𝐵 = 3
2[(𝑟 − 𝑥)2 + 𝑎2 ]2 ½
𝜇0 𝑛 𝐼𝑎2 +𝑙 𝑑𝑥
𝐵= ∫ 3
2 −𝑙 [(𝑟 − 𝑥)2 + 𝑎 2 ]2
½
𝐹or point P, r >> a and 𝑟 ≫ 𝑙

𝜇0 𝑛 𝐼𝑎2 +𝑙 𝜇0 𝑛𝐼2𝑙𝑎2
𝐵= ∫ 𝑑𝑥 =
2 𝑟3 −𝑙 2𝑟 3
𝜇0 2𝑚 ½
𝐵=
4𝜋 𝑟 3
½
2
(b) 𝑀 = 𝑁𝐼𝜋𝑎
22
= 5 × 2 × × 49 × 10−4 ½
7
= 154 × 10−3 ½
= 0.154 𝐴𝑚2
½
⃗⃗ 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 𝑥 − 𝑦 𝑝𝑙𝑎𝑛𝑒 𝑜𝑟 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑡𝑜 𝑍 𝑎𝑥𝑖𝑠
𝑀 ½ 5
OR
(a) Derivation for the force between two current
carrying wires 2
Definition of 1 A 1
(b) Calculation of value of F 1½
Effect on equilibrium if F is withdrawn ½

(a)

½

Magnetic field due to the current 𝐼𝑎 flowing in conductor
‘a’ at any point on conductor ‘b’
𝜇0 𝐼𝑎 ½
𝐵𝑎 =
2𝜋𝑑
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(Acting perpendicular downward )
Therefore force on conductor ‘b’ due to field Ba
⃗⃗⃗ = 𝐼𝑏 ( ⃗⃗⃗
𝐹 𝑙𝑏 × ⃗⃗⃗⃗
𝐵𝑎 ) ½
𝜇0 𝐼𝑎
|𝐹⃗⃗⃗⃗⃗⃗
𝑏𝑎 |=𝐼𝑏 𝑙𝑏 × 2𝜋𝑑
𝜇0 𝐼𝑎 𝐼𝑏 𝑙𝑏
=
2𝜋𝑑
⃗⃗⃗⃗⃗⃗⃗
|𝐹 𝑏𝑎 | 𝜇0 𝐼𝑎 𝐼𝑏
= ½
𝑙𝑏 2𝜋𝑑

Definition of 1 A :
Two straight infinitely long parallel conductors are said
to carry 1 A current each when they interact each other
1
with a force of 2 × 10−7 𝑁𝑚−1 , when kept 1m apart in
vacuum
(b) In equilibrium
Restoring Torque = Deflecting Torque
½
𝐹 × 𝑟 = 𝑚 𝐵 sin 𝜃
𝐹 × 10 × 10−2 = 3 × 0.25 × sin 300 ½
3 × 0.25 × 1
𝐹=
10 × 10−2 × 2 ½
= 3.75 N
The magnet oscillates for sometime but finally aligns along the ½
original direction of the external magnetic field. 5

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