CBSE Class 12 Question Paper 2023 Solution Physics

e s t i o n P a p er
Qu
Solu t i o n
2023

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Senior School Certificate Examination, 2023
SUBJECT: PHYSICS (042) (PAPER CODE 55/1/1)
General Instructions : -

1 You are aware that evaluation is the most important process in the actual and correct assessment of the
candidates. A small mistake in evaluation may lead to serious problems which may affect the future of
the candidates, education system and teaching profession. To avoid mistakes, it is requested that
before starting evaluation, you must read and understand the spot evaluation guidelines carefully.

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3 Evaluation is to be done as per instructions provided in the Marking Scheme. It should not be done
according to one’s own interpretation or any other consideration. Marking Scheme should be strictly
adhered to and religiously followed. However, while evaluating, answers which are based on latest
information or knowledge and/or are innovative, they may be assessed for their correctness
otherwise and due marks be awarded to them. In class-X, while evaluating two competency-
based questions, please try to understand given answer and even if reply is not from marking
scheme but correct competency is enumerated by the candidate, due marks should be awarded.

4 The Marking scheme carries only suggested value points for the answers

These are in the nature of Guidelines only and do not constitute the complete answer. The students can
have their own expression and if the expression is correct, the due marks should be awarded
accordingly.

5 The Head-Examiner must go through the first five answer books evaluated by each evaluator on the
first day, to ensure that evaluation has been carried out as per the instructions given in the Marking
Scheme. If there is any variation, the same should be zero after delibration and discussion. The
remaining answer books meant for evaluation shall be given only after ensuring that there is no
significant variation in the marking of individual evaluators.

6 Evaluators will mark( √ ) wherever answer is correct. For wrong answer CROSS ‘X” be marked.
Evaluators will not put right (✓)while evaluating which gives an impression that answer is correct
and no marks are awarded. This is most common mistake which evaluators are committing.

7 If a question has parts, please award marks on the right-hand side for each part. Marks awarded for
different parts of the question should then be totaled up and written in the left-hand margin and
encircled. This may be followed strictly.

8 If a question does not have any parts, marks must be awarded in the left-hand margin and encircled.
This may also be followed strictly.

1

9 If a student has attempted an extra question, answer of the question deserving more marks should be
retained and the other answer scored out with a note “Extra Question”.

10 No marks to be deducted for the cumulative effect of an error. It should be penalized only once.

11 A full scale of marks 0 - 70(example 0 to 80/70/60/50/40/30 marks as given in Question Paper) has to
be used. Please do not hesitate to award full marks if the answer deserves it.

12 Every examiner has to necessarily do evaluation work for full working hours i.e., 8 hours every day
and evaluate 20 answer books per day in main subjects and 25 answer books per day in other subjects
(Details are given in Spot Guidelines).This is in view of the reduced syllabus and number of questions
in question paper.

13 Ensure that you do not make the following common types of errors committed by the Examiner in the
past:-

● Leaving answer or part thereof unassessed in an answer book.
● Giving more marks for an answer than assigned to it.
● Wrong totaling of marks awarded on an answer.
● Wrong transfer of marks from the inside pages of the answer book to the title page.
● Wrong question wise totaling on the title page.
● Wrong totaling of marks of the two columns on the title page.
● Wrong grand total.
● Marks in words and figures not tallying/not same.
● Wrong transfer of marks from the answer book to online award list.
● Answers marked as correct, but marks not awarded. (Ensure that the right tick mark is correctly
and clearly indicated. It should merely be a line. Same is with the X for incorrect answer.)
● Half or a part of answer marked correct and the rest as wrong, but no marks awarded.
14 While evaluating the answer books if the answer is found to be totally incorrect, it should be marked
as cross (X) and awarded zero (0)Marks.

15 Any un assessed portion, non-carrying over of marks to the title page, or totaling error detected by the
candidate shall damage the prestige of all the personnel engaged in the evaluation work as also of the
Board. Hence, in order to uphold the prestige of all concerned, it is again reiterated that the
instructions be followed meticulously and judiciously.

16 The Examiners should acquaint themselves with the guidelines given in the “Guidelines for spot
Evaluation” before starting the actual evaluation.

17 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title
page, correctly totaled and written in figures and words.

18 The candidates are entitled to obtain photocopy of the Answer Book on request on payment of the
prescribed processing fee. All Examiners/Additional Head Examiners/Head Examiners are once again
reminded that they must ensure that evaluation is carried out strictly as per value points for each
answer as given in the Marking Scheme.

2

MARKING SCHEME: PHYSICS(042)
Code: 55/1/1
Q.No. VALUE POINTS/EXPECTED ANSWERS Marks Total
Marks
SECTION –A

1. F 1 1
(b)
8
2. (d) 1 1

3. (d) 1  1 1
4. (a) 1 1

5. (a) Repelled by both the poles. 1 1
6. (c) 0.19 V 1 1
7. (c) Resistance ( r ) 1 1
8. d E 1 1
(c) o
dt
9. (a) Zero 1 1
10. (c) n2 1 1
11. (d) 95 nm 1 1
12. (d) Independent of A 1 1
13. (c) 1 1

14. (b) it becomes a p-type semiconductor 1 1
15. (d) 0.01 eV 1 1
16. (d) Assertion (A) is false and Reason (R ) is also false. 1 1
17. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct 1 1
explanation of Assertion (A).
18. (b) Both Assertion (A) and Reason ( R ) are true but reason ( R) is the not 1 1
correct explanation of the Assertion (A)
SECTION-B
3

19.

Calculation of acceleration of alpha particle 2

  
F=q(v×B) ½

= q(3 105 i  (0.4i  0.3 j )) N

F = q (0.9  105 k ) N ½
 
F = m a = q (0.9  105 k)N
 q
a  (0.9  105 k )ms 2
½
m
= 4.8 × 107 ×0.9 ×105 k ms-2
½
= 4.32 × 1012 k ms-2 2
Note: Deduct ½ mark if a student does not mention the direction of
acceleration.
20.
Identification 1
Justification 1

Induced electric field due to changing magnetic field is easily observed. 1
Induced electric field due to changing magnetic field can be easily produced
by various ways like rotating/moving a coil in magnetic field, changing the
shape of coil in magnetic field, bringing bar magnet near a coil etc. 1 2
21. (a)
Ray diagram 1
Proof of Snell’s law of refraction 1

1

AB is incident wave front, incident at an angle i. Let  be time taken by the
wave front to travel distance BC.
BC= v1 τ where v1 is speed of wave in medium 1.
To determine shape of refracted wave front, we draw a sphere of radius
v2 τ, where v2 is speed of wave in medium 2.
CE represents a tangent drawn from point C on sphere, CE is the refracted
wave front.
BC v1 ½
sin i = 
AC AC
AE v 2
sin r = 
AC AC

4

sin i v
= 1 = n21 ½
sin r v2

Note: Give full credit if student derives Snell’s law by taking incident
wavefront in denser medium.

OR
(b)

Reason for preferring reflecting type telescope over refracting
telescope ½+½
Justification ½+½

1
No Chromatic Aberration - No refraction in mirrors
2
No Spherical Aberration - Due to use of parabolic reflector
3
Easy mechanical support required - Mirrors weigh less and can be
supported over entire back surface.
4 High resolving power – Due to Mirror with large diameter. 1+1 2
5 Brighter image – Large mirrors gather more light waves.
(Any two)

22.
Finding the ratio of maximum and minimum intensities 2

I max ( I1  I 2 ) 2
= ½
I min ( I1  I 2 ) 2

I1  I 2  2 I1 I 2
= ½
I1  I 2  2 I1 I 2
5I  4 I ½
=
5I  4 I
9
= ½
1
Alternatively
I1 a 2 4 I 4
  
I2 b2 I 1 ½
a 2

b 1 ½

5

I max (a  b)2
 ½
I min (a  b) 2

I max (2  1)2 9
 =
I min (2  1) 2 1 ½ 2

23.
Calculation of potential energy of electron 1
Calculation of kinetic energy of electron 1

13.6
En = eV = Total energy
n2
For third excited state n=4
13.6 13.6
E4 = 2
= = -0.85 eV ½
4 16
Potential Energy = 2  Total Energy = 2  E4
= 2  (-0.85) eV
= - 1.70 eV ½
Kinetic energy = - (Total Energy) = -E4 ½
= 0.85 eV ½ 2
24. (a)

Difference between intrinsic and extrinsic semiconductor 2

Intrinsic semiconductor Extrinsic semiconductor
1. Pure semiconductor. Semiconductor is Doped with
impurities.

2. Low conductivity at room High conductivity at room
temperature. temperature. 1+1
3. ne = nh n e  nh
(Any one)
Note: Give full credit if a student writes any other relevant correct answer.
OR
(b)
Circuit diagram for forward and reverse biased p-n junction diode
½+½
V-I characteristic (Forward and Reverse bias) ½+½

6

½+½

Forward Bias Reverse Bias

½+½

2
Characteristics of silicon Diode

25.
Formation of potential barrier 2

The diffusion current due to concentration gradient at the junction forms a
space charge region consisting of immobile charge carriers. Due to this an
electric field is generated at the junction giving rise to drift current in a
direction opposite to diffusion current.
The potential at which diffusion current becomes equal to drift current is
called potential barrier. 2 2
SECTION –C
26.
a) Finding electric potential at the centre 1
Finding electric field at the centre 1
b) Finding electric potential at the centre 1

(a)
Electric potential due to point charge
kq ½
V=
R
Value of each charge = - q , Total charge = -12q
k ( 12q )
Total potential V =
R
12kq 12q
V = = ½
R 4 0 R
By symmetry the resultant of all electric field vectors becomes zero. 1
So electric field is zero.

7

(b)
Electric potential is a scalar quantity and does not depend on placement of
charges
12kq 12q 1
Therefore V= = 3
R 4 0 R
27. (a)

Difference between resistance and impedance 1
Obtaining expression for impedance 2

1. Resistance is the opposition offered to both alternating current and direct
current while impedance is the opposition offered to alternating current only.

2. Resistance is independent of frequency of source while impedance
depends on frequency.

3. Resistance is opposition offered by material of the conductor while 1
impedance is combined opposition offered by different electrical
components such as resistor, inductor or capacitor.

(Any One)
(Note: Give credit of this part if a student writes any other correct answer.)

½

VR = imR , Vc = im Xc , VL = imXL ½
im = Peak value of current in the circuit.
   
VL VR  VC  Vm
(Vm)2 = VR2 + (VC – VL)2
= ( imR)2 + (imXc - imXL)2 ½
= im [R2 + (Xc – XL)2]
Vm
im =
R  ( X c  X L )2
2

V ½
im = m where Z = R 2  ( X c  X L )2
Z = impedance

8

OR
(b)
Finding condition for resonance 1
Factors affecting resonant frequency 1
Graph 1

½
Z= R 2  ( X L  X c )2
For maximum current, Z should be minimum therefore to minimize Z
XL = XC ½

Alternatively
½
XL = XC
1
ωL =
C
1 ½
ωr =
LC

Resonant Frequency depends on value of Inductance and Capacitance
½+½

1

3

28.
Finding
a) Induced emf 2
b) Mutual inductance between solenoid and coil 1

a) magnetic field produced in the solenoid near the center
B= o nI ½
Flux linked with the coil wound over solenoid
 = NBA= N πr2 B
= N πr2 o nI ½
d dI ½
Induced emf e = = -πr2 Nnμo (i)
dt dt
2
= - μoπr nN Ioω cos ωt ½

9

dI ½
b) comparing Eq (i) with e =  M
dt
M = μoπr2nN ½
3
29.
Explanation of emission of electron 1
a) variation of photocurrent with collector plate potential for
different intensity 1
b) variation of photo current with intensity of incident radiation 1

According to Einstein’s photoelectric equation
An electron absorbs a quantum of energy ‘hυ’ of incident radiation. If the
energy of absorbed quantum exceeds the minimum energy needed by the
electron to escape from the metal surface (work function φo ), the electron is 1
emitted.

Kmax = hυ - φo

a)

1

b)

1 3

10

30.
a)
Energy level diagram for hydrogen atom 1½
Transitions corresponding to ultraviolet region, visible region and
infrared region ½+½+½

1½

½

½

½

Note: Award 1 ½ mark for energy level diagram if the student does not show
the transitions.
OR
b)

Diagram to show variation 1
Two features of diagram ½+½
Reason for nuclear fusion 1

1

(Note: Award full credit even if a student does not mark so many elements
and does not mention the values of Ebn.)

Features of diagram (any two)
1. Binding energy per nucleon is practically independent of atomic number
for nuclei of middle mass number (30 <A < 170)
2. The curve has maximum of about 8.75 MeV for A= 56 and has a value of
7.6 MeV for A= 238
3. Binding energy per nucleon is lower for both light nuclei (A<30) and
heavy nuclei (A>170) ½+½

Two lighter nuclei fuse together to form heavier nuclei as the binding energy
per nucleon of fused heavier nuclei is more than the binding energy per
nucleon of the lighter nuclei. Thus the final system is more tightly bound

11

than initial system.
Alternatively
To attain the stability 1
3
SECTION –D
31. (a)

i) Statement of coulomb’s law and vector form 1+1
ii) Explanation of Gauss’s law based on coulomb’s law 1
iii) Force exerted by charge A on charge B 2

i) Force between two point charges varies inversely with the square of
distance between the charges and is directly proportional to the product of 1
magnitude of the two charges and acts along the line joining the two charges.

1

Alternatively

 1 q1q2 
F12  r12
4 o r12 3

Where r12 is a vector from charge q2 to charge q1.
ii) In derivation of Gauss’s law, flux is calculated using Coulomb’s law and
1
surface area. Here coulomb’s law involves 2 factor and surface area
r 1
involves r2 factor. When product is taken, the two factors cancel out and flux
becomes independent of r.
iii)

½

 
r  AB  ai  a j
 ½
r = AB  a 2  a 2  2a
 1 q1q2 
F r
4 o r 2

12

 1 q  2 q ( ai  a  j)
F  
4 o ( 2 a ) 2
2a
½
 1 
2 q (i  j )
2 
F  2
4 o 2a 2
 1 q 2
F   (i  j)
4 o 2a 2

 q2
F (i   j) ½
4 2 o a 2

Note: Award 1 mark if a student calculates the magnitude of force only.
 1 q2
F 
4 o a 2

Alternatively
Give full credit if a student uses component method to solve the question.
OR
(b)
i) Derivation of electric field 2
ii) Effect on electric field 1
iii) Finding magnitude and direction of electric field 2

i)

½

q 1
E+q =  2
4 o r  a 2
q 1
E-q =  2
4 o r  a 2
½
The components normal to dipole axis cancel away. The components along
the dipole axis add up.
½
Total electric field is opposite to dipole moment.

13


p
= ½
4 o ( r 2  a 2 )3/ 2
Deduct ½ mark if the expression of electric field is not in vector form.
ii) At far off point r >> a

 p
E
4 o r 3
When distance is halved.

 p
E
r ½
4 o ( )3
2

8 p
=
4 o r 3
 ½
E becomes 8 times

iii)

p1  q  2Cm (along OA)
p2  q  2 Cm (along OD)
pnet  p12  p2 2 ½
= 2 2 q Cm
Electric field at centre O
kp
E = 2 net2 3/2
(r +a ) ½
at point O, r = 0 , a = 1 m
k  2 2q 2 2q
E  2 2kq  ½
13
4 o
Along DC ½

14

Alternatively

kq ½
E=
r2
AC= BD= 2m
r = OA =OB= OC=OD=1m
Electric field at O due to charges at B and D
E1 = EB+ ED
kq kq
E1 = 2  2 along OB
1 1 ½
= 2kq
Electric field at O due to charges at A and C
E2 = EA+ EC
kq kq
E2 = 2  2
1 1 ½
= 2kq along OC
Enet = E12  E2 2
2 2q
= 2 2 kq =
4 o ½

Along DC

Alternatively

Considering AB as dipole, electric field at O
2kq  a 2kqa ½
E1 =   2kqa
1 1 1 1
(( ) 2  ( )2 )3/2 (  )3/2
2 2 2 2
Similarly considering DC as another dipole, electric field at O

15

2 kq  a 2 kq a
E2 =   2 kqa
1 2 1 2 3/2 1 1 3/2
(( ) ( ) ) (  ) ½
2 2 2 2
1 1
Enet = E1+ E2 = 4kqa=  4 q ½
4 o 2
2 2q ½
= 2 2 kq 
4 o
Along DC

5

32. (a)

i) Statement of Biot-Savart’s law 1
Expression for magnetic field 2
Diagram for magnetic field lines ½
ii) Finding current by revolving electron 1½
(i)
The magnetic field at a point due to a current carrying element is
proportional to magnitude of current, element length and inversely 1
proportional to the square of the distance from the element.
 
  dl  r
dB  o
I 3
4 r
o Idl sin 
dB 
4 r2

Consider a circular coil of radius a carrying current I.

½

According to Biot-Savart’s law

  Idl sin 
dB  o ½
4 r2
 
At point A I dl  a

16

  90o ,sin 90o  1
 Idl
Hence dB= o 2
4 a
Magnetic field at centre
2 a 2 a
 Idl ½
B=  dB   o 2
o 0
4 a
o I
B=   2 a
4 a 2
o I ½
B=
2a
Note: Give full credit of 2 marks if a student derives the expression for
magnetic field at the axis of the loop and then puts the distance of point as 0
from the centre.

½

ii) q=e , v=107 ms -1 ,r=10-10 m
q
i=
T ½
qv
=
2πr
ev
=
2πr
½
1.6 1019  107
=
2   1010

0.8
=  102 A
 ½
= 0.255 102 A = 2.55 mA

OR
b)
i) Derivation of expression for force 2
Statement of Rule ½
Conditions for maximum and minimum force ½+½
ii) Calculation of magnitude of force 1½

Consider a rod of uniform cross sectional area A and length l. Let the
number density of mobile charge carriers in it be n.
Thus the total number of mobile charge carriers in it is n l A.
 ½
For steady current I, drift velocity of electrons vd , in the presence of

17


external magnetic field B , the force on these carriers is
   ½
F=n l Aq(vd ×B)
 
=  jAl   B ½
 
= I (l  B )
   ½
Where nqvd is current density ( j ) and jA is current (I)
Fleming’s left hand Rule: If forefinger, middle finger and thumb are
stretched in mutually perpendicular directions, such that forefinger indicates
the direction of magnetic field, middle finger indicates the direction of
current in the conductor, then thumb indicates the direction of force on the
conductor.
Alternatively

Right Hand Thumb Rule : If the fingers of right hand are made to rotate from
  ½
l to B through angle θ, the thumb points in the direction of force on the
current carrying conductor.
Condition for maximum force  = 900

F  I l B sin  = I l B ½
Condition for minimum force  = 0o or 1800

F =0 ½

ii)

 o 2 I1 I 2
F= l
4 d ½
107  2  5  2.5
= 2
 10  102 N ½
2.5 10
= 10-5 N
½ 5

18

33. a)

i) (1) Difference between interference pattern and 1+1
diffraction pattern
(2) Two factors affecting fringe width in young’s double ½ + ½
slit experiment
ii) (1) calculation of angular separation 1
(2) calculation of distance between two maxima 1

(i) (1)
(a) The interference pattern has a number of equally spaced bright and dark
bands while diffraction pattern has a central bright maximum which is twice
as wide as the other maxima.
(b) Interference pattern is obtained by superposing two waves originating
from two narrow slits, while diffraction pattern is a superposition of a
continuous family of waves originating from each point on a single slit.
(c) The maxima in interference pattern is obtained at angle  / a , while the
first minima is obtained at same angle  / a for diffraction pattern.
(d) In interference pattern the intensity of bright fringes remain same while 1+1
in diffraction the intensity falls as we go to successive maxima away from
the center on either side.
(any two)

(2) Factors affecting fringes width
Wave length (  ) / distance of screen from slits (D) / separation between ½+½
slits (d).
(any two)
(ii) (1) d sin  = n  ½
n=1

sin  =
d
 1
For small angle sin    = = radian. ½
100 100

D ½
(2)  = = D
d
1
= 50  102
100
= 50 104 m
= 5 mm ½
OR
(b)
i) Derivation of relation between u and v 3
ii) Finding apparent position 2

19

1

Assume that the aperture of the surface is small as compared to other
distance involved, so that small angle approximation can be made.
For small angles
for NOC , i is the exterior angle
 i = NOM  NCM
MN MN ½
i=  (i)
OM MC
Similarly r = NCM  NIM
MN MN
=  (ii)
MC MI
By Snell’s law
n1sin i = n2 sin r
for small angles ½
n1 i = n2 r

substituting i and r from (i) and (ii) we get
n1 n n n
 2  2 1 ½
OM MI MC
Applying Cartesian coordinates
OM= -u, MI= + v , MC= +R
n2 n1 n2  n1
  ½
v u R

n2 n1 n2  n1
(ii)  
v u R
R= - 6 cm, u = - 3cm , n1 =1.5 n2 = 1
1 1.5 1  1.5 ½
 
v 3 6
1 0.5 1.5
 
v 6 3 ½
1 0.5  3
 ½
v 6
1 2.5

v 6
½
v = - 2.4 cm
5
from the left surface inside the sphere

20

SECTION -E
34.
a) Points at same potential 1
b) Current through arm bg 1
c) Potential difference across R3
OR
c) Power dissipated in R2 2
a) Points (a, b, c)
(d, e)
(j, f, g ,h)
are at same potential 1

Note: Give full credit if a student mentions any two points at same potential
from the above.
b)

According to Kirchhoff’s loop rule
for closed loop abgha
- 6 + 10 I2 + I1 = 0
I1 + 10 I2 = 6 (i)
for closed loop acfha
- 6 + 10 (I1 – I2 ) +I1 = 0
11 I1-10 I2 = 6 (ii)
½
Adding (i) and (ii)
12 I1 = 12
I1 = 1 A
I2 = 0.5 A
½
= current through arm bg
Note: Award 1 mark if a student calculates the current by any other method.

c) VR3 = (I1-I2)  R3
1
= 0.5  5
= 2.5 V
1
OR
(c ) P = (I1-I2) 2  R2 = (0.5)2  5
1
= 1.25 W
1
4

21

35.
a) Tracing of path of ray 1
b) Finding velocity of light 1
c) Explanation of two application of TIR 2
OR
c) Definition of TIR 1
Mentioning two conditions of TIR ½+½
a)

1

From fig. angle of incidence on second face i = 600
critical angle ic = 24.5o
(i )  (ic )
 TIR takes place

c
b) n=
v
c 3  108
v= = = 1.24  108 m/s
n 2.41 1

c) Optical Fibre / Brilliance of diamond / mirage (any two)
Note: Give full credit if students mention the names of applications only. 1+1

OR

c) When light travels from optically denser medium to rarer medium at an
interface and gets reflected back into the same medium the phenomenon is
called as total internal reflection. 1

Conditions for TIR
1. Light must travel from optically denser medium to rarer medium.
2. Angle of incidence at the interface must be greater than the critical angle
for the pair of media. ½+½
4

22

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Senior School Certificate Examination, 2023
SUBJECT: PHYSICS (042) (PAPER CODE 55/1/2)
General Instructions : -

1 You are aware that evaluation is the most important process in the actual and correct assessment of the
candidates. A small mistake in evaluation may lead to serious problems which may affect the future of
the candidates, education system and teaching profession. To avoid mistakes, it is requested that
before starting evaluation, you must read and understand the spot evaluation guidelines carefully.

2 “Evaluation policy is a confidential policy as it is related to the confidentiality of the
examinations conducted, Evaluation done and several other aspects. Its’ leakage to public in any
manner could lead to derailment of the examination system and affect the life and future of
millions of candidates. Sharing this policy/document to anyone, publishing in any magazine and
printing in News Paper/Website etc may invite action under various rules of the Board and
IPC.”

3 Evaluation is to be done as per instructions provided in the Marking Scheme. It should not be done
according to one’s own interpretation or any other consideration. Marking Scheme should be strictly
adhered to and religiously followed. However, while evaluating, answers which are based on latest
information or knowledge and/or are innovative, they may be assessed for their correctness
otherwise and due marks be awarded to them. In class-X, while evaluating two competency-
based questions, please try to understand given answer and even if reply is not from marking
scheme but correct competency is enumerated by the candidate, due marks should be awarded.

4 The Marking scheme carries only suggested value points for the answers

These are in the nature of Guidelines only and do not constitute the complete answer. The students can
have their own expression and if the expression is correct, the due marks should be awarded
accordingly.

5 The Head-Examiner must go through the first five answer books evaluated by each evaluator on the
first day, to ensure that evaluation has been carried out as per the instructions given in the Marking
Scheme. If there is any variation, the same should be zero after delibration and discussion. The
remaining answer books meant for evaluation shall be given only after ensuring that there is no
significant variation in the marking of individual evaluators.

6 Evaluators will mark( √ ) wherever answer is correct. For wrong answer CROSS ‘X” be marked.
Evaluators will not put right (✓)while evaluating which gives an impression that answer is correct
and no marks are awarded. This is most common mistake which evaluators are committing.

7 If a question has parts, please award marks on the right-hand side for each part. Marks awarded for
different parts of the question should then be totaled up and written in the left-hand margin and
encircled. This may be followed strictly.

8 If a question does not have any parts, marks must be awarded in the left-hand margin and encircled.
This may also be followed strictly.

1

9 If a student has attempted an extra question, answer of the question deserving more marks should be
retained and the other answer scored out with a note “Extra Question”.

10 No marks to be deducted for the cumulative effect of an error. It should be penalized only once.

11 A full scale of marks 0-70(example 0 to 80/70/60/50/40/30 marks as given in Question Paper) has to
be used. Please do not hesitate to award full marks if the answer deserves it.

12 Every examiner has to necessarily do evaluation work for full working hours i.e., 8 hours every day
and evaluate 20 answer books per day in main subjects and 25 answer books per day in other subjects
(Details are given in Spot Guidelines).This is in view of the reduced syllabus and number of questions
in question paper.

13 Ensure that you do not make the following common types of errors committed by the Examiner in the
past:-

● Leaving answer or part thereof unassessed in an answer book.
● Giving more marks for an answer than assigned to it.
● Wrong totaling of marks awarded on an answer.
● Wrong transfer of marks from the inside pages of the answer book to the title page.
● Wrong question wise totaling on the title page.
● Wrong totaling of marks of the two columns on the title page.
● Wrong grand total.
● Marks in words and figures not tallying/not same.
● Wrong transfer of marks from the answer book to online award list.
● Answers marked as correct, but marks not awarded. (Ensure that the right tick mark is correctly
and clearly indicated. It should merely be a line. Same is with the X for incorrect answer.)
● Half or a part of answer marked correct and the rest as wrong, but no marks awarded.
14 While evaluating the answer books if the answer is found to be totally incorrect, it should be marked
as cross (X) and awarded zero (0)Marks.

15 Any un assessed portion, non-carrying over of marks to the title page, or totaling error detected by the
candidate shall damage the prestige of all the personnel engaged in the evaluation work as also of the
Board. Hence, in order to uphold the prestige of all concerned, it is again reiterated that the
instructions be followed meticulously and judiciously.

16 The Examiners should acquaint themselves with the guidelines given in the “Guidelines for spot
Evaluation” before starting the actual evaluation.

17 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title
page, correctly totaled and written in figures and words.

18 The candidates are entitled to obtain photocopy of the Answer Book on request on payment of the
prescribed processing fee. All Examiners/Additional Head Examiners/Head Examiners are once again
reminded that they must ensure that evaluation is carried out strictly as per value points for each
answer as given in the Marking Scheme.

2

MARKING SCHEME: PHYSICS(042)
Code: 55/1/2
Q.No. VALUE POINTS/EXPECTED ANSWERS Marks Total
Marks
SECTION -A
1. (b) Decreases in both A and B. 1 1
2. d E 1 1
(c) o
dt
3. (d) 4f 1 1
4. Since no option is correct award 1 mark even if student does not attempt. 1 1
5. (d) 95 nm 1 1
6. (c) Number of protons in nucleus. 1 1
7. (c) 1 1

8. (a) repelled by both the poles 1 1
9. (c) 0.19 V 1 1
10. (b) it becomes a p-type semiconductor 1 1
11. (d) 0.01 eV 1 1
12. (a) 1 1

13. Since no option is correct award 1 mark even if student does not attempt. 1 1
14. (c) 1 1


15. F 1 1
(b)
8
16. (a) Both Assertion (A) and Reason (R) are true and reason (R) is the correct 1 1
explanation of the Assertion (A).
17. (b) Both Assertion (A) and Reason (R) are true, but reason (R) is not the 1 1

3

correct explanation of the Assertion (A).
18. (d) Assertion (A) is false and Reason (R) is also false. 1 1
SECTION-B
19.
Formation of potential barrier 2

The diffusion current due to concentration gradient at the junction forms a
space charge region consisting of immobile charge carriers. Due to this an
electric field is generated at the junction giving rise to drift current in a
direction opposite to diffusion current.
The potential at which diffusion current becomes equal to drift current is
called potential barrier. 2 2
20. (a)

Difference between intrinsic and extrinsic semiconductor 2

Intrinsic semiconductor Extrinsic semiconductor
1. Pure semiconductor. Semiconductor is Doped with
impurities.

2. Low conductivity at room High conductivity at room
temperature. temperature. 1+1
3. ne = nh ne  nh

(Any one)
Note: Give full credit if a student writes any other relevant correct answer.
OR

(b)
Circuit diagram for forward and reverse biased p-n junction diode
½+½
V-I characteristic (Forward and Reverse bias) ½+½

½+½
Forward Bias Reverse Bias

4

½+½

Characteristics of silicon Diode
2
21.
Calculation of acceleration of alpha particle 2

  
F=q(v×B) ½

= q(3 105 i  (0.4i  0.3 j )) N

F = q(0.9 105 k ) N ½
 
F = m a = q(0.9 105 k)N
 q
a  (0.9  105 k ) ms 2
½
m
= 4.8 × 107 ×0.9 ×105 k ms-2
½
= 4.32 × 1012 k ms-2 2
Note: Deduct ½ mark if a student does not mention the direction of
acceleration.
22.
a) Mentioning direction of electric field and magnetic field vectors 1
b) Finding ratio of energy densities 1

a) Electric field vector and magnetic field vector are along y-axis and z-axis 1
or vice versa.
1 1 B 2o
b) uE = o E 2 o uB = ½
2 2 o
uE E2
  o o o 2
uB Bo

Eo 1
Using  c and c=
Bo o o
uE
  o o c 2  1
uB ½
Note: Award full credit of 1 mark even if a student finds ratio by taking uE
and uB as equal.
2

5

23. (a)
Ray diagram 1
Proof of Snell’s law of refraction 1

1

AB is incident wave front, incident at an angle i. Let  be time taken by the
wavefront to travel distance BC.
BC= v1 τ where v1 is speed of wave in medium 1.
To determine shape of refracted wave front, we draw a sphere of radius
v2 τ, where v2 is speed of wave in medium 2.
CE represents a tangent drawn from point C on sphere, CE is the refracted
wave front. ½
BC v1
sin i = 
AC AC
AE v 2
sin r = 
AC AC
sin i v
= 1 = n21 ½
sin r v2

Note: Give full credit if student derives Snell’s law by taking incident
wavefront in denser medium.

6

(b)
OR

Reason for preferring reflecting type telescope over refracting
telescope ½+½
Justification ½+½

1 No Chromatic Aberration - No refraction in mirrors
2 No Spherical Aberration - Due to use of parabolic reflector
3 Easy mechanical support required - Mirrors weigh less and can be
supported over entire back surface.
4 High resolving power – Due to Mirror with large diameter are better. 1+1
5 Brighter image – Large mirrors gather more light waves
(Any two) 2
24.
Calculation of radius of curvature 2

P= 4D, n=1.5
1 1 1 ½
P=  ( n  1)(  )
f R1 R2
R1 = R , R2 = - R
1 2 ½
P=  (n  1)( )
f R
2
4= (1.5-1) ( )
R ½
0.5  2
4= ( )
R
1
R= = 0.25m =25 cm ½
4 2

25.
Calculation of wave length of second line of Lyman series 2

1  1 1 
 R 2  2 ½
  n1 n2 
For second line of Lyman series
n1= 1 , n2= 3
1 1 1  ½
 1.1 107  2  2 
 1 3 
 1
= 1.1 10 7 1  
 9
8
= 1.1 10 7 
9

7

8.8
=  107 m 1 ½
9
9
  10 7 m
8.8
= 1.023 107 m ½ 2
= 1023 Å

SECTION-C
26. a)
Energy level diagram for hydrogen atom 1½
Transitions corresponding to ultraviolet region, visible region and
infrared region ½+½+½

1½

½

½

½

Note: Award 1 ½ mark for energy level diagram if a student does not show
the transitions.
OR
b)

Diagram to show variation 1
Two features of diagram ½+½
Reason for nuclear fusion 1

1

(Note: Award full credit even if a student does not mark so many elements
and does not mention the values of Ebn.)

Features of diagram (any two)
1. Binding energy per nucleon is practically independent of atomic number
for nuclei of middle mass number (30 <A < 170)

8

2. The curve has maximum of about 8.75 MeV for A= 56 and has a value of
7.6 MeV for A= 238
3. Binding energy per nucleon is lower for both light nuclei (A<30) and
heavy nuclei (A>170)
½+½
Two lighter nuclei fuse together to form heavier nuclei as the binding energy
per nucleon of fused heavier nuclei is more than the binding energy per
nucleon of the lighter nuclei. Thus the final system is more tightly bound
than initial system.
Alternatively

To attain the stability 1

3
27.
Calculation of wave length of incident light 3

o  3315Å
Kmax = 1.25 eV = 1.25  1.6  10 19 J
 2  1019 J
 0.2  10 18 J

hc
Work function = o  ½
o
6.63  1034  3 108 ½
= 10
= 0.6 10 18 J
3315 10
Using hv = o  K max
hc ½
 o  K max

= (0.6 1018  0.2  1018 )J
= 0.8  10 18 J ½

hc 6.63  1034  3  108
  ½
0.8 1018 0.8 1018
= 24.86  10-8 m
= 2486  10-10 m = 2486 Å ½
3

28.
a) Calculation of Voltage 1
b) Calculation of current 1
c) Calculation of average power 1

Np =650, Ns = 25 ,Vp =240V, Ip =1.5A
Vs N s ½
a) 
Vp N p

9

Ns 25
Vs =  Vp =  240
Np 650
120
=
13 ½
= 9.23V
N I ½
b) s  p
N p Is
Np  Ip 650 1.5
Is =   39 A ½
Ns 25

c) Pav = VsIs ½
120
=  39  360 W ½
13
3
29. (a)

Difference between resistance and impedance 1
Obtaining expression for impedance 2

1. Resistance is the opposition offered to both alternating current and direct
current while impedance is the opposition offered to alternating current only.

2. Resistance is independent of frequency of source while impedance
depends on frequency.

3. Resistance is opposition offered by material of the conductor while 1
impedance is combined opposition offered by different electrical
components such as resistor, inductor or capacitor.

(Any One)
(Note: Give credit of this part if a student writes any other correct answer.)

½

VR = imR , Vc = im Xc , VL = imXL
im = Peak value of current in the circuit.
    ½
VL VR  VC  Vm
(Vm)2 = VR2 + (VC – VL)2
= ( imR)2 + (imXc - imXL)2

10

= im [R2 + (Xc – XL)2] ½
Vm
im =
R 2  ( X c  X L )2
Vm ½
im = where Z = R 2  ( X c  X L )2
Z = impedance
OR
(b)
Finding condition for resonance 1
Factors affecting resonant frequency 1
Graph 1

Z= R 2  ( X L  X c )2 ½
For maximum current, Z should be minimum therefore to minimize Z ½
XL = XC

Alternatively

XL = XC ½
1
ωL =
C
1
ωr =
LC ½

Resonant Frequency depends on value of Inductance and Capacitance ½+½

1

3
30.
a) Calculation of electrostatic energy stored by the capacitor 1
b) Calculation of electrostatic energy stored by system 2

C= 100  F  100  106 F , V  12V
1
a) U = CV 2 ½
2

11

1
=  100  10 6  (12) 2
2
1
=  104  144 ½
2
= 72  10 4 J = 7.2 m J
b) Ceq = C1 + C2 ½
= 200  F
Q  CV
= 100  10 6  12 ½
= 12  10 4 C
Q2 (12  104 ) 2
U = ½
2Ceq 2  200  106
144
=  10 4
4 3
= 36×10-4 J = 3.6 m J ½

SECTION-D
31 (a)

i) Statement of Biot-Savart’s law 1
Expression for magnetic field 2
Diagram for magnetic field lines ½
ii) Finding current by revolving electron 1½
(i)
The magnetic field at a point due to a current carrying element is 1
proportional to magnitude of current, element length and inversely
proportional to the square of the distance from the element.
 
  dl  r
dB  o
I 3
4 r
 Idl sin 
dB  o
4 r2

Consider a circular coil of radius a carrying current I.

½

According to Biot-Savart’s law

12

  Idl sin 
dB  o
4 r2  
At point A I dl  a
½
  90o ,sin 90o  1
 Idl
Hence dB= o 2
4 a
Magnetic field at centre
2 a 2 a
 Idl
B=  dB   o 2
4 a
o 0
½
 I
B= o  2  2 a
4 a
 I
B= o
2a ½
Note: Give full credit of 2 marks if a student derives the expression for
magnetic field at the axis of the loop and then puts distance of point as 0
from the centre.

½

ii) q=e , v=107 ms -1 ,r=10-10 m
q
i=
T ½
qv
=
2πr
ev
=
2πr
1.6  1019 107 ½
=
2    10 10

0.8
=  10 2 A
 ½
= 0.255 10 2 A = 2.55 mA

OR
b)
i) Derivation of expression for force 2
Statement of Rule ½
Conditions for maximum and minimum force ½+½
ii) Calculation of magnitude of force 1½

13

Consider a rod of uniform cross sectional area A and length l. Let the
number density of mobile charge carriers in it be n. ½
Thus the total number of mobile charge carriers in it is n l A.

For steady current I, drift velocity of electrons vd , in the presence of external

magnetic field B , the force on these carriers is
   ½
F=n l Aq(vd ×B)
 
=  jAl   B
  ½
= I (l  B ) ½
  
Where nqvd is current density ( j ) and jA is current (I)
Fleming’s left hand Rule: If forefinger, middle finger and thumb are
stretched in mutually perpendicular directions, such that forefinger indicates
the direction of magnetic field, middle finger indicates the direction of
current in the conductor, then thumb indicates the direction of force on the
conductor.
Or

Right Hand Thumb Rule : If the fingers of right hand are made to rotate from ½
 
l to B through angle θ, the thumb points in the direction of force on the
current carrying conductor.
Condition for maximum force  = 900 ½

F  I l B sin  = I l B
Condition for minimum force  = 0o or 1800 ½

F =0

ii)

o 2 I1 I 2
F= l ½
4 d
107  2  5  2.5 ½
= 2
10 102 N
2.5 10

= 10-5 N ½ 5

14

32. a)
i) (1) Difference between interference pattern and 1+1
diffraction pattern
(2) Two factors affecting fringe width in young’s double ½ + ½
slit experiment
ii) (1) calculation of angular separation 1
(2) calculation of distance between two maxima 1

(i) (1)
(a) The interference pattern has a number of equally spaced bright and dark
bands while diffraction pattern has a central bright maximum which is twice
as wide as the other maxima.
(b) Interference pattern is obtained by superposing two waves originating
from two narrow slits, while diffraction pattern is a superposition of a
continuous family of waves originating from each point on a single slit.
(c) The maxima in interference pattern is obtained at angle  / a , while the
first minima is obtained at same angle  / a for diffraction pattern.
(d) in interference pattern the intensity of bright fringes remain same while 1+1
in diffraction the intensity falls as we go to successive maxima away from
the center on either side.
(any two)

(2) Factors affecting fringes width
Wave length (  ) / distance of screen from slits (D) / separation between ½+½
slits (d).
(any two)
(ii) (1) d sin  = n  ½
n=1

sin  =
d
 1
For small angle sin    = = radian. ½
100 100

D ½
(2)  = = D
d
1
=  50  102
100
= 50 10 4 m ½
= 5 mm
OR
(b)
i) Derivation of relation between u and v 3
ii) Finding apparent position 2

15

1

Assume that the aperture of the surface is small as compared to other
distance involved, so that small angle approximation can be made.
For small angles
for NOC , i is the exterior angle
 i = NOM  NCM
MN MN
i=  (i)
OM MC
Similarly r = NCM  NIM
MN MN ½
=  (ii)
MC MI
By Snell’s law
n1Sin i = n2 Sin r
for small angles ½
n1 i = n2 r

substituting i and r from (i) and (ii) we get
n1 n n n
 2  2 1 ½
OM MI MC
Applying Cartesian coordinates
OM= -u, MI= + v , MC= +R
n2 n1 n2  n1
  ½
v u R

n2 n1 n2  n1
(ii)   ½
v u R
R= - 6 cm, u = - 3cm , n1 =1.5 n2 = 1
1 1.5 1  1.5 ½
 
v 3 6
1 0.5 1.5
 
v 6 3
1 0.5  3 ½

v 6
1 2.5

v 6 ½
v = - 2.4 cm 5
from the left surface inside the sphere

16

33. (a)

i) Statement of coulomb’s law and vector form 1+1
ii) Explanation of Gauss’s law based on coulomb’s law 1
iii) Force exerted by charge A on charge B 2

i) Force between two point charges varies inversely with the square of
distance between the charges and is directly proportional to the product of
magnitude of the two charges and acts along the line joining the two charges. 1

1

Alternatively

 1 q1q2 
F12  r12
4 o r12 3

Where r12 is a vector from charge q2 to charge q1.
ii) In derivation of Gauss’s law, flux is calculated using Coulomb’s law and
1
surface area. Here coulomb’s law involves 2 factor and surface area
r 1
involves r2 factor. When product is taken, the two factors cancel out and flux
becomes independent of r.
iii)

½

 
r  AB  ai  a j ½

r = AB  a 2  a 2  2a
 1 q1q2 
F r
4 o r 2

 1 q  2q (ai  a j)
F  
4 o ( 2a ) 2
2a ½
 1 2q (i  
2
j)
F  2
4 o 2a 2

17

 1 q2
F   (i  
j)
4 o 2a 2

 q2
F (i  
j)
4 2 o a 2 ½

Note: Award 1 mark if a student calculates the magnitude of force only.
 1 q2
F 
4 o a 2

Alternatively
Give full credit if a student uses component method to solve the question.
OR

(b)
i) Derivation of electric field 2
ii) Effect on electric field 1
iii) Finding magnitude and direction of electric field 2

i)

½

q 1
E+q =  2
4 o r  a 2
q 1
E-q =  2
4 o r  a 2 ½
The components normal to dipole axis cancel away. The components along
the dipole axis add up. ½
Total electric field is opposite to dipole moment.

 ½
p
=
4 o (r 2  a 2 )3/2
Deduct ½ mark if the expression of electric field is not in vector form.
ii) At far off point r >> a

18


 p
E
4 o r 3
When distance is halved.
 ½
 p
E
r
4 o ( )3
2

8 p
=
4 o r 3

E becomes 8 times
½

iii)

p1  q  2 Cm (along OA)
p2  q  2 Cm (along OD) ½
pnet  p  p2
1
2 2

= 2 2 q Cm
Electric field at centre O ½
kp
E = 2 net2 3/2
(r +a )
at point O, r = 0 , a = 1 m ½
k  2 2q 2 2q
E  2 2kq 
13
4 o ½
Along DC

Alternatively

kq
E=
r2

19

AC= BD= 2m
r = OA =OB= OC=OD=1m ½
Electric field at O due to charges at B and D
E1 = EB+ ED
kq kq
E1 = 2  2 along OB
1 1
= 2kq ½
Electric field at O due to charges at A and C
E2 = EA+ EC
kq kq
E2 = 2  2 ½
1 1
= 2kq along OC
Enet = E12  E22
2 2q
= 2 2 kq =
4 o
½
Along DC

Alternatively

Considering AB as dipole, electric field at O
2kq  a 2 kqa
E1 =   2 kq a
1 2 1 2 3/2 1 1 3/2
(( ) ( ) ) (  )
2 2 2 2
Similarly considering DC as another dipole, electric field at O ½
2 kq  a 2kqa
E2 =   2 kq a
1 2 1 2 3/2 1 1
(( ) ( ) ) (  )3/2
2 2 2 2
1 1 ½
Enet = E1+ E2 = 4kqa=  4 q
4 o 2
2 2q
= 2 2kq  ½
4 o
Along DC ½ 5

20

SECTION - E
34.

a) Points at same potential 1
b) Current through arm bg 1
c) Potential difference across R3
OR
c) Power dissipated in R2 2

a) Points (a, b, c)
(d, e) 1
(j, f, g ,h)
are at same potential

Note: Give full credit if a student mentions any two points at same potential
from the above.
b)

According to Kirchhoff’s loop rule
for closed loop abgha
- 6 + 10 I2 + I1 = 0
I1 + 10 I2 = 6 (i)
½
for closed loop acfha
- 6 + 10 (I1 – I2 ) +I1 = 0
11 I1-10 I2 = 6 (ii)
Adding (i) and (ii)
12 I1 = 12
½
I1 = 1 A
I2 = 0.5 A
= current through arm bg
Note: Award 1 mark if a student calculates the current by any other method.

c) VR3 = (I1-I2)  R3
1
= 0.5  5
= 2.5 V
1
OR
P = (I1-I2) 2  R2 = (0.5)2  5
1
= 1.25 W
1 4

21

35.

a) Tracing of path of ray 1
b) Finding velocity of light 1
c) Explanation of two application of TIR 2
OR
c) Definition of TIR 1
Mentioning two conditions of TIR ½+½
a)

1

From fig. angle of incidence on second face i = 600
critical angle ic = 24.5o
(i )  (ic )
 TIR takes place

c
b) n=
v 1
c 3 108
v= = = 1.24  108 m/s
n 2.41

c) Optical Fibre / Brilliance of diamond / mirage (any two) 1+1
Note: Give full credit if students mention the names of applications only.

OR

c) When light travels from optically denser medium to rarer medium at an
interface and gets reflected back into the same medium the phenomenon is 1
called as total internal reflection.

Conditions for TIR
1. Light must travel from optically denser medium to rarer medium.
2. Angle of incidence at the interface must be greater than the critical angle ½+½
for the pair of media. 4

22

Marking Scheme
Strictly Confidential
(For Internal and Restricted use only)
Senior School Certificate Examination, 2023
SUBJECT : PHYSICS (042) (PAPER CODE 55/1/3)
General Instructions : -

1 You are aware that evaluation is the most important process in the actual and correct assessment of the
candidates. A small mistake in evaluation may lead to serious problems which may affect the future of
the candidates, education system and teaching profession. To avoid mistakes, it is requested that
before starting evaluation, you must read and understand the spot evaluation guidelines carefully.

2 “Evaluation policy is a confidential policy as it is related to the confidentiality of the
examinations conducted, Evaluation done and several other aspects. Its’ leakage to public in any
manner could lead to derailment of the examination system and affect the life and future of
millions of candidates. Sharing this policy/document to anyone, publishing in any magazine and
printing in News Paper/Website etc may invite action under various rules of the Board and
IPC.”

3 Evaluation is to be done as per instructions provided in the Marking Scheme. It should not be done
according to one‟s own interpretation or any other consideration. Marking Scheme should be strictly
adhered to and religiously followed. However, while evaluating, answers which are based on latest
information or knowledge and/or are innovative, they may be assessed for their correctness
otherwise and due marks be awarded to them. In class-X, while evaluating two competency-
based questions, please try to understand given answer and even if reply is not from marking
scheme but correct competency is enumerated by the candidate, due marks should be awarded.

4 The Marking scheme carries only suggested value points for the answers

These are in the nature of Guidelines only and do not constitute the complete answer. The students can
have their own expression and if the expression is correct, the due marks should be awarded
accordingly.

5 The Head-Examiner must go through the first five answer books evaluated by each evaluator on the
first day, to ensure that evaluation has been carried out as per the instructions given in the Marking
Scheme. If there is any variation, the same should be zero after delibration and discussion. The
remaining answer books meant for evaluation shall be given only after ensuring that there is no
significant variation in the marking of individual evaluators.

6 Evaluators will mark( √ ) wherever answer is correct. For wrong answer CROSS „X” be marked.
Evaluators will not put right (✓) while evaluating which gives an impression that answer is correct
and no marks are awarded. This is most common mistake which evaluators are committing.

7 If a question has parts, please award marks on the right-hand side for each part. Marks awarded for
different parts of the question should then be totaled up and written in the left-hand margin and
encircled. This may be followed strictly.

8 If a question does not have any parts, marks must be awarded in the left-hand margin and encircled.
This may also be followed strictly.

9 If a student has attempted an extra question, answer of the question deserving more marks should be

1

retained and the other answer scored out with a note “Extra Question”.

10 No marks to be deducted for the cumulative effect of an error. It should be penalized only once.

11 A full scale of marks __________(example 0 to 80/70/60/50/40/30 marks as given in Question Paper)
has to be used. Please do not hesitate to award full marks if the answer deserves it.

12 Every examiner has to necessarily do evaluation work for full working hours i.e., 8 hours every day
and evaluate 20 answer books per day in main subjects and 25 answer books per day in other subjects
(Details are given in Spot Guidelines).This is in view of the reduced syllabus and number of questions
in question paper.

13 Ensure that you do not make the following common types of errors committed by the Examiner in the
past:-

● Leaving answer or part thereof unassessed in an answer book.
● Giving more marks for an answer than assigned to it.
● Wrong totaling of marks awarded on an answer.
● Wrong transfer of marks from the inside pages of the answer book to the title page.
● Wrong question wise totaling on the title page.
● Wrong totaling of marks of the two columns on the title page.
● Wrong grand total.
● Marks in words and figures not tallying/not same.
● Wrong transfer of marks from the answer book to online award list.
● Answers marked as correct, but marks not awarded. (Ensure that the right tick mark is correctly
and clearly indicated. It should merely be a line. Same is with the X for incorrect answer.)
● Half or a part of answer marked correct and the rest as wrong, but no marks awarded.
14 While evaluating the answer books if the answer is found to be totally incorrect, it should be marked
as cross (X) and awarded zero (0)Marks.

15 Any un assessed portion, non-carrying over of marks to the title page, or totaling error detected by the
candidate shall damage the prestige of all the personnel engaged in the evaluation work as also of the
Board. Hence, in order to uphold the prestige of all concerned, it is again reiterated that the
instructions be followed meticulously and judiciously.

16 The Examiners should acquaint themselves with the guidelines given in the “Guidelines for spot
Evaluation” before starting the actual evaluation.

17 Every Examiner shall also ensure that all the answers are evaluated, marks carried over to the title
page, correctly totaled and written in figures and words.

18 The candidates are entitled to obtain photocopy of the Answer Book on request on payment of the
prescribed processing fee. All Examiners/Additional Head Examiners/Head Examiners are once again
reminded that they must ensure that evaluation is carried out strictly as per value points for each
answer as given in the Marking Scheme.

2

MARKING SCHEME: PHYSICS(042)
Code: 55/1/3
Q.No. VALUE POINTS/EXPECTED ANSWERS Marks Total
Marks
SECTION -A
1. (d) 0.01 eV 1 1
2. (a) 1 1

3. 1 1
(c)

4. (d) 95 nm 1 1
5. dE 1 1
(c) o
dt
6. (a) 3.0 eV 1 1
7. (c) 0.19 V 1 1
8. (d) 1:1 1 1
9. F 1 1
(b)
8
10. (d) 5.3 A 1 1
11. (b) it becomes a p-type semiconductor 1 1
12. (a) repelled by both the poles 1 1
13. (d) Diamond to air 1 1
14. (c) Less than g 1 1
15. (c) 1 1

16. (b) Both the assertion (A) and Reason (R ) are true , but Reason (R) is not 1 1
the correct explanation of the Assertion (A).
17. (d) Assertion (A) is false and Reason (R) is also false. 1 1

3

18. (a) Both Assertion (A) and Reason (R) are true and Reason (R) is the 1 1
correctly explanation of the Assertion (A).
SECTION-B
19.
Finding refractive index of the glass 2

25 25
R1 =20 cm, R2 =-30 cm P = D cm1
6 600
1 1
P= (n-1) (  ) ½
R1 R2
25 1 1 ½
 (n  1)(  )
600 20 30
25 50
 (n  1)( )
600 20  30
25= (n-1)50 ½
1
n-1=
2
3
n= ½
2
2
20.
Formation of potential barrier 2

The diffusion current due to concentration gradient at the junction forms a
space charge region consisting of immobile charge carriers. Due to this an
electric field is generated at the junction giving rise to drift current in a
direction opposite to diffusion current.
The potential at which diffusion current becomes equal to drift current is
called potential barrier. 2 2
21.
Calculation of acceleration of alpha particle 2

F=q(v×B) ½

= q(3 105 i  (0.4i  0.3 j )) N
F = q(0.9 105 k ) N ½
F = m a = q(0.9 105 k )N
q ½
a  (0.9 105 k )ms 2
m
= 4.8 × 107 ×0.9 ×105 k ms-2
½
= 4.32 × 1012 k ms-2 2
Note: Deduct ½ mark if a student does not mention the direction of
acceleration.

4

22.

Calculating wavelength of spectral line 2

E  E2  E1 ½
= (-1.51-(-3.4))eV
= 1.89eV
=1.89  1.6  10-19 J ½
hc
E  ½

hc 6.63 1034  3 108
  m
E 1.89 1.6 1019
= 6.58 107 m
= 658nm ½ 2
23. (a)
Ray diagram 1
Proof of Snell‟s law of refraction 1

1

AB is incident wave front, incident at an angle i. Let  be time taken by the
wave front to travel distance BC.
BC= v1 τ where v1 is speed of wave in medium 1.
To determine shape of refracted wave front, we draw a sphere of radius
v2 τ, where v2 is speed of wave in medium 2.
CE represents a tangent drawn from point C on sphere, CE is the refracted ½
wave front.
BC v1
sin i = 
AC AC
AE v2
sin r = 
AC AC
sin i v ½
= 1 = n21
sin r v2

Note: Give full credit if student derives Snell‟s law by taking incident wave
front in denser medium.

5

OR
(b)

Reason for preferring reflecting type telescope over refracting
telescope ½+½
Justification ½+½

1
No Chromatic Aberration - No refraction in mirrors
2
No Spherical Aberration - Due to use of parabolic reflector
3
Easy mechanical support required - Mirrors weigh less and can be
supported over entire back surface. 1+1 2
4 High resolving power – Due to Mirror with large diameter are better.
5 Brighter image – Large mirrors gather more light waves.
(Any two)
24.

(a )Identification and use ½+½
(b) Identification and use ½+½

a) Infrared Rays ½
Uses: -Muscular pain therapy
(any one) -Remote control
- Photography in foggy conditions ½
b) X-rays
Uses: -To study crystal structure ½
(any one) -Detection of fracture in bones
-Cancer treatment ½
Any other correct use. 2
25. (a)
Difference between intrinsic and extrinsic semiconductor 2

Intrinsic semiconductor Extrinsic semiconductor
1. Pure semiconductor. Semiconductor is Doped with
impurities.

2. Low conductivity at room High conductivity at room
temperature. temperature.
3. ne = nh ne  nh 1+1

6

(Any one)
Note: Give full credit if a student writes any other relevant correct answer.
OR

(b)
Circuit diagram for forward and reverse biased p-n junction diode
½+½
V-I characteristic (Forward and Reverse bias) ½+½

½+½

Forward Bias Reverse Bias

½+½

2
Characteristics of silicon Diode

SECTION-C
26.
(a) Calculation of reactance of capacitor 1
(b) Calculation of amplitude of current 1
Writing expression of current 1

V= Vm sin  t C = 15  106 F
V= 310 sin 100  t
1
i) Xc = ½
C
1
=
100 15 106
104
= = 212 
15 ½

7

Vm
ii) im = ½
Xc
310
= = 1.46 A ½
212
Equation of current

i= im sin(t  )
2
 1 3
= 1.46 sin (100  t  )
2
27. a)
Energy level diagram for hydrogen atom 1½
Transitions corresponding to ultraviolet region, visible region and
infrared region ½+½+½

1½

½

½

½

Note: Award 1 ½ mark for energy level diagram if a student does not show
the transitions.
OR
b)

Diagram to show variation 1
Two features of diagram ½+½
Reason for nuclear fusion 1

1

(Note: Award full credit even if a student does not mark so many elements
and does not mention the values of Ebn.)

Features of diagram (any two)
1. Binding energy per nucleon is practically independent of atomic number

8

for nuclei of middle mass number (30 <A < 170)
2. The curve has maximum of about 8.75 MeV for A= 56 and has a value of
7.6 MeV for A= 238
3. Binding energy per nucleon is lower for both light nuclei (A<30) and ½+½
heavy nuclei (A>170)

Two lighter nuclei fuse together to form heavier nuclei as the binding energy
per nucleon of fused heavier nuclei is more than the binding energy per
nucleon of the lighter nuclei. Thus the final system is more tightly bound
than initial system.
Alternatively
1
To attain the stability 3

28.
Calculating Electrostatic potential Energy 3

1 Q1Q2 ½
UQ1Q2 =
4 o r12
9 109  (15 106 )  (10 106 )
=
3 102 ½
= - 45 J
1 Q2Q3
UQ2Q3 =
4 o r23
9 109  (10 106 )  (16 106 )
=
4 102 ½
= 36 J
1 Q1Q3
UQ1Q3 =
4 o r13
9 109  (15 106 )  (16 106 )
=
5 102 ½
= - 43.2 J ½
Unet = UQ1Q2 + UQ2Q3 + UQ1Q3
= -45+36-43.2 ½
= -52.2 J
Alternatively
QQ Q Q QQ
U  k( 1 2  2 3  1 3 ) 1
r12 r23 r13

9 -6 -6 9 -6 -6 9 -6 -6
= 9×10 ×(-15×10 -2)×(10×10 ) + 9×10 ×(10×10 -2)×(16×10 ) + 9×10 ×(-15×10 -2)×(16×10 )
3×10 4×10 5×10 1
=(-45 +36-43.2) J
= 52.2 J 1
3

9

29. (a)

Difference between resistance and impedance 1
Obtaining expression for impedance 2

1. Resistance is opposition offered to both alternating current and direct
current while impedance is opposition offered to alternating current only.

2. Resistance is independent of frequency of source while impedance
depends on frequency.

3. Resistance is opposition offered by material of the conductor while 1
impedance is combined opposition offered by different electrical
components such as resistor, inductor or capacitor.

(Any One)
(Note: Give credit of this part if a student writes any other correct answer.)

½

VR = imR , Vc = im Xc , VL = imXL
im = Peak value of current in the circuit.
VL VR  VC  Vm ½
(Vm)2 = VR2 + (VC – VL)2
= ( imR)2 + (imXc - imXL)2 ½
= im [R2 + (Xc – XL)2]
Vm
im =
R  ( X c  X L )2
2

V
im = m where Z = R 2  ( X c  X L )2 ½
Z = impedance
OR
(b)
Finding condition for resonance 1
Factors affecting resonant frequency 1
Graph 1

Z= R 2  ( X L  X c )2 ½
For maximum current, Z should be minimum therefore to minimize Z
XL = XC ½

10

Alternatively

XL= XC
1 ½
ωL =
C
1
ωr =
LC ½
Resonant Frequency depends on value of Inductance and Capacitance ½+½

1

3

30.
Calculating wavelength of wave 3

h ½

2mK
½
h
λ' =
2m×4K
1 h 1
=
2 2mK
 ½
=
2
1.2nm
=
2
= 0.6nm ½ 3
SECTION-D
31 a)
i) (1) Difference between interference pattern and 1+1
diffraction pattern
(2) Two factors affecting fringe width in young‟s double ½ + ½
slit experiment
ii) (1) calculation of angular separation 1
(2) calculation of distance between two maxima 1

11

(i) (1)
(a) The interference pattern has a number of equally spaced bright and dark
bands while diffraction pattern has a central bright maximum which is twice
as wide as the other maxima.
(b) Interference pattern is obtained by superposing two waves originating
from two narrow slits, while diffraction pattern is a superposition of a
continuous family of waves originating from each point on a single slit.
I The maxima in interference pattern is obtained at angle  / a , while the
first minima is obtained at same angle  / a for diffraction pattern.
(d) in interference pattern the intensity of bright fringes remain same while 1+1
in diffraction the intensity falls as we go to successive maxima away from
the center on either side.
(any two)

(2) Factors affecting fringes width
Wave length (  ) / distance of screen from slits (D) / separation between ½+½
slits (d).
(any two)
(ii) (1) d sin  = n  ½
n=1

sin  =
d
 1
For small angle sin    = = radian. ½
100 100

D
(2)  = = D ½
d
1
= 50  10 2
100
= 50 104 m ½
= 5 mm
OR

(b)
i) Derivation of relation between u and v 3
ii) Finding apparent position 2

1

Assume that the aperture of the surface is small as compared to other
distance involved, so that small angle approximation can be made.
For small angles

12

for NOC , I is the exterior angle
 I = NOM  NCM
MN MN
i=  (i)
OM MC
Similarly r = NCM  NIM
MN MN
=  (ii)
MC MI ½
By Snell‟s law
n1Sin I = n2 Sin r
for small angles
n1 I = n2 r ½
substituting I and r from (i) and (ii) we get
n1 n n n
 2  2 1
OM MI MC
Applying Cartesian coordinates ½
OM= -u, MI= + v , MC= +R
n2 n1 n2  n1
 
v u R

n2 n1 n2  n1 ½
(ii)  
v u R
R= - 6 cm, u = - 3cm , n1 =1.5 n2 = 1
1 1.5 1  1.5 ½
 
v 3 6
1 0.5 1.5
  ½
v 6 3
1 0.5  3

v 6
1 2.5 ½

v 6
v = - 2.4 cm 5
from the left surface inside the sphere ½
32. (a)

i) Statement of coulomb‟s law and vector form 1+1
ii) Explanation of Gauss‟s law based on coulomb‟s law 1
iii) Force exerted by charge A on charge B 2

i) Force between two point charges varies inversely with the square of
distance between the charges and is directly proportional to the product of
magnitude of the two charges and acts along the line joining the two charges. 1

1

Alternatively

13

1 q1q2
F12  r12
4 o r123

Where r12 is a vector from charge q2 to charge q1.
ii) In derivation of Gauss‟s law, flux is calculated using Coulomb‟s law and
1 1
surface area. Here coulomb‟s law involves 2 factor and surface area
r
involves r2 factor. When product is taken, the two factors cancel out and flux
becomes independent of r.
iii)

½

½
r  AB  ai  a j
r = AB  a 2  a 2  2a
1 q1q2
F r
4 o r 2

1 q  2q (ai  a j )
F   ½
4 o ( 2a)2 2a
1 2q 2 (i  j )
F  2
4 o 2a 2
1 q2
F   (i  j )
4 o 2a 2
q2
F (i  j ) ½
4 2 o a 2

Note: Award 1 mark if a student calculates the magnitude of force only.
1 q2
F 
4 o a 2

14

Alternatively
Give full credit if a student uses component method to solve the question.
OR

(b)
i) Derivation of electric field 2
ii) Effect on electric field 1
iii) Finding magnitude and direction of electric field 2

i)

½

q 1
E+q =  2
4 o r  a 2
q 1
E-q =  2
4 o r  a 2 ½
The components normal to dipole axis cancel away. The components along
the dipole axis add up. ½
Total electric field is opposite to dipole moment.

p ½
=
4 o (r 2  a 2 )3/2
Deduct ½ mark if the expression of electric field is not in vector form.
ii) At far off point r >> a
p
E
4 o r 3
When distance is halved.
p
E ½
r
4 o ( )3
2

15

8 p
=
4 o r 3
E becomes 8 times
½
iii)

p1  q  2 Cm (along OA)
p2  q  2 Cm (along OD) ½
pnet  p  p2
1
2 2

= 2 2 q Cm
Electric field at centre O
kp ½
E = 2 net2 3/2
(r +a )
at point O, r = 0 , a = 1 m
½
k  2 2q 2 2q
E  2 2kq 
13
4 o ½
Along DC

Alternatively

kq
E=
r2
AC= BD= 2m
r = OA =OB= OC=OD=1m
Electric field at O due to charges at B and D
E1 = EB+ ED
kq kq ½
E1 = 2  2 along OB
1 1
= 2kq
Electric field at O due to charges at A and C

16

E2 = EA+ EC
kq kq
E2 = 2  2
1 1
= 2kq along OC ½
½
Enet = E12  E2 2
2 2q
= 2 2 kq =
4 o
½
Along DC

Alternatively

Considering AB as dipole, electric field at O
2kq  a 2kqa
E1 =   2kqa
1 2 1 2 3/ 2 1 1 3/ 2
(( ) ( ) ) (  ) ½
2 2 2 2
Similarly considering DC as another dipole, electric field at O
2kq  a 2kqa
E2 =   2kqa
1 2 1 2 3/ 2 1 1 3/ 2
(( ) ( ) ) (  ) ½
2 2 2 2
1 1
Enet = E1+ E2 = 4kqa=  4 q
4 o 2 ½
2 2q
= 2 2kq 
4 o
½
Along DC

5

17

33. (a)

i) Statement of Biot-Savart‟s law 1
Expression for magnetic field 2
Diagram for magnetic field lines ½
ii) Finding current by revolving electron 1½
(i)
The magnetic field at a point due to a current carrying element is 1
proportional to magnitude of current, element length and inversely
proportional to the square of the distance from the element.

o dl  r
dB  I 3
4 r
 Idl sin 
dB  o
4 r2

Consider a circular coil of radius a carrying current I.

½

According to Biot-Savart‟s law

o Idl sin 
dB 
4 r2
At point A I dl  a
½
  90 ,sin 90  1
o o

 Idl
Hence Db= o 2
4 a
Magnetic field at centre
2 a 2 a
 Idl
B=  dB   o 2 ½
o 0
4 a
 I
B= o  2  2 a
4 a
I ½
B= o
2a

18

Note: Give full credit of 2 marks if a student derives the expression for
magnetic field at the axis of the loop and then puts distance of point as 0
from the centre.

½

ii) q=e , v=107 ms-1 ,r=10-10 m
q
i=
T ½
qv
=
2πr
ev
=
2πr
1.6 1019 107 ½
=
2   1010

0.8
= 102 A
 ½
= 0.255 102 A = 2.55 Ma

OR
b)
i) Derivation of expression for force 2
Statement of Rule ½
Conditions for maximum and minimum force ½+½
ii) Calculation of magnitude of force 1½

Consider a rod of uniform cross sectional area A and length l. Let the
number density of mobile charge carriers in it be n. ½
Thus the total number of mobile charge carriers in it is n l A.
For steady current I, drift velocity of electrons vd , in the presence of external
magnetic field B , the force on these carriers is
½
F=n l Aq(vd ×B)
=  jAl   B ½
½
= I (l  B)
Where nqvd is current density ( j ) and jA is current (I)
Fleming‟s left hand Rule: If forefinger, middle finger and thumb are
stretched in mutually perpendicular directions, such that forefinger indicates

19

the direction of magnetic field, middle finger indicates the direction of
current in the conductor, then thumb indicates the direction of force on the
conductor.
Or

Right Hand Thumb Rule : If the fingers of right hand are made to rotate from
l to B through angle θ, the thumb points in the direction of force on the ½
current carrying conductor.
Condition for maximum force  = 900
½
F  I l B sin  = I l B
Condition for minimum force  = 0o or 1800
½
F =0

ii)

o 2I1I 2
F= l ½
4 d
107  2  5  2.5
= 10 102 N ½
2.5 102 5
= 10-5 N ½
SECTION - E
34.
a) Points at same potential 1
b) Current through arm bg 1
c) Potential difference across R3
OR
c) Power dissipated in R2 2
a) Points (a, b, c)
(d, e)
(j, f, g ,h)
are at same potential 1

Note: Give full credit if a student mentions any two points at same potential
from the above.

20

b)

According to Kirchhoff‟s loop rule
for closed loop abgha
- 6 + 10 I2 + I1 = 0
I1 + 10 I2 = 6 (i)
for closed loop acfha
- 6 + 10 (I1 – I2 ) +I1 = 0
½
11 I1-10 I2 = 6 (ii)
Adding (i) and (ii)
12 I1 = 12
I1 = 1 A
I2 = 0.5 A
½
= current through arm bg
Note: Award 1 mark if a student calculates the current by any other method.

c) VR3 = (I1-I2)  R3
1
= 0.5  5
= 2.5 V
1
OR
(c) P = (I1-I2) 2  R2 = (0.5)2  5
1
= 1.25 W
1 4
35.
a) Tracing of path of ray 1
b) Finding velocity of light 1
c) Explanation of two application of TIR 2
OR
c) Definition of TIR 1
Mentioning two conditions of TIR ½+½

21

a)

1

From fig. angle of incidence on second face i = 600
critical angle ic = 24.5o
(i)  (ic )
 TIR takes place

c
b) n=
v
c 3  108 1
v= = = 1.24  108 m/s
n 2.41

c) Optical Fibre / Brilliance of diamond / mirage (any two) 1+1
Note: Give full credit if students mention the names of applications only.

OR

c) When light travels from optically denser medium to rarer medium at an
interface and gets reflected back into the same medium the phenomenon is 1
called as total internal reflection.

Conditions for TIR
1. Light must travel from optically denser medium to rarer medium. 4
2. Angle of incidence at the interface must be greater than the critical angle ½+½
for the pair of media.

22