CBSE Class 12 Physics Question Paper 2020 Set 55-3-2 Solutions

Strictly Confidential: (For Internal and Restricted use only)
Senior School Certificate Examination-2020
Marking Scheme – PHYSICS THEORY (042)
(55/3/2)
General Instructions: -

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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: 55/3/2
Q.No. Value Points/Expected Answer Marks Total
Marks
SECTION A
1 (C) 1 1

Zero
2 D 1 1

-F
3 (B) 1 1

𝑄
∈𝑜

4 (A) 1 1

𝑉
2

5 (D) 1 1

Helix
6 (D) 1 1

1 : n2
7 (C) 1 1

heavily doped n-side as well as p-side
8 (D) 1 1

1
𝑛2

9 (B) 1 1

Mobility
10 (A) 1 1

1
∈𝑜

11 White 1 1
12 Intensity 1 1

OR

ℎ(𝜐 − 𝜐𝑜 )
13 Zero 1 1
14 Divergent lens/ Concave lens 1 1
15 √3 1 1
16 Inductor 1 1
Page 3 of 17

17 Z=R 1 1
Impedance = Resistance
18 J.C. Bose observed / produced electromagnetic waves of short 1 1
wavelength / did very significant work in the production of em
waves.
19 Zero 1 1
Or
Eddy currents are produced in metal block / block gets heated
20 Frequency 1 1
SECTION B
21
Formulae ½ + ½ mark
Calculation of R ½ mark
Calculation of terminal voltage ½ mark

𝐸 ½
𝐼=
𝑅+𝑟
12
0.5 =
𝑅+4

𝑅 = 20Ω ½
𝑉 = 𝐸 − 𝐼𝑟 or V=IR ½
𝑉 = 12 − 0.5 × 4 = 10 𝑣𝑜𝑙𝑡 ½ 2
22
Determining power of the combination 1 ½ mark
Nature of combination ½ mark

1 1 1
= −
𝑓 𝑓1 𝑓2
½
1 𝑓2 − 𝑓1
=
𝑓 𝑓1 𝑓2 ½
𝑓2 − 𝑓1 ½
∴𝑃=
𝑓1 𝑓2

Because 𝑓2 < 𝑓1 ∴ P is negative
½ 2
∴ nature is diverging lens

OR

Writing the formula 1 mark
(a) effect of wavelength on Resolving power ½ mark
(b) effect of diameter of lens on Resolving power ½ mark

Resolving power of compound microscope is
1
Page 4 of 17

2𝜇 𝑠𝑖𝑛𝜃
𝑅𝑒𝑠𝑜𝑙𝑣𝑖𝑛𝑔 𝑃𝑜𝑤𝑒𝑟 =
1.22𝜆

Justification of the following is based on the above formula: ½
a) If 𝜆 decreases, Resolving Power increases. ½ 2
b) If diameter of objective lens is increased, 𝑠𝑖𝑛𝜃 increases,
Resolving Power increases
23
a) Finding the change in inductance of solenoid 1 mark
b) Finding the final energy stored in the inductor 1 mark

a)
𝐿 = 𝜇0 𝜇𝑟 𝑛2 𝐴𝑙 ½
or alternatively, 𝐿 ∝ 𝑛2
so, L becomes 4 times ½
1
b) Energy stored == 𝐿𝐼 2
2
As L increases 4 times, energy also increases 4 times. ½
½ 2
24
a) Stating the number of spectral lines ½ mark
Showing the transitions in energy level diagram 1 mark
b) Stating the transition for the shortest wave length emission
½ mark
½
a) number of spectral lines =6
energy level diagram

1

b) n=4 to n=1
½ 2
25
Modification in magnetic field pattern by paramagnetic
material 1 mark
Modification in magnetic field pattern by diamagnetic
material 1 mark

1+1 2

diamagnetic paramagnetic

Page 5 of 17

26
Formula 1 mark
Proving that the kinetic energy of electron is greater that of
proton 1 mark

ℎ ℎ
𝜆= = ½+½
𝑝 √2𝑚𝐸𝑘
where Ek= kinetic energy
for equal 𝜆
1
𝐸𝑘 ∝
𝑚𝑎𝑠𝑠
mass of electron <mass of proton ½ 2
∴Kinetic Energy of electron is more than kinetic energy of proton ½

27
Radiation of electromagnetic wave by an oscillating charge
1 mark
Relation between the frequency of radiated wave and the
frequency of oscillating charge 1 mark

An oscillating charge produces an oscillating electric field in space,
which produces an oscillating magnetic field, which in turn, is a
source of oscillating electric field, and so on. The oscillating electric
and magnetic fields thus regenerate each other, as the wave 1
propagates through the space.

The frequency of the electromagnetic wave equals the frequency of 1 2
oscillation of the charge.

OR

a) Explaining the fact that e.m waves carry energy
1 mark
b) Correct Explanation 1 mark

a) Consider a plane perpendicular to the direction of propagation of
the electromagnetic wave. If there are, on this plane, electric
charges, they will be set and sustained in motion by the electric and 1
magnetic fields of the electromagnetic wave. The charges thus
acquire energy and momentum from the waves.

b) When the sun shines on your hand, you feel the energy being
absorbed from the electromagnetic waves (your hands get warm). 1 2
Electromagnetic waves also transfer momentum to your hand but
Page 6 of 17

because c is very large, the amount of momentum transferred is
extremely small and you do not feel the pressure.

[For any other alternative correct explanation also, award full 2
marks]
SECTION C
28
a) Relationship between Mobility and drift velocity 1 mark
b) Formula 1 mark
Finding the ratio 1 mark

a)
𝑉𝑑
𝜇= 1
𝐸
(Alternatively, if a student writes that mobility 𝜇 is defined as drift
velocity per unit electric field award full marks)

b) 3
𝑒𝜏𝐸 𝑒𝜏𝑉 1
𝑉𝑑 = =
𝑚 𝑚𝑙

𝑉𝑑1 𝑙2 3
= = 1
𝑉𝑑2 𝑙1 2
29
a) Stating the reason for adding impurity atoms ½ mark
b) Naming the two processes 1 mark
Explaining the two processes 1 mark
Creation of potential barrier ½ mark

a) To increase the electrical conductivity / to increase the number
density of charge carriers ½

b) Diffusion and Drift ½+½

Explanation
Diffusion: During the formation of p-n junction, due to the
concentration gradient across the p and n sides, the motion of ½
majority charge carriers give rise to diffusion current.

Drift: Due to the electric field developed at the junction, the
motion of the minority charge carriers due to electric field is called ½
drift.

With the passage of time, diffusion current decreases whereas drift
current increases and balance each other. This, creates a potential ½
barrier. 3

Page 7 of 17

30
a) Explaining the high nuclear density 1 mark
b) Explaining the non-Colombian nature 1 mark
c) Drawing the graph 1 mark
a) Volume of Nucleus is very small but its mass is almost the total
mass of the atom
𝑀𝑎𝑠𝑠
𝑁𝑜𝑤 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 =
𝑉𝑜𝑙𝑢𝑚𝑒

That is why density of nucleus is very high. 1
Alternatively, the matter consisting of atoms, has a very large
amount of empty space.

b) Nuclear forces are very strong, attractive and independent of 1
charge and are short ranged.
Whereas Colombian Force are charge dependent and long range.
(Accept any one point of difference)

1 3

31
Meaning of Matter Waves 1 mark
Finding the ratio of de Broglie wavelengths associated with
proton and alpha particle, when both:-
(a) accelerated through same potential difference 1 mark
(b) have same velocity 1 mark

The de Broglie waves associated with moving particles are called 1
matter waves.
a) (i)
ℎ
𝜆=
√2𝑚𝐸𝑘 ½

ℎ
𝜆𝛼 =
√2𝑚𝛼 𝑞𝛼 𝑉

ℎ
𝜆𝑝 =
√2𝑚𝑝 𝑞𝑝 𝑉

𝜆𝑝 𝑚𝛼 𝑞𝛼 4𝑚𝑝 2𝑞𝑝
=√ =√
𝜆𝛼 𝑚𝑝 𝑞𝑝 𝑚𝑝 𝑞𝑝
½
2√2
=
1
b)
Page 8 of 17

ℎ
𝜆= ½
𝑚𝑣

ℎ ℎ
𝜆𝑝 = & 𝜆𝛼 =
𝑚𝑝 𝑣 𝑚𝛼 𝑣

𝜆𝑝 𝑚 𝛼 4
= = ½ 3
𝜆𝛼 𝑚 𝑝 1

32
a) Ray diagram for concave mirror ½ mark
derivation of mirror formula 2 marks
b) Correct explanation ½ mark

a) Ray diagrams for concave mirror

½

Derivation of Mirror Formula

From the diagram,
∆𝐴′ 𝐵′ 𝐹 & ∆𝑀𝑃𝐹 are similar

𝐵′ 𝐴′ 𝐵′ 𝐹
∴ =
𝑃𝑀 𝐹𝑃

𝐵′ 𝐴′ 𝐵′ 𝐹 ½
= (∵ 𝑃𝑀 = 𝐴𝐵) − − − − − − − 𝑒𝑞1
𝐵𝐴 𝐹𝑃

Since
∠𝐴𝑃𝐵 = ∠𝐴′ 𝑃𝐵′

∆𝐴′ 𝐵′ 𝑃 & ∆𝐴𝐵𝑃 are also similar

𝐵′ 𝐴′ 𝐵′ 𝑃 ½
= − − − − − − − − − 𝑒𝑞 2
𝐵𝐴 𝐵𝑃

Comparing eq. 1 and eq. 2

𝐵′ 𝑃 𝐵′ 𝑃 − 𝐹𝑃
=
𝐵𝑃 𝐹𝑃

Page 9 of 17

As per the sign convention
𝐵′ 𝑃 = −𝜈, 𝐹𝑃 = −𝑓, BP = -u

−𝑣 + 𝑓 −𝑣 𝑣
= =
−𝑓 −𝑢 𝑢
−𝑣𝑢 + 𝑢𝑓 = −𝑣𝑓
Dividing by uvf
1 1 1 1
⇒+ = 3
𝑣 𝑢 𝑓
b) Magnification is different for different object distances ½
33
Labelled circuit diagram 1½ mark
Explanation 1½ mark

1½

Explanation
During positive half of the AC input, diode D1 gets forward biased ½
and conducts and diode D2 gets reverse biased.

During negative half of the AC input, diode D2 gets forward
biased and conducts; and diode D1 gets reverse biased. ½

So, output is obtained during both positive and negative half of the ½
cycle in the same direction. 3

34
a) writing the formula for resonant angular frequency
½ mark
calculating this angular frequency 1 mark
b) writing the formula for Q value ½ mark
calculating Q value 1 mark

a)
1 ½
𝜔𝑜 =
√𝐿𝐶

1 ½
=
√2 × 32 × 10−6

=125 rad/s ½

Page 10 of 17

b)
1 𝐿 𝐿𝜔
𝑄= √ 𝑜𝑟 𝑄=
𝑅 𝐶 𝑅 ½

1 2 2 125
𝑄= √ = 25 Alternatively Q=  25 1 3
10 32×10−6 10
OR

a) Calculating rms value of current 1 mark
calculating peak value of current 1 mark
b) Phase difference between current through inductor and
applied voltage ½ mark
change in phase difference ½ mark

a)
𝑋𝐿 = 𝜔𝐿 = 2𝜋𝜈𝐿
5
∴ 𝑋𝐿 = 2𝜋 × 50 × = 500 Ω
𝜋 ½
200 2
𝐼𝑟𝑚𝑠 = = = 0.4𝐴
500 5
½
𝐼0 = √2 𝐼𝑟𝑚𝑠

= √2 × 0.4 ½
= 0.56 𝐴
[Even if student expresses the answer as (0.4√2)𝐴 give the last ½
marks] ½

𝜋
b) 𝑜𝑟 900
2
decreases
½ 3
½
SECTION D
35
a) Diagram showing direction ½ mark
Derivation of expression 1 ½ marks
Definition of 1 ampere 1 mark
b) Magnetic fields of the wires ½ mark
Net magnetic field and its direction 1 + ½ mark

a)

½

Page 11 of 17

𝜇𝑜 𝐼1
𝐵1 =
2𝜋𝑑 ½

𝐹⃗ = 𝐼(𝑙⃗ × 𝐵
⃗⃗ )

𝐹21 = 𝐼2 𝑙2 𝐵1 𝑠𝑖𝑛90𝑜

𝜇𝑜 𝐼1
= 𝐼2 𝑙2 ½
2𝜋𝑑
Force per unit length
𝐹21 𝜇𝑜 𝐼1 𝐼2
𝑓21 = = ½
𝑙2 2𝜋𝑑

Definition of 1 ampere – One ampere is defied as that steady current
which, when maintained in each of the two very long, straight
parallel conductors of negligible cross section, and placed at a
distance 1 meter apart in vacuum, will produce on each of the 1
conductors a force equal to 2 × 10−7 𝑁 per metre of length.
𝐹
Alternatively, 𝐼1 = 𝐼2 = 𝐼𝐴, 𝑑 = 1𝑚, = 2 × 10−7 𝑁/𝑚
𝑙

b)

⃗⃗ = ⃗⃗⃗⃗⃗
𝐵 𝐵1 + ⃗⃗⃗⃗⃗
𝐵2
½
𝜇𝑜 𝐼1 𝜇𝑜 𝐼2
𝐵= +
2𝜋𝑟1 2𝜋𝑟2

𝜇𝑜 3 3
= ( −2
+ )
2𝜋 6 × 10 6 × 10−2

4𝜋 × 10−7 × 3
=
𝜋 × 6 × 10−2

= 2 × 10−5 𝑡𝑒𝑠𝑙𝑎
Direction of 𝐵⃗⃗ at midpoint is perpendicular to the plane containing 1
the two conductors and pointing downwards.
(Note: give full credit of this direction if student takes direction ½
opposite to the shown in fig and answer accordingly) 5

Page 12 of 17

OR

a) Diagram 1 mark
explaining the shape of the path 2 marks
b) formula ½ mark
calculation 1 mark
result ½ mark

a)

1

Inside the dee, the magnetic field makes the charged particle to ½
move in semi-circular path.
Electric field between the dees accelerates the charged particle. ½
The sign of Electric field is changed in tune with the circular
motion of the particle. ½
Each time, the acceleration increases the energy of the particle.
As the energy increases, radius of circular path increases.
So, the path is spiral. ½

b)
𝑉
𝑅= −𝐺 ½
𝑖𝑔
2𝑉
𝑅1 = − 𝐺 = 𝑅𝑜 − 𝐺
𝑖𝑔
𝑅1 + 𝐺 = 2𝑅𝑜 ½
𝑉
[𝑊ℎ𝑒𝑟𝑒 𝑅𝑜 = ]
𝑖𝑔
Similarily
𝑅2 + 𝐺 = 𝑅𝑜 5
𝑅3 + 𝐺 = 𝑅𝑜 /2 ½
From the above equations,
𝑅1 − 𝑅2 = 2(𝑅2 − 𝑅3
𝑅1 − 3𝑅2 + 2𝑅3 = 0 ½

Page 13 of 17

36
a) Meaning of plane polarised light 1 mark
Diagram ½ mark
Derivation of the relationship between 𝜇 𝑎𝑛𝑑 𝜃 1 ½ marks
b) Each graph 1+1 marks

a) A light whose electric vector direction does not change with
time is a plane polarised light.
Alternatively, if electric vector is confined to one particular plane,
containing direction of propagation it is referred to as plane 1
polarized light.

½

sin 𝑖
𝜇=
sin 𝑟 ½

𝑠𝑖𝑛𝜃
= 𝜋 = 𝑖𝑓 𝑖 = 𝜃
sin ( − 𝜃) ½
2

𝑠𝑖𝑛𝜃 ½
= = tan 𝜃
𝑐𝑜𝑠𝜃

b) (i) (ii)

1+1 5

[Note: also accept if a student plots (ii) graph as follows ]

Page 14 of 17

OR

a) description of experiment with diagram 1 mark
derivation of the expression for fringe width 2 marks
b) finding the wavelength of refracted light 1 mark
finding the speed of refracted light 1 mark

a)

½

S is a monochromatic source of light. S1 and S2 are two pinholes
separated by a distance d. GG' is the screen placed at the distance
½
D from the pinholes.
P is a general point on the screen.
Derivation
2 2 2
𝑑 2 2
𝑑 2 ½
(𝑆2 𝑃) − (𝑆1 𝑃) = [𝐷 + (𝑥 + ) ] − [𝐷 + (𝑥 − ) ]
2 2
𝑑2 𝑑2
= 𝐷2 + 𝑥 2 + + 𝑥𝑑 − 𝐷2 − 𝑥 2 − + 𝑥𝑑
4 4
= 2𝑥𝑑
2𝑥𝑑 2𝑥𝑑
𝑝𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 = 𝑆2 𝑃 − 𝑆1 𝑃 = ≈ ½
𝑆2 𝑃 + 𝑆1 𝑃 2𝐷
𝑥𝑑
𝑃𝑎𝑡ℎ 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 =
𝐷
For maxima
𝑥𝑑 ½
= 𝑛𝜆, 𝑛 = 0,1,2 …
𝐷
𝑛𝜆𝐷
𝑜𝑟 𝑥𝑛 =
𝑑
(𝑛 + 1)𝜆𝐷
𝑥𝑛+1 =
𝑑
𝛽 = 𝑥𝑛+1 − 𝑥𝑛
𝜆𝐷 ½
𝛽=
𝑑
b) 5
𝑐0 𝜐𝜆0 𝜆0
𝜇𝑤 = = = ½
𝑐𝑤 𝜐𝜆𝑤 𝜆𝑤
𝜆0 588 × 3
𝜆𝑤 = = = 441 𝑛𝑚
𝜇𝑤 4 ½
𝑐0 3 × 108 × 3
𝑐𝑤 = = = 2.25 × 108 𝑚/𝑠 ½+½
𝜇𝑤 4

Page 15 of 17

37
a) Diagram ½ mark
Derivation 1 ½ mark
Orientation for maximum and half of the maximum
torque ½ + ½ mark

b) Formula ½ mark
Calculation 1 mark
Result ½ mark

a)

½

From diagram
½
Magnitude of Torque= (𝑞𝐸)(2𝑎 𝑠𝑖𝑛𝜃)
= (2𝑞𝑎)(𝐸 𝑠𝑖𝑛𝜃)
½
= 𝑝𝐸𝑠𝑖𝑛𝜃
For direction

𝜏⃗ = 𝑝⃗ × 𝐸⃗⃗ ½

i) for maximum Torque, dipole should be placed perpendicular to
the direction of electric field
𝜋
𝜃 = 900 = ½
2

ii) For the torque to be half the maximum,
𝜋
𝜃 = 300 =
6 ½
(b)

𝑘𝑞
𝐸𝑃𝐴 = 𝐸𝑃𝐵 ; 𝐸=
𝑟2

𝑘𝑞𝐴 𝑘𝑞𝐵
2
=
𝑥 (2 − 𝑥)2 ½
1 4
2
=
𝑥 2 − 𝑥)2 ½

1 2
= ½
𝑥 2−𝑥
2
𝑥= 𝑚 ½ 5
3

Page 16 of 17

OR

a) Derivation of expression for energy stored 2 ½ marks
Form of energy stored ½ mark
b) formula 1 mark
calculation ½ mark
result ½ mark

a) Work done in adding a charge dq = 𝑑𝑊
= 𝑉𝑑𝑞 ½
𝑞 ½
= 𝑑𝑞
𝑐
∴Total Amount of work(W )in charging a capacitor

1 𝑄
𝑊 = ∫ 𝑑𝑊 = ∫ 𝑞𝑑𝑞
𝐶 0 ½

𝑄2
𝑊=
2𝐶
½
2
(𝐶𝑉) 1
= = 𝐶𝑉 2 ½
2𝐶 2

The electrostatic Energy/ potential energy is stored in the electric
field between the plates. ½
b) 𝐶 = 1𝜇𝐹 = 1 × 10−6 𝐹; 𝑉 = 10 𝑣𝑜𝑙𝑡
𝑄 = 𝐶𝑉
= 1 × 10−6 × 10 1
= 10−5 𝑐𝑜𝑢𝑙𝑜𝑚𝑏 ½
½ 5

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