CBSE Class 12 Physics Question Paper 2020 Set 55-5-3 Solutions

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

(55/5/3)
General Instructions: -

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Page 1 of 21

answer.)
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Page 2 of 21

MARKING SCHEME: PHYSICS
QUESTION PAPER CODE: 55/5/3
Q. No. Value Points/Expected Answer Marks Total
Marks
SECTION A
1. (b) P/2 1 1
2. (a) 1 1 1
3. (b) 3:4 1 1
4. (a) Net Charge enclosed and permittivity of the medium 1 1
5. (b) 2 r 1 1
6. (a) Forward bias and energy gap of the semiconductor 1 1
7. (b)

1 1

8. (c) L is large and R is small 1 1
9. (d) Optical Signals 1 1
10. (a) v tan  = c 1 1
11. Rectify 1 1
12. 4:1 1 1
13. Conductivity/ Resistivity
( Also give full credit if a student writes semiconducting nature) 1 1
14. Electric current 1 1
15. Electrostatic potential difference/ Electric potential 1 1
16.

1 1

Alternatively

17. (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

Page 3 of 21

18. 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 ½
expression of induced emf ]
(Note: Award this one mark if a student draws the graph in first quadrant as
shown below.)
1

OR
Z = R 2 + (X L − XC )2 ½

[Award one full mark even if the
student just draws the graph ½
without writing expression of
impedance]
1

19. Angular deflection of the galvanometer coil per unit current./deflection per 1 1
unit current
Alternatively
 NAB
Is = alternatively
I K
20.

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
Page 4 of 21

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
2 2 ½
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
SECTION B
21.
Shortcomings of Rutherford model 1
Use of Bohr’s postulates to explain these short comings 1

Rutherford’s Model
(i) Could not explain the stability of the atom ½
(ii) Could not explain the observed line spectrum of hydrogen ½
Bohr’s Atomic Model postulates
(i) Explained the stability of the nucleus by postulating that the electron can
orbit the nucleus in only those special orbits (given by mvr = nh / 2 )
in which it does not radiates energy. ½
(ii) Explained the line spectrum by postulating that the electron emits
photons of well defined unique energies when it undergoes a transition ½
from one permitted orbit to another.
[Note: (i) Award one mark for this part even if the student writes only the
first of the two short comings
(ii) Award one mark for this part even if the student just writes
Bohr’s two postulates: mvr = nh / 2 and h = 2 − E1 ] 2

22.
Finding Planck’s constant from the graph 1
Effect on stopping potential ½
Justification ½
According to Einstein’s Photo electric equation
h = 0 + eVs
eVs = h − 0
h 0
Vs = − ½
e e
since  = c / 

Page 5 of 21

hc 0
 Vs = −
e e
 hc  1  −0 
=  + ½
 e    e 
Comparing with the equation of straight line y=mx +c ½
hc me
(a) The slope of the line m = . Hence, Planck’s constant h =
e c
(b) Stopping potential will remain same ½ 2
Justification
Variation of distance of light source from the metal surface will alter the
intensity while the stopping potential however depends only on the
frequency and not on the intensity of the incident light.
23.
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.

1
2

Page 6 of 21

24.
Calculation of energy per atom ½
Calculations of atom in 2kg deuterium ½
Calculation of total energy released ½
Calculation of time ½

3.27
Energy released per atom = 1.6 10−13 J ½
2
6.023 1023  2000
No. of atoms in 2kg of deuterium = = 6.023 1026
2 ½
3.27
Total energy released = 6.023 1026  1.6 10−13 = 15.76  1013 J
2 ½
total energy 15.76 1013
Time required = = = 1.97 1011 s
power of lamp 800
1.97 1011 ½
Time required = = 2.28 106 years
365  24  60  60
[Note: Give full credit if student calculates time in seconds. Also give full
credit if student make calculations by taking approximation of values up to 2
nearest place of decimals. ]
25.
Finding Planck’s constant from the graph 1
Effect on stopping potential ½
Justification ½
According to Einstein’s Photo electric equation
h = 0 + eVs
eVs = h − 0
h 0
Vs = − ½
e e
since  = c / 
hc 0
 Vs = −
e e
 hc  1  −0 
=  +
 e    e 
Comparing with the equation of straight line y=mx +c
hc me
(a) The slope of the line m = . Hence, Planck’s constant h = ½
e c
(b) Stopping potential will remain same
½
Justification
Variation of distance of light source from the metal surface will alter the
intensity while the stopping potential however depends only on the
½ 2
frequency and not on the intensity of the incident light.

Page 7 of 21

26.
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
dW = Vdq ½
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
q Q2  1 1 
= dq =  CV = QV 
2 ½
0
C 2C  2 2 
Since Energy stored = work done
Q2  1 1  ½
U =  CV = QV 
2

2C  2 2 
Energy density: Electrical energy stored per unit volume is known as
energy density. ½
Alternatively:
1 1 2
Energy density =  0E 2 = 2
2 2 0

Page 8 of 21

27.
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)
SECTION C
28.
(a) Definition of half life 1
(b) Calculation of activity 2

(a) The half life of a radioactive sample is the time in which half of the
original number of its nuclei gets decayed.
Alternatively 1
ln 2 0.6931
T1/2 = =
 
Where 𝜆 is a decay constant of the given radioactive sample.
Alternatively
Half life of a radioactive sample is the time in which the activity of the
sample becomes half of its original activity.
Alternatively
Half life = ( ln 2)( Mean life)
= (0.6932)( Mean life)
(b) Activity, R =  ½
0.69312 0.6931
Now  = =  10−9 year −1
T1/2 4.5 ½
5
92U = N =
Number of atoms in 5g sample of 238  6.023 1023 atoms
238 ½
 0.6931 5 
R =  10−9   6.023  1023  atoms/year
 4.5 238 
=0.0195 10 atoms/year
16

=1.95 1014 atoms/year
0.0195 1016
= atoms/second
86400  365
=6.2 106 atoms/second ½ 3
(Note: Award the last 1/2 mark even if the student writes the answer in
units of atoms/year)

Page 9 of 21

29.
Formation of Potential Barrier 1
Formation of Depletion Region 1
Effect of applying forward bias 1

When a p-n junction is formed, there is a diffusion of the majority charge
carriers, from either side, across the junction. ½
This results in making the p-side acquire a negative charge and the n-side
acquire positive charge. The resulting electric field finally stops (decreases)
the diffusion effect and a stage of equilibrium comes in.
The potential difference developed across the junction is called the barrier
potential. ½
Due to diffusion, the diffusing electrons from the n-side combine with the
holes on the p-side. Similarly the diffusing holes from the p-side combine ½
with the electrons on the n-side.
The resulting charge free region on either side of the junction, is called the ½
depletion region
When a forward bias is applied, the width of the depletion region 1 3
decreases.
OR

Photodiode 1
Working of photodiode 1
V-I characteristics of photodiode 1

Photodiode is a special type of p-n junction diode fabricated with a
transparent window to allow light to fall on the diode. 1
Alternatively
A photodiode is a special purpose p-n junction diode that enables it to
detect optical signals or to observe change in light intensity.
Working
When a photodiode is illuminated with light, (whose photon energy is
greater than its energy gap) electron hole pairs get generated. The ½
magnitude of the resulting photocurrent depends on the intensity of the
incident light.
A photodiode is preferably used under reverse bias. This is because, under
reverse bias, it is easier to observe the change in photocurrent with changes
in light intensity. ½
It therefore becomes an efficient detector of optical signals.
V-I characteristics

1 3

Page 10 of 21

30.
Calculating the total energy of the electron in the second excited state
1
Calculating associated kinetic energy 1

Calculating associated de-Broglie wavelength 1

−13.6 ½
Energy of the electron in the second excited state = eV
32
= −1.51eV ½
Kinetic energy (K) in this state = eV
1
h h
de-Broglie wavelength  = = ½
p 2mK
−34
6.63 10
= m
(2  9.110 1.511.6 10−19 )1/2
−31

6.63
= 1/2
10−9 m ½ 3
(43.97)
= 10−9 m =nm
31.
(a) Difference between electrical resistance and resistivity 2
(b) Obtaining the expression for effective resistivity 1

(a) Electrical resistance (R) of a conductor equals the ratio of the potential
difference (V) applied across it, to the resulting current (I) flowing
V 1
through it. (Alternatively: R = )
I
The resistivity of a conductor equals the resistance of a wire of unit
length and unit area of cross section, drawn from the material of that
l RA 1
conductor. (Alternatively: R =  or  = )
A l
( or any other one relevant difference)

(b) For the parallel combination equivalent resistance is given by
1 1 1
= +
R R1 R2 ½
A1 + A2 A A
= 1 + 2
eq L 1L 2L
Where (A1 + A2 )is the effective area of cross section
of combined rod in parallel combination of the rods.
12 eq
=
( 2 A1 + 1A2 ) ( A1 + A2 )

Page 11 of 21

12 (A1 + A2 ) ½
 eq =
( 2 A1 + 1A2 ) 3
(Note :If a student just writes the expression of equivalent resistance,
award half mark of this part)
32.
(a) Calculation of work done 1½

(b) Calculation of the required distance 1½

(a) Work done in placing the charge Q3
½
= Q3  net potential at the mid point

 1  Q1  1  −Q 2  
= Q3   +  
  0  r / 2   0  r / 2   ½
1 2Q3
= Q1 -Q2 
 0 r ½

(b) Let the required distance be x, to the right of Q1 . We then have net
potential at this point = zero
½
1  Q1 Q2 
− =0
 0  x ( x − r ) 

½ 3
( x − r )Q1 = Q2 x
r
x= Q1
(Q1 − Q 2 )
½
Alternatively
Let the required distance be x to the left of Q1 . We then have
1  Q1 Q2 
− =0
 0  x ( x + r ) 

( x + r )Q1 = Q2 x
− r Q1 r Q1
x= =
(Q1 − Q2 ) (Q 2 − Q1 )

33.
Identifying the optical instrument 1
Calculation of angular magnification produced 2

1
Focal length of objective lens = m = cm
100
Page 12 of 21

1
Focal length of eye piece = m = cm
40
Both the objective, as well as the eye piece, are convex lenses of short focal
lengths. Hence the given optical instrument is a compound microscope 1
(Note: Award this one mark if a student writes directly compound
microscope)
L D 1
When the final image is formed at infinity, total magnification = 
f0 fe
L = 20 cm, f0 = 1cm
D = cm, fe = 2.5cm 1
20 25 3
 Magnification =  = 200
1 2.5
34.
Finding
(i) The charge passed through the loop 1
(ii) Change in magnetic flux through the loop 1
(iii) Magnitude of the magnetic field applied 1

(i) Total charge passed through the loop (Q)

Q = area under the I-t graph ½
1 ½
=  0.4  1 coulomb = 0.2C
2
(ii) Change in magnetic flux
 change in magnetic flux 
Total charge passing =   ½
 R 
Change in magnetic flux = R  0.2C 
= 10  0.2 Wb
½
= Wb
(iii) Magnitude of magnetic field applied
Let B be the magnitude of the magnetic field applied
Initial magnetic flux =   (10 10−4 )Wb
Final magnetic flux = zero
Change in magnetic flux = ( −3 − 0) = 2 ½

 B = 2  103 Wb / m 2 ½ 3
(Note: Award two marks to a student who only calculates charge and not
able to calculate correctly the remaining two parts of this question)
SECTION D

Page 13 of 21

35.
(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 −   )
The instantaneous power supplied to the inductor
PL = IV
= I0sin(t −   )(V0 sin t ) ½

= −I0 V0costsint ½
I V
= − 0 0 sin2t
2

Now average power over a completer cycle,
½
I V
PL = − 0 0 sin2t
2
I V
= − 0 0 sin2t = 0
2
 Average value of sin2t over a complete cycle is zero.
½
Thus average power dissipated over a complete cycle is zero.
(b) (i )X L = 2 fL ½

XL 40 ½
L= = = 0.1/  henry = 0.032H
2 f   200

Maximum power dissipation takes place at resonance ½

1 ½
=
2 LC
½
1
C = F
L  3002  4 2

C= F = 8.8 F ½
0.1 9  10  4
4 2
5

OR

Page 14 of 21

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

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

½+½

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

Magnetic flux linked with linked with the smaller coil = MI

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

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

Alternatively

Page 15 of 21

d
e.m. f . induced = −
dt
d dBc ½
= − (Bc Ai ) = − Ai
dt dt
d  0 I 
= − Ai  
dt  2R 
A  dI
=− i 0
2R dt ½
− −4  4  10−7  5 5
= V
2  ( 20  10−2 )  10−3
20 2  10−6
=− V
2  20
½
= −5  10−6V
36.
(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
1+1
points
(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
= dE =
4 0 (x + r 2 )
2
½
This is directed along AP
Its component along the axis OP of the ring is
x
=dE cos = dE
x +r2
2

Page 16 of 21

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
E =  dE cos ½

1 dq x
4  x + r
=
0 ( 2 2
) x +r2
2

1 x
4 0 (x + r 2 )3/2 
= 2
 dl

1 x
=  2 r ½
4 0 (x + r 2 )3/2
2

Q x ½
=
4 0 (x + r 2 )3/2
2

Where Q =   2 r = total charge on the ring
This field is directed along the axis.

When x muchlarger thanr, we have

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

This corresponds to the expression for the electric field due to a point 5
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
E = 1
0

Page 17 of 21

½

½

Let the charge be uniformly distributed on a wire
𝜙 = ∮ 𝑑𝜙 = ∫ 𝐸⃗ ⋅ ds1 + ∫ 𝐸⃗ ⋅ ds2 + ∫ 𝐸⃗ ⋅ ds3
𝑠1 𝑠2 𝑠3

= ∫ 𝐸𝑑𝑠1 𝑐𝑜𝑠 00 + ∫ 𝐸𝑑𝑠2 𝑐𝑜𝑠 9 00 + ∫ 𝐸𝑑𝑠3 𝑐𝑜𝑠 9 00
𝑠1 𝑠2 𝑠3 ½
= 𝐸 ∫ 𝑑𝑠1 = 𝐸. 2𝜋𝑟𝑙
𝑠1
𝑏𝑦𝐺𝑎𝑢𝑠𝑠′𝑠𝑙𝑎𝑤 ½
𝑞
= 𝐸 ⋅ 2𝜋𝑟𝑙
𝜀0
𝑞 1 𝜆 ½
E= =
2𝜋𝜀0 𝑟𝑙 2𝜋𝜀0 𝑟

(b)
E = 10r+5

dV = −E  dr
10 ½
 dV = −  E dr
1
10
= −  (10r+5)dr
1

10 10
 ½
V = −   10r dr +  5dr 
1 1 
10
r 2 
V = 10   + 5(r )101
 2 1
5
V = −[100 − 1] + 5[10 − 1]
½
V = −  99 + 5  9 = −540V
37.
(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
Page 18 of 21

(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

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

Page 19 of 21

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

The two positions of the lens are as shown

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

Page 20 of 21

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 ½
f = cm = 18.75cm
8

Alternatively

 D 2 − x 2  802 − 202 1+1
F = = = 18.75 cm
 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 5
 = − =
f 50  −30  150 ½
f =  cm ½

Page 21 of 21