CBSE Class 12 Physics Question Paper 2020 Set 55-1-3 Solutions

Strictly Confidential: (For Internal and Restricted use only)
Senior School Certificate Examination-2020
Marking Scheme – PHYSICS THEORY (042)
(55/1/3)
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Page 1 of 17

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awarded.

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Page 2 of 17

MARKING SCHEME: PHYSICS
QUESTION PAPER CODE: 55/1/3
Q.No. Value Points/Expected Answer Marks Total
Marks
SECTION A
1 (A) 1.47 1 1
2 (A) Red colour 1 1
3 (D) The stability of atom was established by the model 1 1
4 (C) 1:3 1 1
5 (D)Material of the turns of the coil 1 1
6 (C) Decrease in relaxation time 1 1
7 (C) Potential difference across the bigger resistor is greater 1 1
8 (D) F/8 1 1
9 (A) No net charge is enclosed in the surface 1 1
10 (B) lesser than the focal length of the eyepiece 1 1
11 Four times 1 1
12 Integral 1 1
OR
Nucleon
13 √3 1 1
14 Attracted/ Concentrated 1 1
15 Eddy 1 1
16 4.8 fm 1 1
OR
1/1836
17 M2 1 1
18 Si & Ge cannot be used for fabrication of visible LED because 1 1
their energy gap is less 1.8eV
19 Conduction current is due to the flow of charges whereas the 1 1
displacement current due to the change of electric field/ flux
between the capacitor plates or due to changing electric field/ flux.
20 Decreases 1 1

SECTION B
21
Explanation of depletion layer and potential barrier
½ + ½ mark
Effect on depletion layer ½ mark
Effect on Potential barrier ½ mark

The small region in the vicinity of the junction which is depleted of ½
free charge carrier and has only immobile ions is called depletion
region/ layer.

The accumulation of negative charges in p - region and positive ½
charges in n- region set up a potential difference across the junction,
which acts as a barrier and is called barrier potential.
½
In forward bias (a) width of depletion layer decreases
(b) value of potential decreases ½ 2

Page 3 of 17

22
Equation of Bohr’s postulates ½ +½
Expression of velocity 1
Centripetal force is provided by electrostatic force
𝐹𝑐 = 𝐹𝑒
𝑚𝑒 𝑣𝑛 2 𝑘𝑍𝑒 2
= ½
𝑟𝑛 𝑟𝑛 2
𝑚𝑒 𝜈𝑛2 𝑟𝑛 = 𝑘𝑍𝑒 2
𝑏𝑢𝑡 ½
𝑛ℎ
𝑚𝑒 𝑣𝑛 𝑟𝑛 =
2𝜋
2 kZe2 ½
vn 
nh
For H atom Z  1
2 ke2
vn  ½ 2
nh
𝑒2
=
2∈𝑜 𝑛ℎ
OR
Two differences between process of emission of photoelectrons
and process of emission of β particles ½+½+½+½

Emission of photoelectron takes place due to the incident photons
on the surface of metal whereas emission of β particles takes place ½+½
in radioactive nucleus spontaneously.
Photoelectrons are not accompanied by emission of any other
particle whereas β particles accompanied by emission of followed ½+½
by neutrino or antineutrino. 2
(Note: award full marks for any other two relevant differences.)

23
Magnetic field at point P 1 ½ mark
Curve ½ mark

a)
𝜇𝑜 𝐼
𝐵= ½
2𝜋𝑥

𝜇𝑜 𝐼 𝜇0 𝐼 𝜇𝑜 𝐼(𝑑 − 2𝑥)
𝐵𝑃 = 𝐵1 − 𝐵2 = − = 1
2𝜋𝑥 2𝜋(𝑑 − 𝑥) 2𝜋(𝑑 − 𝑥)𝑥

b)

½ 2

Page 4 of 17

24
Definition of wavefront ½ mark
Figure ½ mark
Derivation of law of refraction 1 mark

Wavefront is defined as the surface of constant phase;
Alternatively
It is a locus of all the points in the same phase of disturbance ½

½

𝐵𝐶 𝜐1 𝑡
sin 𝑖 = = ½
𝐴𝐶 𝐴𝐶
𝐴𝐸 𝜐2 𝑡
sin 𝑟 = =
𝐴𝐶 𝐴𝐶
sin 𝑖 𝜐1
= ½ 2
sin 𝑟 𝜐2

OR

Lens Maker’s formula 1 mark
Derivation of focal length of three lenses 1 mark

1 1 1 1
∵ − = (𝑛 − 1) ( − ) − − − − − −1
𝜐 𝑢 𝑅1 𝑅2
1
When u=∞ and 𝜐 = 𝑓
1 1 1
= (𝑛 − 1) ( − ) − − − − − − − − − 2
𝑓 𝑅1 𝑅2
½
𝑛2
[𝑛 = ]
𝑛1
From Eq 1 and 2
1 1 1
= − 𝑡ℎ𝑒𝑛 𝑙𝑒𝑛𝑠 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
𝑓 𝜐 𝑢
Page 5 of 17

1 1 1 ½ 2
[Even if the student derives = − for biconvex lens, award
𝑓 𝜐 𝑢
1 ½ marks]
25
Definition of half life 1 mark
Determination of ratio R1 and R2 1 mark

The time interval in which the number of radioactive nuclei
reduced / disintegrated to half of initial value
1
Let R1 and R2 be their activities then

𝑅1 = 𝜆1 𝑁1

𝑅2 = 𝜆2 𝑁2 ½
𝑁1
𝑅1 𝜆1 𝑁1 𝑇 𝑁1 𝑇2
= = 1 =
𝑅2 𝜆2 𝑁2 𝑁2 𝑁2 𝑇1 ½ 2
𝑇2
26.
Calculation of capacitance in case 1 ½
Calculation of capacitance in case 2 ½+½
Relationship between K, K1 and K2 ½
𝐾𝜀0 𝐴
𝐶1 =
𝑑 ½
𝐶2 = 𝐶 ′ + 𝐶 ′′
𝐾1 𝜀0 𝐴
𝑤ℎ𝑒𝑟𝑒 𝐶 ′ =
2𝑑
′′
𝐾2 𝜀0 𝐴 ½
𝐶 =
2𝑑
𝐾𝜀0 𝐴 𝐾1 𝜀0 𝐴 𝐾2 𝜀0 𝐴 ½
= +
𝑑 2𝑑 2𝑑 2
𝐾1 +𝐾2
𝐾= ½
2

27
(a) Principle 1 mark
(b) Circuit diagram for determining unknown resistance
of meter bridge 1 mark

½

Page 6 of 17

Meter bridge works on the principle of a balanced wheatstone
bridge. ½
𝑅1 𝑅
= 3 at null point when Ig=0
𝑅2 𝑅4
(unknown)

1 2

SECTION C
28
(a) Speed of light in material medium 1 mark
(b) (i) Identification and Range ½ + ½ mark
(ii) Identification and Range ½ + ½ mark

(a) Speed of light in medium
1 1
𝜐= = 1
√𝜇𝜖 √𝜇0 𝜇𝑟 𝜖0 𝜖𝑟
(b) (i) Microwave range 0.1mt – 1mm ½+½
(10 m – 10 m)
-3 -1

½+½ 3
(ii) Infrared waves range 1 𝑚𝑚 − 700𝑛𝑚
29
Reason for (a) ,(b) and (c) 1+1+1
(a) The points where the two waves meet out of phase or with
the path difference of an integral multiple of 𝜆/2, they
1
cancel out the contribution of each other.
Alternatively
Due to the destructive interference of two waves.
(b) When light is passed through a polaroid the vibrations of
electric field vector which are perpendicular to the pass
axis of the polaroid are absorbed and only those parallel to
1
pass axis pass, therefore the intensity of transmitted light is
less than that of the incident light.
(c) At any point away from the central bright fringe, path
difference is different for different colours, therefore
1 3
maxima for different colours are formed at different points
and coloured fringes are seen.
30
Circuit diagram of full wave rectifier 1
Working 1
Input and output waveform ½+½

1

Page 7 of 17

Working: For positive cycle of input ac source, diode D1 gets
forward biased and conducts while D2 being reverse biased is not ½
conducting.
For the negative cycle of the ac source, diode D1 would not
conduct but diode D2 would. Thus, we get output voltage during ½
both the positive as well as the negative half of the cycle.

½

½

3
31
a) Internal resistance 1 ½ mark
b) Voltage across R 1 ½ mark

(a)

Current drawn from cell -1
𝐸1 − 𝑉
𝐼1 = ½
𝑟1
Current drawn from cell -2
𝐸2 − 𝑉
𝐼2 =
𝑟2
Resultant current 𝐼 = 𝐼1 + 𝐼2

On solving
𝐸1 𝑟2 + 𝐸2 𝑟1 𝑟2 + 𝑟1
∴𝐼= −𝑉( )
𝑟1 𝑟2 𝑟1 𝑟2

𝐸1 𝑟2 + 𝐸2 𝑟1 𝑟1 𝑟2
∴𝑉= −𝐼( )
𝑟1 𝑟2 𝑟2 + 𝑟1

𝑉 = 𝐸𝑒𝑞 − 𝐼𝑟𝑒𝑞

𝐸1 𝑟2 + 𝐸2 𝑟1 ½
𝐸𝑒𝑞 =
𝑟1 + 𝑟2
Page 8 of 17

𝑟1 𝑟2 ½
𝑟𝑒𝑞 =
𝑟2 + 𝑟1

𝑟1 𝑟2 2×2 ½
𝑟𝑒𝑓𝑓 = = = 1Ω
𝑟1 + 𝑟2 2 + 2
Current through R
𝐸𝑒𝑓𝑓𝑒𝑐𝑡 5 5
𝐼= = = 𝐴 ½
𝑅 + 𝑟𝑒𝑓𝑓 10 + 1 11

P.D across R
5
= × 10 = 4.54 𝑣𝑜𝑙𝑡 ½ 3
11
32
(a)Obtaining the expression for the instantaneous current 2
(b)Graphs showing the variations of V and I with ωt 1

(a)

½

½

The length of these phasors or the amplitude of VR, and VC are:
VRm = im R, VCm = im XC
𝑉𝑜 𝑉𝐶
𝐼𝑜 = =
𝑍 [𝑅2 + ( 1 )2 ]1/2
𝜔𝑐

In RC circuit current leads the voltage by ½
an angle 𝜙 and given by
I=I0 𝑆𝑖𝑛(𝜔𝑡 + 𝜙)
From the phasor diagram the Phase angle ½
𝑋
𝜙 = 𝑡𝑎𝑛−1 𝐶 , currents leads voltage
𝑅
1
𝑊ℎ𝑒𝑟𝑒𝑋𝐶 =
𝜔𝐶

Page 9 of 17

(b)

½+½

3

[Note: A student may draw two graphs separately provided they
are co-related]
33
a) Writing expression for magnetic moment ½ mark
b) Figure ½ mark
Magnetic field and calculation 2 mark

(a) magnetic moment = M= NIA ½
𝑀 = 𝑁𝐼𝜋𝑟 2

½

According to Biot-sevart law
⃗⃗⃗ × 𝑟|
𝜇0 𝐼 |𝑑𝑙
⃗⃗⃗⃗⃗
𝑑𝐵 = ½
4𝜋 𝑟 3
𝜇0 𝐼 𝑑𝑙
𝑑𝐵 =
4𝜋 𝑟 2
𝑑𝐵⊥ components due to diametrically opposite components cancel
out. Only 𝑑𝐵𝑥 components refrain

𝜇0 𝐼𝑑𝑙 ½
𝑑𝐵𝑥 = . 𝑐𝑜𝑠𝜃
4𝜋𝑟 2

𝐵 = ∫ 𝑑𝐵𝑥

𝜇0 𝐼𝑅2 1 3
𝐵= (𝑎𝑙𝑜𝑛𝑔 𝑥 𝑎𝑥𝑖𝑠)
2(𝑅2 + 𝑥 2 )3⁄2

Page 10 of 17

OR

a) Definition and expression 1 mark
b) Conversion of Galvanometer
(i) into ammeter 1 mark
(ii) Effective resistance 1 mark

a) Deflection per unit current
𝜃 𝐵𝑁𝐴 ½
𝐼𝑠 = =
𝐼 𝐾
b) (i) By connecting a low resistance (Rs) in parallel to ½
galvanometer such that ½

(𝐼0 − 𝐼𝑔 )𝑅𝑠 = 𝐼𝑔 𝐺
(ii) effective resistance ½
1 1 1 𝐺 + 𝑅𝑠
= + =
𝑅𝐴 𝑅𝑠 𝐺 𝑅𝑠 𝐺
𝑅𝑠 𝐺 3
∴ 𝑅𝐴 = 1
𝐺+𝑅𝑠

34
KE of α particle 1 mark
Calculation 2 marks

KE of α particle 𝐸𝑘𝛼 = (𝑚𝑦 − 𝑚𝑥 − 𝑚𝛼 )𝑐 2 ½

= 𝑚𝑦 𝑐 2 − 𝑚𝑥 𝑐 2 − 𝑚𝛼 𝑐 2 ½
= (235 × 7.8 – 231 × 7.835 – 4 × 7.07) MeV ½
= 1833 – 1809.885 – 28.28 ½
= 1833 – 1838.165 = -5.165 MeV 1 3
Ek < 0 wrong information
[Award full marks till this step]
SECTION D
35.
a) Labelled diagram 2 marks
Figure
Expression for resolving power 1 mark
b) Calculation of angular magnification 1 mark
Diameter of image formed by objective lens
1 mark

a)

2

Page 11 of 17

1

1

½
𝐷
Resolving power of telescope =
1.22𝜆
½ 5
𝛽 𝑓𝑜 20𝑚
b) (i) Angular magnification 𝑚 = = = = 2000
𝛼 𝑓𝑒 10−2 𝑚
(ii)
𝐷 𝑥
=
𝑑 𝑓𝑜
𝐷𝑓0 3.5 × 106 × 20
𝑑= = = .18m
𝑥 3.8 × 108

OR

(a) Labelled diagram 1 mark
Derivation of mirror relation 2 marks
(b) Position of image 1 ½ marks 1
Nature of image 1 ½ marks

½

∆𝐴𝐵𝑃 ~𝐴′ 𝐵′ 𝑃
′ ′
𝐴𝐵 𝑃𝐵′
= −−−−−−−−−1
𝐴𝐵 𝑃𝐵

Also ∆𝐴′ 𝐵′ 𝐹 ~𝑀𝑁𝑃 (for small curvature)
𝐴′ 𝐵′ 𝐵′ 𝐹
∴ =
𝑀𝑃 𝑃𝐹 ½

𝐴′ 𝐵′ 𝐵′ 𝐹
= −−−−−−−−−−−2
𝐴𝐵 𝑃𝐹
½
From 1 and 2
𝑃𝐵′ 𝐵′ 𝐹
= −−−−−−−−−−−3
𝑃𝐵 𝑃𝐹

𝑃𝐵′ 𝐵′ 𝑃 + 𝑃𝐹
= − − − − − − − −4
𝑃𝐵 𝑃𝐹 ½
𝑃𝐵 = −𝑢 𝑃𝐵′ = 𝜈 𝑃𝐹 = −𝑓

Page 12 of 17

𝜈 𝜈−𝑓
=
−𝑢 −𝑓
−𝜈𝑓 = −𝜈𝑢 + 𝑢𝑓 ½
1 1 1
= +
𝑓 𝑣 𝑢
(b) According to lens maker’s formula
½
1 1 1
= (𝜇 − 1) ( − )
𝑓 𝑅1 𝑅2
for plano convex lens R1 R and R2∞

1 (𝜇 − 1) 1.5 − 1
= =
𝑓 𝑅 20

∴f=40 cm ½
1 1 1 ½ 5
= −
𝑓 𝜐 𝑢
1 1 1
= −
40 𝜐 −30

𝜐 = −12 𝑐𝑚
Nature: virtual
36.
(a) (i) Electric Field inside hollow sphere 1½ mark
(ii) Electric Field outside hollow sphere 1½ mark
(b) (i) The net outward flux through cylinder 1 mark
(ii) The net charge present inside the cylinder 1 mark

(a)
(i)

½

According to Gauss’s Law

𝑞𝑖𝑛
∮ 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 = ½
∈0

∵ 𝑖𝑛𝑠𝑖𝑑𝑒 ℎ𝑜𝑙𝑙𝑜𝑤 𝑠𝑝ℎ𝑒𝑟𝑒
qin = 0
∴ ∮ 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 = 0
E=0
(ii) ½

½
Page 13 of 17

½

𝑞 = 𝜎4𝜋𝑅2
𝑞
∮ 𝐸⃗ ⃗⃗⃗⃗⃗⃗⃗
. 𝑑𝐴 =
∈0
𝑞
𝐸 ∮ 𝑑𝐴 = ½
∈0
2
𝜎4𝜋𝑅
𝐸. 4𝜋𝑥 2 =
𝜖0
2
𝜎𝑅
𝐸=
𝜖0 𝑥
b)

½

(i) The net outward flux through cylinder ½
𝜙 = 𝐸𝐴 + 𝐸𝐴 = 2𝐸𝐴 𝐴 = 𝜋𝑟 2
= 2 ×200 × 3.14 × 0.05 × 0.05
𝑁 ½
= 3.14 𝑚2
𝐶
½ 5
(ii) The net charge present inside the cylinder
𝑞 = 𝜖0 𝜙
q= 8.854 × 3.14 × 10-12
= 2.78×10-11 C

OR

a) Expression for potential energy 3 marks
b) Equipotential surface due to isolated -ve charge
1 mark
1
c) Work done in assembling the charge 1 mark

(a) Work done in bringing q from infinity against the field
E= 𝑞1 𝑉|𝑟⃗⃗⃗1 | 1

Work done on q2 against the field E= 𝑞2 𝑉|𝑟⃗⃗⃗2 |
Work done on q2 against the field due to q1
𝑞1 𝑞2 ½
=
4𝜋𝜖0 (𝑟12 )
½

Page 14 of 17

Potential energy of the system= Total work done in assembling the
system
𝑞1 𝑞2
= 𝑞1 𝑉(𝑟⃗⃗⃗1 ) + 𝑞2 𝑉(𝑟⃗⃗⃗2 ) +
4𝜋𝜖0 𝑟12

1
b)

½
c) Work done= charge in potential energy

𝑘𝑞1 𝑞2 𝑘𝑞1 𝑞3 𝑘𝑞2 𝑞31
= + +
𝑟12 𝑟13 𝑟23 ½ 5

9 × 109 × 10−12
= [1 × −1 + −1 × 2 + 1 × 2]
0.1
= 9 × 10−2 [−1 − 2 + 2]
= −9 × 10−2 𝐽

37
a) Labelled diagram 1 mark
Derivation for torque 1 mark
Justification of radial magnetic field 1 marks
(b) Calculation of radius of the path 2 marks

1

Magnetic forces of AB and CD are equal and opposite and have
different line of action so constitute torque
Force acting on current carrying arms AB and CD
𝐹1 = 𝐹2 = 𝐵𝐼𝑙 = 𝐹 (𝑠𝑎𝑦)
∴ 𝜏 = 𝐹 × 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑙𝑎𝑟 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑏𝑒𝑡𝑤𝑒𝑒𝑛 𝑡𝑤𝑜 𝑓𝑜𝑟𝑐𝑒 𝑎𝑟𝑚 ½
∴ 𝜏 = 𝐵𝐼𝑙𝑏𝑠𝑖𝑛𝜃
𝑙𝑏 = 𝐴
𝜏 = 𝐵𝐼𝐴 𝑠𝑖𝑛𝜃
Page 15 of 17

For N turn
𝜏 = 𝐵𝐼𝑁𝐴 𝑠𝑖𝑛𝜃 ½
Radial fields always produce maximum torque and removes the
dependence of torque on 𝜃 1
𝑚𝜐 √2𝑚𝐸𝑘
(b) Radius of circular path = = 1
𝐵𝑞 𝐵𝑞

1 2𝑚𝑞𝑉
= √ 2
𝐵 𝑞

1 2𝑚𝑉 1
= √ =
𝐵 𝑞 2 × 10−3
1 5
r = 10m

OR
(a) Labelled diagram 1 mark
Working 1 mark
(i) & (ii) Reason/justification ½ + ½ mark
(b) (i) External force required 1 mark
(ii) Power required 1 mark

1

[Note: Diagram with different windings can also be drawn]
When an alternating voltage is applied to the primary, the resulting
current produces an alternating magnetic flux which links the
secondary and induces an emf

Induced emf across primary coil ½
𝑑∅
𝑒𝑃 = −𝑁𝑝
𝑑𝑡

Induced emf across secondary coil
𝑑∅
𝑒𝑠 = −𝑁𝑠 ½
𝑑𝑡
𝑒𝑠 𝑁𝑠
= =𝑟 ½
𝑒𝑝 𝑁𝑝 ½

(i) to minimise the eddy currents
(ii) To reduce the heat loss
(b)(i)
F=BIl ½

Page 16 of 17

𝐸 𝐵𝜐𝑙
𝐼= =
𝑅 𝑅
𝐵2 𝜐𝑙2
𝐹= ½
𝑅
0.4 × 0.4 × 0.1 × 0.2 × 0.2
=
0.1
= 6.4 × 10-3 N ½
𝑃 = 𝐹. 𝜐 = 6.4 × 10−3 × 0.1 ½ 5
= .64 × 10−3 𝑊

Page 17 of 17