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

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

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Page 1 of 18

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the X for incorrect answer.)
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awarded.

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should be marked as cross (X) and awarded zero (0)Marks.

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Page 2 of 18

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

no net charge is enclosed by the surface
2 (C) 1 1

−𝑞𝐿𝐸
3 (C) 1 1

No current flows in the potentiometer wire at balance
4 (B) 1 1

3:2
5 (D) 1 1

material of the turns of the coil
6 (A) 1 1

increases the resolving power of telescope
7 (A) 1 1

1.47
8 (A) 1 1

red colour
9 (D) 1 1

The stability of atom was established by the model
10 (C) 1 1

1:3
11 0.15G 1 1
12 Eddy 1 1
13 Four times 1 1
14 Integral 1 1
OR
Nucleons
15 √3 1 1
16 1 1
∮ 𝐵. 𝑑𝑙 = 𝜇0 (𝑖𝑐 + 𝑖𝑑 )
17 Decreases or reduce 1 1
18 4.8 fermi 1 1
OR
1
1836
19 M2 1 1
20 Si & Ge cannot be used for fabrication of visible LED because 1 1
their energy gap is less 1.8eV

Page 3 of 18

SECTION B
21
(a) Principle 1 mark
(b) Circuit diagram for determining unknown resistance
of meter bridge 1 mark

½

Meter bridge works on the principle of a balanced wheatstone
bridge.
𝑅1 𝑅 ½
= 𝑅3 at null point when Ig=0
𝑅
2 4
(unknown)

1 2

22
Formula for parallel plate ½ mark
Calculation of effective capacitance of the combination
1 mark
Relation K, K1 and K2 ½ mark

𝑘𝜖𝑂 𝐴
𝐶1 = ½
𝑑
Capacitor are connected in series
𝐶 ′ 𝐶 ′′ 2𝐾1 𝐾2 ∈𝑂 𝐴
𝐶2 = ′ = ( ) 1
𝐶 + 𝐶 ′′ 𝐾1 + 𝐾2 𝑑

C1=C2

2𝐾1 𝐾2 ½
𝐾=
𝐾1 + 𝐾2 2

Page 4 of 18

23
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
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

Page 5 of 18

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

1 1 1 1
∵ − = (𝑛 − 1) ( − ) − − − − − −1 1
𝜐 𝑢 𝑅1 𝑅2

When u=∞ and 𝜐 = 𝑓
1 1 1 ½
= (𝑛 − 1) ( − ) − − − − − − − − − 2
𝑓 𝑅1 𝑅2

𝑛2
[𝑛 = ]
𝑛1
From Eq 1 and 2
1 1 1 ½ 2
= − 𝑡ℎ𝑒𝑛 𝑙𝑒𝑛𝑠 𝑓𝑜𝑟𝑚𝑢𝑙𝑎
𝑓 𝜐 𝑢
1 1 1
[Even if the student derives = − for biconvex lens, award
𝑓 𝜐 𝑢
1 ½ marks]
25
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 6 of 18

26
Electrostatic force= centripetal force ½ mark
𝑛ℎ
Angular momentum= ½ mark
2𝜋

Formula for radius of nth orbit 1 mark

𝐹𝑐 = 𝐹𝐸

𝑚𝑒 𝜐𝑛2 𝐾𝑧𝑒 2 ½
= 2
𝑟𝑛 𝑟𝑛

𝑚𝑒 𝑣𝑛2 𝑟𝑛 = 𝐾𝑧𝑒 2
By Bohr’s second postulate

𝑛ℎ
𝐿 = 𝑚𝑒 𝜐𝑛 𝑟𝑛 = ½
2𝜋

𝑛2 ℎ2
𝑟𝑛 =
4𝜋 2 𝑚𝑒 𝑘𝑒 2 𝑍
2 2
𝑛 ℎ
𝑟𝑛 = 2 (∵ 𝑍 = 1) 1 2
4𝜋 𝑚𝑒 𝑘𝑒 2

OR

Two observations 1 mark
Diagram 1 mark

a) ½
(i) There exists a threshold frequency below which no
photoelectron is ejected.
(ii) KE of electron depends linearly on frequency and is ½
independent of intensity of radiation.
[or any other correct observation]
b)

1 2

[only curve is essential to draw]

Page 7 of 18

27
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
SECTION C
28
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

𝑟1 𝑟2 ½
𝑟𝑒𝑞 =
𝑟2 + 𝑟1
Page 8 of 18

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
29
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 𝐼𝑅 2 1 3
𝐵= (𝑎𝑙𝑜𝑛𝑔 𝑥 𝑎𝑥𝑖𝑠)
2(𝑅 2 + 𝑥 2 )3⁄2

Page 9 of 18

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

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

𝑠 𝑅 𝐺 1 3
∴ 𝑅𝐴 = 𝐺+𝑅
𝑠

30
(a) Peak value of current and phasor 1 mark
Potential across R ½ mark
Potential across C ½ mark
(b) Phase difference ½ mark
Identification ½ mark

(a)

½

Page 10 of 18

Peak value of current
𝑉0 𝑉0 ½
𝐼0 = =
𝑍 √𝑋𝑐2 + 𝑅 2
1
𝑋𝑐 =
𝜔𝐶
𝑉0 𝑅
(𝑖) 𝑉𝑅 = 𝐼0 𝑅 = ½
√𝑋𝑐2 + 𝑅 2

𝑉0 ½
(𝑖𝑖) 𝑉𝑐 = 𝐼0 𝑋𝑐 = ( ) 𝑋𝑐
√𝑋𝐶2 + 𝑅 3
(b) From phasor
𝑋𝑐 ½
𝑡𝑎𝑛𝜙 =
𝑅
Current leads the applied voltage by phase 𝜙 ½ 3
31
a) Dependence on distance D from slit 1 mark
b) Dependence on slit separation d 1 mark
c) Dependence on distance between source and slit
1 mark

(a) Fringe width increases, 𝛽 ∝ 𝐷
1 ½+½
(b) Fringe width decreases, 𝛽 ∝ 𝑑
𝑠 𝜆 ½+½
(c) Fringes disappear because 𝑆 < 𝑑 not satisfied ½+½ 3

32
(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
(ii) Infrared waves range 1 𝑚𝑚 − 700𝑛𝑚 3
33
KE of α particle 1 mark
Calculation 2 marks

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

= 𝑚𝑦 𝑐 2 − 𝑚 𝑥 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
Page 11 of 18

Ek < 0 wrong information
[Award full marks till this step]
34
(a) Circuit diagram 1 mark
Working of Zener diode as DC voltage regulator
1 mark
V-I graph ½ mark
(b) Reason of heavy doping ½ mark

(a)

1

If the input voltage increases, the current through Rs and Zener
diode also increases. This increases the voltage drop across Rs 1
without any changes in the voltage across the Zener diode. This is
because in the breakdown region, Zener voltage remains constant
even though the current through that Zener diode changes.

½

½ 3
(b) To decrease the width of depletion region which increases
electric field at the junction.
SECTION D
35
(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

Page 12 of 18

(a)
(i)

½

According to Gauss’s Law

𝑞𝑖𝑛 ½
∮ 𝐸⃗ . ⃗⃗⃗⃗⃗
𝑑𝐴 =
∈0

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

½

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

(i) The net outward flux through cylinder
𝜙 = 𝐸𝐴 + 𝐸𝐴 = 2𝐸𝐴 𝐴 = 𝜋𝑟 2 ½
= 2 ×200 × 3.14 × 0.05 × 0.05
𝑁
= 3.14 𝑚2
𝐶 ½

(ii) The net charge present inside the cylinder
𝑞 = 𝜖0 𝜙 ½
q= 8.854 × 3.14 × 10-12
= 2.78×10-11 C ½ 5

Page 13 of 18

OR

a) Expression for potential energy 3 marks
b) Equipotential surface due to isolated -ve charge
1 mark
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
= 1
4𝜋𝜖0 (𝑟12 )

Potential energy of the system= Total work done in assembling the
system ½
𝑞1 𝑞2
= 𝑞1 𝑉(𝑟⃗⃗⃗1 ) + 𝑞2 𝑉(𝑟⃗⃗⃗2 ) + ½
4𝜋𝜖0 𝑟12

b)

1

c) Work done= charge in potential energy

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

9 × 109 × 10−12
= [1 × −1 + −1 × 2 + 1 × 2]
0.1
= 9 × 10−2 [−1 − 2 + 2]
= −9 × 10−2 𝐽 ½ 5

Page 14 of 18

36
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
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
r = 10m 1 5

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

Page 15 of 18

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

𝑑∅ ½
𝑒𝑃 = −𝑁𝑝
𝑑𝑡

Induced emf across secondary coil

𝑑∅
𝑒𝑠 = −𝑁𝑠
𝑑𝑡
𝑒𝑠 𝑁𝑠
= =𝑟 ½
𝑒𝑝 𝑁𝑝

(i) to minimise the eddy currents ½
(ii) To reduce the heat loss ½

(b)
(i)
F=BIl
𝐸 𝐵𝜐𝑙
𝐼= =
𝑅 𝑅
𝐵 2 𝜐𝑙 2 ½
𝐹=
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
= .64 × 10−3 𝑊 ½ 5

Page 16 of 18

37
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

𝐷
Resolving power of telescope = 1.22𝜆 1

𝛽 𝑓 20𝑚 1
b) (i) Angular magnification 𝑚 = 𝛼 = 𝑓𝑜 = 10−2𝑚= 2000
𝑒
(ii)
𝐷 𝑥
= ½
𝑑 𝑓𝑜
𝐷𝑓0 3.5 × 106 × 20
𝑑= = = .18m
𝑥 3.8 × 108 ½ 5

OR

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

1

∆𝐴𝐵𝑃 ~𝐴′ 𝐵 ′ 𝑃
′ ′
𝐴𝐵 𝑃𝐵 ′ ½
= −−−−−−−−−1
𝐴𝐵 𝑃𝐵

Also ∆𝐴′ 𝐵 ′ 𝐹 ~𝑀𝑁𝑃 (for small curvature)

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𝐴′ 𝐵 ′ 𝐵 ′ 𝐹
∴ =
𝑀𝑃 𝑃𝐹

𝐴′ 𝐵 ′ 𝐵 ′ 𝐹
= −−−−−−−−−−−2
𝐴𝐵 𝑃𝐹

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

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

𝜈 𝜈−𝑓
=
−𝑢 −𝑓
−𝜈𝑓 = −𝜈𝑢 + 𝑢𝑓
1 1 1 ½
= +
𝑓 𝑣 𝑢
(b) According to lens maker’s formula

1 1 1
= (𝜇 − 1) ( − ) ½
𝑓 𝑅1 𝑅2
for plano convex lens R1→ R and R2→∞

1 (𝜇 − 1) 1.5 − 1
= = ½
𝑓 𝑅 20

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

𝜐 = −12 𝑐𝑚 ½
Nature: virtual ½ 5

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