CBSE Class 12 Physics Question Paper 2020 Set 55-5-2 Solutions

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

(55/5/2)
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Page 1 of 19

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MARKING SCHEME: PHYSICS
QUESTION PAPER CODE: 55/5/2
Q.No. Value Points/Expected Answer Marks Total
Marks
SECTION A
1. (a) Forward bias and energy gap of the semiconductor 1 1
2. (d) 1 1

3. (c) L is large and R is small 1 1
4. (d) Optical Signals 1 1
5. (b) / (c) / (b) and (c) v tan  = c 1 1
6. (c) 2f 1 1
7. (d) zero 1 1
8. (c) 3:4 1 1
9. (a) Net Charge enclosed and permittivity of the medium 1 1
10. (b) 2 r 1 1
11. 4:1 1 1
12. Conductivity/ Resistivity 1 1
( Also give full credit if a student writes semiconducting nature)
13. Rectify 1 1
14. Electrostatic potential difference/ Electric potential 1 1
15. Electric current 1 1
16. (i) for constructive interference path difference, p = n  ½

(ii) for destructive interference path difference,p = (2 n + 1) , n = 0,1, 2,3 − −
2
½ 1
Alternatively

p = (2 n − 1) , n = 1, 2,3 − −
2
17. dI
Induced e.m.f. in a coil,  = −L ½
dt

[Award one full mark even if the student just draws
the graph without writing the expression of induced emf ] ½

(Note: Award this one mark if a student draws the graph in first quadrant as 1
shown below.)

Page 3 of 19

OR
Z = R 2 + (X L − XC )2 ½

1

[Award one full mark even if the ½
student just draws the graph
without writing expression of
impedance]

18.

1 1

Alternatively

19.

1

Alternatively: Circular path in the X-Y plane in clockwise sense. ½+½
[ Note: If the student just writes, force on the electron will be along
negative Y axis, i.e.
F = −e (v iˆ)  (B (−kˆ ) = evB (− ˆj ) award 1 mark only )
2
OR
Magnitude of force on side NO is = F
Alternatively
Let force on side MP be = F1 ½
F
Force on side NO = 1
2
F F
Magnitude of net force = F1 − 1 = 1 = F ½
2 2

Page 4 of 19

F1
Therefore force on side NO = =F
2
Give full credit if a student calculates the force as shown below.
 1
F = 0 I1I 2
2
20. Angular deflection of the galvanometer coil per unit current/deflection per 1 1
unit current
Alternatively
 NAB
Is = alternatively
I K
SECTION B
21.
Definition of becquerel 1
Calculation of half life 1

(a) One becquerel (Bq) equals the activity of a radioactive sample that is
disintegrating at the rate of on disintegration per second. 1
Alternatively
one disintegration
1 Bequerel =
one second
(b) R1 = −1 N and R2 = −2 N
R = R1 + R2 = −(1 + 2 ) N
R = −
  = 1 + 2 ½
0.6931
T1/2 = ½ 2
1 + 2
22.
Effect and justification ½+½
Effect and justification ½+½
(i) On increasing the width of the slit, the size of the central bright band
will decrease ½
2
(ii) Justification: Angular width = , i.e. angular width is inversely
a
proportional to the width of the slit ½
(iii)The intensity of central bright band will increase ½
Justification: The amplitude/intensity of light passing through slit has
increased. ½ 2
23.
(a) Relevant explanation 1
(b) Relevant explanation 1

(a) According to wave theory photoelectrons can be emitted using high intensity
incident light of any frequency. 1
Hence, there need not be any threshold frequency for any given
photosensitive surface.
(b) Increase in intensity of light increases the number of incident photons. 1

Page 5 of 19

Hence the number of photoelectrons emitted and therefore, the photoelectric 2
current increases.
24.
Origin of gamma rays and radio waves ½+½
Main application of each ½+½
Gamma rays are emitted by radioactive nuclei/produced in nuclear
reactions. ½
Radio waves are produced by accelerated /oscillating charges/LC circuit. ½
Gamma rays are used for the treatment of cancer/in nuclear reactions. ½
Radio waves are used in communication systems/radio or television 2
communication systems/cellular phones. ½
(or any other correct applications)
25.
Expression for angular momentum ½
Expression for magnetic moment 1
Relation between the two ½
According to Bohr’s model
nh ½
L = Angular momentum =mvr =
2
 = agnetic moment = current  area of the orbit ½

e vr ½
 = e    r 2 =
2
L mvr  2 2m
 = =
 e vr e
e ½ 2
= L
2m
26.
Effect and justification ½+½
Effect and justification ½+½

(i) Intensity of light transmitted by P1 remains unaffected when P1 is
rotated about the direction of propagation of light. ½
Justification: The intensity of unpolarized light transmitted by a Polaroid
does not depend on the orientation of the Polaroid with respect to the
direction of propagation of light. ½
(ii) The intensity of light transmitted by P2 will vary from I1 to zero. ½
Justification: As per Malus’ Law I = I 0 cos2 
Where θ is the angle between the pass axis of the polaroid P2 and the pass
axis of polaroid P1.
As  varies from 00 to    2 will vary from I1 to zero. ½ 2
OR
Definition of wave front 1
Obtaining refracted wave front 1
1
The wave front is a surface of constant phase.
Alternatively
The wave front is the locus of all points that are oscillating in phase.

Page 6 of 19

1

2

27.
Derivation of the expression for the torque 1½
Identification of the orientation of stable equilibrium ½

½

Force on q is qE and a force on –q is –qE.
Hence torque
 = qE  2a sin  ½
 = PE sin 
½
 = P E
For stable equilibrium 𝜃 = 00 ½
OR
Obtaining the expression for the energy stored 1½
Definition of energy density ½
Let the charge on the capacitor plates at any instant, during charging
process be q, amount of work done to supply further charge dq to the
capacitor ½
dW = Vdq
q
where V is the potential difference and equals to
C
Total work doneto charge the capacitor upto charge Q
Q

W =  Vdq
0
Q ½
q Q2  1 1 
=  dq =  CV = QV 
2

0
C 2C  2 2 
Since Energy stored = work done
½
Q2  1 1 
U =  CV = QV 
2

2C  2 2 
½

Page 7 of 19

Energy density: Electrical energy stored per unit volume is known as
energy density.
Alternatively:
1 1 2 2
Energy density =  0E 2 =
2 2 0
SECTION C
28.
(a) Deducing the expression for potential energy 1½
(b) Expression for energy in the presence of an external electric field 1½

(a) Work done in bringing the charge q2 , from infinity, to a point
= q2  potential at the point due to charge q1
½
1 q1
= q2 
4 0 r12
1 q1q 2
 potential energy of the system =
4 0 r12 1
(b)Let the potentials, at two points, due to an external electric field (E) be
V1 and V2 respectively.
Now the total energy of the system is:
 1 q1q 2 
q1V1 + q 2V2 + 
 4 0 r12  1½ 3

29.
Solar cell 1
V-I characteristics ½
Three processes involved ½+½+½

A solar cell is basically a p-n junction which generates emf when solar
radiation falls on the p-n junction. 1
Alternatively:
A solar cell works on the same principle as the photodiode, however, no
external bias applied to it and its junction area is much larger than that of a
photodiode.
V-I Characteristics

½

Three processes involved in the working of the solar cell are
Generation: Light (h  g ) generates electron-hole pairs. ½
Separation: Electric field, of the depletion region, separates the electron
and the holes. ½

Page 8 of 19

Collection: The front contact collects the electrons reaching the n-side and ½
back contact collects holes reaching the p-side.
[Note: For the last part, award one mark if the student just writes the three
names of three processes without giving any explanation.] 3
OR
Circuit diagram 1
Working 1
Input and output waveform 1

1

During one half cycle of the input a.c. signal, only diode 1 is forward biased
and conducts. ½
During the next half cycle of the input ac signal only diode 2 is forward
biased and conducts.
However, due to the use of the centre tapped transformer, the current in the ½
load flows in the same direction during both these half cycles. The current
through the load is therefore unidirectional.

Input waveform
½

Output waveform

½
3

30.
(a) Definition of internal resistance 1
(b) (i) Relevant formula +Graph ½+½
(ii) Relevant formula +Graph ½+½

(a) The resistance offered due to electrolyte of a cell, is called its internal
resistance. 1
(b) V = E − Ir ½

Page 9 of 19

 E  ½
V = IR =  R
r+R

½+½

3

31.
Finding the energy in the first excited state 1
Finding the associated kinetic energy 1
Finding the associated de-Broglie wavelength 1
Energy of the electron in the first excited state
13.6
E1 = − 2 eV = −3.4eV ½
2
= −3.4  1.6  10−19 J
= −5.44  10−19 J ½
Associated kinetic energy =Negative of total energy ½
K = 5.44 10−19 J ½
de-Broglie wavelength,  = h/p
h
= ½
2mK
6.63  10−34
= m
(2  9.1  10−31  5.44  10−19 )1/2
6.63  10−34
= m
(99.008)1/2  10−25
 0.663  10−9 m = 0.663nm = 6.63A
½ 3
32.
Finding
(i) The charge passed through the loop 1
(ii) Change in magnetic flux through the loop 1
(iii) Magnitude of the magnetic flux applied 1

(i) Total charge passed through the loop
Q = area under the I-t graph ½
1
=  0.5  2 = 0.5C ½
2
(ii) Change in magnetic flux = resistance × charge passing ½
= 10× 0.5=5 Wb ½
(iii) change in magnetic flux = change in magnetic field× area of the loop ½
5 Wb = ( B -0)× 10 × 10−4 m2
∴ B =5 × 103 Wb/ m2
(Note: Award two marks for the first part if a student unable to do (ii) and ½
(iii) parts of this question.)

Page 10 of 19

33.
B.E. per nucleon versus mass number curve 1
(i) Explanation of fusion of nuclei on the ascending part 1
(ii) Explanation of fission of nuclei on the descending part 1

1

Explanation for fusion
On the ascending part of the curve B.E./A increases with mass number (A).
So, B.E. of the resultant will be greater than that of the nuclei which are fused
together. Hence the fusion of the nuclei on the ascending part, results in an 1
increase in binding energy i.e. a more stable nucleus.
Explanation for fission
On the descending part of the curve, the B.E. of the heavier nuclei is lower .
Hence the fission of the heavier nuclei, on the descending part, will again 1 3
cause an increase in B.E. and, therefore, more stable nucleus.

34.
(a) Identification of the optical instrument 1
(b) Calculation of magnification 2
(a) Compound microscope
(b) 1

L D
M= 1
f0  fe
L D
= 
1 1
    ½
 P0   Pe 
   
 20 10−2   25 10−2 
=  
 1   1 
 100   50 
= 20 12.5
= 250 ½ 3

Page 11 of 19

SECTION D
35.
(a) Two important characteristics 1+1
(b) Derivation of the expression of the electric field 2½
Showing the behaviour as point charge ½

(a) For equipotential surfaces
(i) Potential has the same value at all points on the surface.
(ii) Electric field is normal to the equipotential surface at all
points 1+1
(iii) Work done in moving any charge between any two points on
the equipotential surface is zero ( any two)

(b)

½

Electric field due to any elemental (point) charge dq, at point P.

1 dq ½
= dE =
4 0 (x + r 2 )
2

This is directed along AP
Its component along the axis OP of the ring is
x
=dE cos = dE
x2 + r 2

The component, perpendicular to the axis gets cancelled by the elemental
electric field due to another elemental charge symmetrically located on the
other side of the axis.
Hence total electric field
½
E =  dE cos
1 dq x
4 0  ( x 2 + r 2 )
=
x +r2
2

1 x
4 0 (x + r 2 )3/2 
= 2
 dl

1 x ½
=  2 r
4 0 (x + r 2 )3/2
2

Q x ½
=
4 0 (x 2 + r 2 )3/2
Where Q =   2 r = total charge on the ring

Page 12 of 19

This field is directed along the axis.

When x muchlarger thanr, we have

Q x 1 Q ½ 5
E = =
4 0 (x )
2 3/2
4 0 x 2

This corresponds to the expression for the electric field due to a point
charge. Thus at large distances the ring behaves like a point charge.

(Note: Award these three marks even if a student tries to attempt this part)

OR
(a) Statement of Gauss’s law 1
Derivation of the expression of the electric field 2½
(b) Finding the increase in potential 1½

1
(a) Gauss law: Electric flux through of a closed surface is times the
0
charge enclosed by the surface.
Alternatively
q 1
E =
0

½

Let the charge be uniformly distributed on a wire
 =  d =  E  ds1 +  E  ds 2 +  E  ds 3 ½
s1 s2 s3

=  Eds1 cos 00 +  Eds2 cos 900 +  Eds3 cos 900
s1 s2 s3

= E  ds1 = E .2 rl ½
s1

by Gauss ' s law
q
= E  2 rl ½
0
q 1 
E= = ½
2 0rl 2 0 r

Page 13 of 19

(b)
E = 10r+5

dV = −E  dr ½
10

 dV = −  E dr
1
10
= −  (10r+5)dr
1
½
10 10

V = −   10r dr +  5dr 
1 1 
10
r 2 
V = 10   + 5(r )101
 2 1
V = −[100 − 1] + 5[10 − 1] ½ 5
V = −  99 + 5  9 = −540V
36.
(a)Definition of focal length 1
Obtaining the relation between focal length and radius of curvature 1½
(b)Calculation of angle of emergence 2
Qualitative change in the angle of emergence ½

(a) Focal length of mirror: It is the distance of the point from the pole of
mirror through which ray of light moving parallel to its principle axis
passes (or appear to come from). 1
Alternatively: It is half of the distance of its centre of curvature from the
pole of a mirror.

½

Let C be the centre of curvature of mirror, MD be the perpendicular
from M to the principal axis.
MCP =  and MFP = 2
MD MD
tan  = , tan 2 = (1) ½
CD FD
For small angles, tan   and tan2  2

Page 14 of 19

MD MD
From equation 1, =2
FD CD
CD
FD = − − − −equation (2)
2
For small   the point D is very close to the point P
∴ FD ≃ FP = 𝑓 and CD ≃ CP = 𝑅
R
 from equation 2, we get f = ½
2
(b) Applying Snell’s law at face AB, we get
3 sin 30 = 1  sin e ½
1
3 = sin e ½
2
3
= sin e
2
sin 60 = sin e
½
e = 600 ½
When the medium (the air) in which the prism is kept is replaced with a liquid
of refractive index 1.3 the angle of emergence would decrease. It is
because bending in the ray of light will be lesser. ½

OR

(a) Definition of resolving power 1
(i) Effect and justification ½+½
(ii) Effect and justification ½+½
(b) Calculation of focal length 2

Resolving power of a telescope is defined as the reciprocal of the smallest
angular separation between two distinct objects whose image can be just 1
resolved by it.

1 D
Alternatively: Resolving power = =
d 1.22

Alternatively

It is the reciprocal of the limit of resolution.

(i) As  increases, R.P. decreases ½
½
D 1
Reason R.P.= i.e. R.P .
1.22  ½
(ii ) As D increases, R.P. increases
D ½
Reason R.P.= i.e. R.P . D
1.22

Page 15 of 19

The two positions of the lens are as shown

for position 1
u = −a cm ½
v = b = +(80 − a ) cm
for position 2
½
u = −(a + 20) cm
v = + (b − 20 ) = +( − a ) cm

By lens formula
1 1 1
− = ½
v u f
1 1 1 1
 + = +
(80 − a ) a (60 − a ) (20 + a )
This gives
a(80-a)=(a+20)(60-a)
or 80a-a 2 = 40a + 1200 − a 2
or a = cm
1 1 1 3+5 8
 = + = =
f 50 30 150 150
150
f = cm = 18.75cm
8 ½

Alternatively

 D 2 − x 2  802 − 202
F = = = 18.75 cm 1+1
 4D  4  80

Alternatively The values of

a = u and b = v simply get interchanged in the two positions. ½
b+a = cm
b-a = cm ½
This gives b = 50cm and a = 30cm
1 1  1  8
 = − =
f 50  −30  150 ½
f =  cm ½ 5

Page 16 of 19

37.
(a) Showing No dissipation of power 2
(b) (i) Calculation of self inductance 1
(ii) Calculation of capacitance 2

(a)

V=V0 sin t
I =  0 sin(t −   )
½
The instantaneous power supplied to the inductor
PL = IV ½
= I0sin(t −   )(V0 sin t )
= −I0 V0costsint
I V
= − 0 0 sin2t ½
2

Now average power over a completer cycle,
I V
PL = − 0 0 sin2t
2
I V
= − 0 0 sin2t = 0
2 ½
 Average value of sin2t over a complete cycle is zero.
Thus average power dissipated over a complete cycle is zero.
(b) (i )X L = 2 fL
½
XL 40
L= = = 0.1/  henry = 0.032H ½
2 f   200

Maximum power dissipation takes place at resonance
½
1
= ½
2 LC
1 ½
C = F
L  3002  4 2

C= F = 8.8 F
0.1 9  104  4 2 ½ 5

OR
(a) Formula 1
Plot of two graphs ½+½
(b) (i) Finding the coefficient of mutual induction 1½
(ii)Finding the induced emf 1½

Page 17 of 19

d ½
e.m.f. induced in the rod = −
dt
1 2
e = Bl  ½
2

½+½

(i) Imagine a current I to flow through the larger coil.

Magnetic flux linked with linked with the smaller coil = MI
½
= Bcentre    10 Wb −4

0 I
=    10−4 Wb
2  20  10 −2
½
0 −3
M=  10 H
4
4  10−7   10 −3
= H
4
=9.9  10−10 H
½
=10−9H

(ii) ½
dI
e.m. f . induced = − M
dt ½
5
= −10−9  −3 V ½
10
−6
= −5 10 V

Alternatively
d
e.m. f . induced = −
dt
d dBc
= − (Bc Ai ) = − Ai ½
dt dt

Page 18 of 19

d  0 I 
= − Ai  
dt  2R 
A  dI
=− i 0
2R dt
− −4  4  10−7  5 ½
= V
2  ( 20  10−2 )  10−3 5

20 2  10−6
=− V
2  20

= −5  10−6V ½

Page 19 of 19