CBSE Class 12 Physics Question Paper 2020 Set 55-B

Strictly Confidential: (For Internal and Restricted use only)
Senior School Certificate Examination-2020
Marking Scheme – PHYSICS THEORY (042)
(55/B)
General Instructions: -

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Page 1 of 16

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awarded.

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Page 2 of 16

MARKING SCHEME: PHYSICS
QUESTION PAPER CODE: 55(B)
Q.No. Value Points/Expected Answer Marks Total
Marks
SECTION A
1 (B) 1 1
𝜎
∈𝑜

2 (A) 1 1

Converging lens of f=100 cm
3 (D) 1 1

𝐾
>
2

4 (D) 1 1

𝐶1 = 12𝜇𝐶; 𝐶2 = 12𝜇𝐶
5 (C) 1 1

n2R
6 (A) 1 1

𝑚𝑑
(𝑚 + 1)2

7 (D) 1 1

Photoelectric emission will not take place.

8 (C) 1 1

The complete image will be formed with decrease in intensity.
9 (A) 1 1

V/8
10 (A) 1 1

Motion of electrons from n to p side and holes from p to n side.
11 Less than one 1 1
12 Full marks even if not attempted [Question is wrong] 1 1
13 Equilibrium 1 1
14 50% 1 1
15 -27.2 eV 1 1
16 If they cross at a point then there will be two directions of the 1 1
resultant field at that point which is not possible.
17 The horizontal component of Earth magnetic field is zero at the 1 1
magnetic pole/ angle of dip at the magnetic pole is 900.
18 Most of the shorter wavelengths (violet and blue light) are 1 1
scattered.

Page 3 of 16

OR
Magnifying power of compound microscope is more than that of
the simple microscope.
19 At the resonance maximum power is dissipated in the circuit 1 1
(through resistance)
20 Spherical /converging 1 1
OR
1. High resolving power
2. No chromatic aberration ½+½
3. Less spherical aberration
4. Sharper and brighter image [Any two]
SECTION B
21
𝐸1 ∝ 𝑙1 𝑞
Formula 𝜙=𝜖 ½ mark
𝑜

Substitution ½ mark
Calculation and result 1 mark
𝑞 ½
𝜙= ∴ 𝑞 = 𝜙𝜖0
𝜖𝑜

𝑞 = −4𝜋 × 103 𝜖0 = −4𝜋𝜖0 × 103 ½

−1
× 103
9 × 109
1 2
= −1.1 × 10−7 C

22
Formula 𝐹⃗ = 𝑞(𝜐⃗ × 𝐵
⃗⃗ ) ½ mark
Definition 1 ½ mark

𝐹⃗ = 𝑞(𝜐⃗ × 𝐵
⃗⃗ ) ½

𝐹 = 𝑞𝜐𝐵 𝑠𝑖𝑛𝜃

𝐵 = 𝐹 ⁄𝑞𝜐 𝑠𝑖𝑛𝜃
𝐵 = 𝐹 (𝑞 = 1𝐶, 𝜐 = 1 𝑚⁄𝑠 𝑎𝑛𝑑 𝜃 = 900 ½
When a force 1N acts on a unit charge moving perpendicular to a
magnetic field with a unit speed, then the magnetic field is said to 1 2
be of 1 tesla.
OR

Page 4 of 16

Two differences 1 +1 marks

Diamagnetic Paramagnetic
Susceptibility Negative (less than Positive (Between
zero) 0 and 1)
Permeability Between 0 and 1 More than one
Behaviour in Weekly repelled Weekly attracted
external magnetic
field
Effect of No effect Intensity of
temperature magnetisation 1+1 2
decreases
[Any two]
23
Maximum wavelength 𝜆𝑚𝑎𝑥 ½ mark
Minimum wavelength 𝜆𝑚𝑖𝑛 ½ mark
𝜆
Ratio 𝜆𝑚𝑎𝑥 1 mark
𝑚𝑖𝑛

1 1 1 ½
= 𝑅 [ 2 − 2]
𝜆 𝑛1 𝑛2

For minimum wavelength 𝜆𝑚𝑖𝑛 - Electron jumps from n=3 to n=1

1 1 1 8𝑅 ½
= 𝑅 ⌈ 2 − 2⌉ =
𝜆𝑚𝑖𝑛 1 3 9

9
𝜆𝑚𝑖𝑛 =
8𝑅
For maximum wavelength electron jumps from n=3 to n=2

1 1 1 1 1 5𝑅
= 𝑅 ⌈ 2 − 2⌉ = 𝑅 ⌈ − ⌉ =
𝜆𝑚𝑎𝑥 2 3 4 9 36

36
𝜆𝑚𝑎𝑥 =
5𝑅 ½
𝜆𝑚𝑎𝑥 36 × 8𝑅 32
∴ = = ½ 2
𝜆𝑚𝑖𝑛 5𝑅 × 9 5
24
Circuit diagram ½ mark
Equivalent Resistance 1 mark
Value of electric current ½ mark

Page 5 of 16

½

𝑅𝑠 = 1 + 2 = 3Ω

3×3
𝑅𝑝 = = 1.5Ω 1
3+3
𝑅𝑛𝑒𝑡 = 1.5 + 0.5 = 2Ω
𝐸 2
𝐼= = = 1𝐴 ½ 2
𝑅𝑛𝑒𝑡 2
25
Principle ½ mark
Circuit diagram ½ mark
Working 1mark

Principle
p-n junction allows the current to pass only when it is forward
bias.
OR
p-n junction diode offers very low resistance in forward bias and ½
very high resistance in reverse bias.

½

Page 6 of 16

For 1st half cycle, one of the diode (say D1) is in forward bias and
D2 will be in reverse bias. In 2nd half cycle, diode D1 will be in 1 2
reverse bias and D2 will be in forward bias.

26
Formula ½ mark
Calculation 1 ½ mark

1 𝑛
𝑁 = 𝑁0 × (2) ½
7 𝑁0
𝑁 = 𝑁0 − 𝑁0 =
8 8

𝑁0 1 𝑛
∴ = 𝑁0 × ( ) 1
8 2
n=3

∴ Time = n ×half life = 3×20 = 60 days
Alternatively ½
½
𝑁 = 𝑁0 𝑒 𝜆𝑡
𝑁0
𝑁=
8
𝑁0
∴ = 𝑁0 𝑒 −𝜆𝑡
8

⇒ 8 = 𝑒 −𝜆𝑡
⇒ 3 log 𝑒 2 = 𝜆𝑡
𝜆
⇒3= 𝑡
log 𝑒 2
𝜆 1
3= 𝑡
0.693
0.693
∵ 𝑇1⁄ = 20 𝑑𝑎𝑦𝑠 =
2 𝜆
𝑡
⇒3=
20 ½ 2
t=60 days

OR

Formula ½ mark
Calculation 1 ½ mark

1 𝑛
𝑁 = 𝑁0 × (2) ½

3.125 𝑁0
𝑁= 𝑁0 =
100 32

𝑁0 1 𝑛
= 𝑁0 × ( )
32 2

Page 7 of 16

n=5, T1/2 =12.5 years 1

Time taken = n × T1/2

=5 × 12.5=62.5 years ½ 2
[alternate method using the formula 𝑁 = 𝑁0 𝑒 −𝜆𝑡 is also accepted]
27
Diagram ½ mark
Derivation 1 ½ mark

½

𝐹1 = 𝐹2 = 𝐼𝑏𝐵
𝑎 𝑎 ½
𝜏 = 𝐹1 𝑠𝑖𝑛𝜃 + 𝐹2 𝑠𝑖𝑛𝜃
2 2

𝜏 = 𝐼𝑎𝑏𝐵𝑠𝑖𝑛𝜃
𝜏 = 𝐼𝐴𝐵𝑠𝑖𝑛𝜃 ½
⃗⃗⃗ = 𝐼𝐴⃗
∴𝑚
⇒ 𝜏⃗ = 𝑚 ⃗⃗
⃗⃗⃗ × 𝐵 ½ 2
SECTION C
28
a) Energy stored in 900 PF capacitor 1 mark
b) Energy stored in the system after disconnecting 2 marks

a) C=900 PF=900 × 10−12 𝐹
V=100V
1 1
∴ 𝑈𝑖 = 𝐶𝑉 2 = × 900 × 10−12 × 1002 1
2 2
= 4.5 × 10−6 𝐽

b) Common Potential
𝐶1 𝑉1 + 𝐶2 𝑉2
𝑉=
𝐶1 + 𝐶2

Page 8 of 16

900 × 100 + 0 1
𝑉= = 50𝑉
900 + 900
∴Energy stored in the system
1
𝑈𝑓 = (𝐶1 + 𝐶2 )𝑉 2
2
1
= (900 + 900) × 10−12 × 502
2 1 3
= 2.25 × 10−6 𝐽
29
(a) Angle of minimum deviation 1 ½ mark
(b) Refractive index of material 1 ½ mark

a) A=600
3
⇒𝑖= × 60 = 450
4
For ray passes symmetrically angle of deviation should be ½
minimum
𝛿 = 2𝑖 − 𝐴 ½
= 2 × 45 − 60 = 300 ½
b)
𝐴 60 ½
𝑟= = = 300
2 2
For Snell’s law
sin 𝑖 sin 45 ½
𝜇= =
sin 𝑟 sin 30
1
= × 2 = √2
√2
=1.414 ½ 3
30
𝜋
(a) To prove that current lags behind the voltage by 2

1 ½ mark
(b) Reason 1 ½ mark

a) 𝑉 = 𝑉𝑜 𝑠𝑖𝑛𝜔𝑡

from Kirchoff’s Law
𝑑𝐼 ½
𝑉−𝐿 =0
𝑑𝑡

𝑑𝐼 𝑉 𝑉0
⇒ = = 𝑠𝑖𝑛𝜔𝑡
𝑑𝑡 𝐿 𝐿
𝑉0
𝑑𝐼 = 𝑠𝑖𝑛𝜔𝑡 𝑑𝑡
𝐿
𝑉0 ½
⇒ ∫ 𝑑𝐼 = ∫ 𝑠𝑖𝑛𝜔𝑡𝑑𝑡
𝐿
𝑉𝑜
𝐼= (− cos 𝜔𝑡) + 𝑐
𝜔𝐿

Page 9 of 16

𝑉𝑜 𝜋
= sin (𝜔𝑡 − )
𝜔𝐿 2
½
𝜋
⇒ 𝐼 = 𝐼0 sin (𝜔𝑡 − )
2
b) In a.c. circuit
when iron rod is inserted, then opposition (reactance) ½
𝑋𝐿 = 𝜔𝐿 𝑖𝑛𝑐𝑒𝑎𝑠𝑒𝑠
𝜇𝑟 𝜇𝑜 𝑁 2 𝐴 ½
∵𝐿= 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑠
𝑙
∴current decreases ½ 3
Hence brightness of the bulb decreases
31
a) Definition of resolving power 1 mark
b) Expression of magnifying power 1 mark
c) Objective Lens: 1D 1 mark

a) It is the ability of compound microscope to form separate image
of two closely lying point objects. 1

b)
𝑓𝑜 𝑓0
𝑚= 𝑜𝑟 1
−𝑓𝑒 |𝑓𝑒 |
c)
1
𝑓=
𝑃
1
∴ 𝑓1 = = 0.1𝑚
10
1
𝑓2 = = 1𝑚
1
𝑓2 > 𝑓1 ⇒ 𝑓2 𝑤𝑖𝑙𝑙 𝑏𝑒 𝑢𝑠𝑒𝑑 𝑎𝑠 𝑜𝑏𝑗𝑒𝑐𝑡𝑖𝑣𝑒 𝑙𝑒𝑛𝑠 1 3

32
a) Definition of threshold frequency 1 mark
b) Einstein equation ½ mark
Calculation of frequency 1 ½ mark

a) The minimum frequency of incident radiation/ cut off
frequency, below which no photoelectric emission takes place, is 1
called threshold frequency.

b) 𝜙 = 2.5 𝑒𝑉
𝑉0 = 4.1𝑉
Einstein photoelectric equation
𝑒𝑉0 = ℎ𝜐 − 𝜙𝑜 ½

ℎ𝜐 = 𝑒𝑉0 + 𝜙𝑜

= 4.1𝑒𝑉 + 2.5 𝑒𝑉
½
= 6.6 𝑒𝑉

Page 10 of 16

6.6𝑒𝑉 6.6 × 1.6 × 10−19
∴𝜐= =
ℎ 6.63 × 10−34
≈ 1.6 × 1015 𝐻𝑧 1 3

33
a) Principle ½ mark
working 1 ½ mark
b) Necessity of radial and magnetic field 1 mark

(a) Principle: A current carrying tool experience the torque in the ½
magnetic field
Working:
When current flows through the coil, the torque act on it in the
radial magnetic field
𝜏 = 𝑁𝐼𝐴𝐵 − − − − − − − − − −1 ½
The restoring torque provided by spring
𝜏 = 𝐾𝜙 − − − − − − − − − − − 2 ½
K=torsional constant
At equilibrium
𝑁𝐼𝐴𝐵 = 𝐾𝜙
𝐾
𝐼=( )𝜙
𝑁𝐴𝐵
𝐼 = 𝐺𝜙 ½
where G is equals to Galvanometer constant

(b) It makes the deflecting torque independent of orientation of
coil in the magnetic field (i.e. 𝜃 = 900 )/ 1 3

Increase the strength of magnetic field
34
a) Formation of barrier potential 2 mark
b) Effect on the width
(i) decrease in forward bias ½ mark
(ii)increase in reverse bias ½ mark

(a) Diffusion process and drift continues until diffusion current
equals drift current. At this equilibrium potential barrier is formed, 2
which prevent the flow of charge.
(b) (i) Decreasing in forward bias
(ii) Increase in Reverse bias ½
½ 3

Page 11 of 16

SECTION D
35
(a) Definition of potential gradient 1 mark
(b) Principle of Potentiometer 1 mark
[award ½ mark if student writes only Potentiometer]
(c) Method of increasing sensitivity 1 mark
(d) Advantage of potentiometer over voltmeter 1+1

(a) The variation of potential with distance / length is called
potential gradient. 1
(b) The potential difference with the length of wire is directly
proportional to length of the portion of uniform wire/ potential 1
gradient is constant.
(c)(i) By increasing the length 1
(ii) By connecting resistance in series
[any one]

(d) (i) Potentiometer measures the e.m.f. in open circuit so its 1
reading is always accurate.
(ii) Voltmeter always measure the potential difference in
closed circuit hence reading will never be accurate. 1

OR

(a) Balance condition in wheatstone Bridge

(circuit + derivation) ½ + 1 ½ mark
(b) Circuit diagram of metre Bridge 1 mark
Mathematical expression 1 mark
(c) Precaution in metre Bridge 1 mark

(a)

½

By applying Kirchoff’s law in loops ADBA and CBDC

Page 12 of 16

−𝐼1 𝑅1 + 0 + 𝐼2 𝑅2 = 0 (𝐼𝑔 = 0) − (1) ½
And
𝐼2 𝑅4 + 0 − 𝐼1 𝑅3 = 0 − (2) ½
𝐼1 𝑅2
⇒ = 𝑓𝑟𝑜𝑚 (1)
𝐼2 𝑅1
And
𝐼1 𝑅4
= 𝑓𝑟𝑜𝑚 (2)
𝐼2 𝑅3
From (1) and (2)
𝑅2 𝑅4 ½
=
𝑅1 𝑅3

(b)

1

The four arms AB, BC, DA and CD [with resistances R, S, 𝑅𝑐𝑚𝑙1
and 𝑅𝑐𝑚(100−𝑙1) ] obviously form wheatstone bridge with AC as the
battery arm and BD the galvanometer arm.

The balance condition of wheatstone bridge.

𝑅 𝑅𝑐𝑚 𝑙1 𝑙1
= = ½
𝑆 𝑅𝑐𝑚 (100 − 𝑙1 ) (100 − 𝑙1 )
By finding the length l1, the unknown resistance 'R' is known in
terms of standard known resistance S
𝑙1
𝑅=𝑆 ½
(100 − 𝑙1 )

(c) The balance point should be obtained at the midpoint of metre 1 5
bridge wire.
36
(a) Construction ½ mark
Principle ½ mark
(b) Expression of secondary voltage 1½ marks
Expression of secondary current 1½ marks
(c) Factors for energy losses ½ +½ mark

(a) Construction: ½
Description of transformer
Mutual Induction ½

Page 13 of 16

(b) Induced e.m.f in primary coil
𝑑𝜙
𝑒𝑝 = −𝑁𝑝 − − − − − − − − − −1 ½
𝑑𝑡
Induced e.m.f in secondary coil
𝑑𝜙
𝑒𝑠 = −𝑁𝑝 −−−−−−−−−−−2
𝑑𝑡 ½
From 1 and 2
𝑒𝑠 𝑁𝑠
= −−−−−−−−−−−3
𝑒𝑝 𝑁𝑝
If there is no loss of power ½
Input power = Output power
𝑒𝑝 𝐼𝑝 = 𝑒𝑠 𝐼𝑠 ½
𝑒𝑠 𝐼𝑝
⇒ = −−−−−−−−−−−4
𝑒𝑝 𝐼𝑠
½
𝐼𝑝 𝑁𝑠
⇒ = = − − − − − − − − − − −5
𝐼𝑠 𝑁𝑝 ½
(c) copper loss/ flux leakage / Iron loss (any two) ½+½ 5

OR
(a) Principle 1 mark
(b) Working and explanation 2 marks
(c) frequency of cyclotron and its independency
1+1 marks

(a) Principle: A charged particle can be accelerated by using small
oscillating electric field and strong perpendicular magnetic field
OR
A charged particle can be accelerated by applying crossed electric
and magnetic field and its frequency is independent of energy. 1
(b) When charged particle is released between the two Dee’s it
acquires circular path due to perpendicular magnetic field. The 1
charged particle is accelerated again and again due to oscillating
electric field applied across the Dee’s perpendicular to magnetic
field. Every time it accelerates acquire the path of longer radius
½
Magnetic force = centripetal force
𝑚𝜐 2
𝑞𝜐𝐵 =
𝑟
𝑚𝜐
𝑟=
𝑞𝐵
(c) Time period ½
2𝜋𝑟
𝑇=
𝜐
1 𝑣
∴𝜐= = 1
𝑇 2𝜋𝑟
𝑣𝑞𝐵 𝑞𝐵
𝜐= = = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 ½
2𝜋𝑚𝑣 2𝜋𝑚
It is independent of velocity i.e. energy.
½ 5

Page 14 of 16

37
(a) Diffraction pattern 1 mark
(b) Effect on (i) Angular width 1 mark
(ii) Linear width 1 mark
(c) Difference between interference and diffraction (2 points)
1 + 1 marks

(a) Diffraction pattern due to bending of light through the edge of
slit 1
(b) (i)
2𝜆 ½
𝜃=
𝑎
No effect on angular width as it is independent from distance (D) ½

(ii)
2𝜆𝐷
𝛽=
𝑎 ½
Linear width increases with increase in the distance (D) ½

(c)
Interference Diffraction
(1) It is obtained due to (1) It is obtained due to
superposition of two light waves superposition of two light
coming from coherent sources. waves originating from two
different parts of the same
wave front.
(2) Number of equally spaced (2) Centrally bright maxima
bright and dark bands. which is twice as wide as
other maxima.
(3) Intensity of all bright fringes (3) Intensity goes on 2 5
are same. decreasing away from
central maxima.
(any two)
OR

(a) Definition of coherent sources 1 mark
Necessity of coherent source 1 mark
(b) Derivation of resultant intensity 2 marks
Condition of dark and bright fringe ½ + ½ mark

(a) Coherent sources:
Two sources which emit continuous light of same frequency nearly
same amplitude and have constant or zero phase difference are 1
called coherent.
In absence of coherent sources, the phase difference between two
sources will change with time and no stable pattern which will be 1
obtained

Page 15 of 16

(b) Two coherent sources are
𝑦1 = 𝑎 𝑐𝑜𝑠𝜔𝑡

And
𝑦2 = 𝑎 cos (𝜔𝑡 + 𝜙)

the resultant displacement will be given by y = y1 + y2

𝑦 = 𝑎 cos 𝜔𝑡 + 𝑎 cos (𝜔𝑡 + 𝜙) 1
𝜙 𝜙
𝑦 = 2𝑎 cos ⁄2 . cos (𝜔𝑡 + ⁄2)

𝜙
∴ the amplitude of resultant displacement is 2𝑎 cos ⁄2 and
resultant intensity
𝜙 1
𝐼 = 4𝑎2 𝑐𝑜𝑠 2 ⁄2
𝜙
⇒ 𝐼 = 4𝐼0 𝑐𝑜𝑠 2 ⁄2
Condition for bright fringe
𝜙 = 2𝑛𝜋 where n= 0,1,2,3…. ½
Condition for dark fringe 5
𝜋 ½
𝜙 = (2𝑛 + 1) n=1,2,3….
2

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