HBSE Class 12 Sample Paper 2025 Answers Physics

Board Of School Education Haryana

मॉडल पेपर
उत्तर
2025

BSEH MODEL PAPER (2024-25)
Marking Scheme

section-A
17
1. (c) O8. 1
2. (b) magnitude of magnetisation. 1
3. (b) speed. 1
4. (d) first decreases to become zero and then increases. 1
I0
5. (b) . 1
2
6. (b) p  1


R
7. (b) . 1
2
8. (a) 0°. 1
9. (c) 25. 1
10. Zero. 1
11. Remains same. 1
12. 7X3 due to more neutrons. 1
13. 0 K. 1
14. increases. 1
15. Photoelectric effect. 1
16. (c) Assertion (A) is true, but Reason (R) is false. 1
17. (b) Both A and R are true and R is not the correct explanation of A. 1
18. (a) Both A and R are true and R is the correct explanation of A. 1

section-B

19. Production of infrared waves. 1
Reason of Calling heat waves. 1
Infrared waves are produced by hot bodies and vibrations of molecules. 1
They are referred as heat waves because they are readily absorbed by water molecules and
increases their thermal energy and heat them. 1
Or
Production of X-rays 1
Two uses ½+½

Page-1

When fast moving electrons strike a heavy target like tungsten, X-rays are produced. 1
Two uses:
1. To study crystal structure. ½
2. Used as diagnostic tool in medical. ½
20. P = +5 D fe = –100 cm
mg = 1.5 ml = ?
1 1
fa = = = 0.2 m = 20 cm ½
P 5


1
fa
= ( m − 1)  R1 − R1 
a
g
1 2

1  1 1 
= ( 1.5 − 1)  −  ...(1) ½
20 R
 1 R2 

1  mg  1 1 
=  − 1  − 
fl  ml   R1 R2 
1  1.5  1 1 
=  − 1  −  ...(2) ½
( −100 )  ml   R1 R2 
From (1) and (2), on solving
5
ml = = 1.67. ½
3
21. Meaning of ionization energy 1
Value for H-atom 1
Ionization energy is the minimum energy required to remove an electron from an isolated
atom of an element. 1
The Ionization energy for hydrogen atom is 13.6 eV. 1
Or
Def. of mass defect 1
Relation with stability 1
Mass defect is the difference between the actual mass of the nucleus and the sum of the
masses of its nucleons. 1
Greater the mass defect, greater will be the binding energy and more stable will be the
nucleus. 1
22. n-type 1
p-type 1

Page-2

C.B. C.B.

0.045 eV
Donor energy level
Accepter energy level

0.04 eV

V.B. V.B.

(n-type) (p-type)

23. Def. of magnetic susceptibility 1
Identification of A and B ½+½
Magnetic susceptibility is a property which determines how easily a specimen can be
magnetised when placed in the magnetic field. 1
0.96 – Diamagnetic ½
500 – Ferro magnetic ½
24. Statement of coulomb’s law 1
Vector form 1
Two like charges attract or repel each other with a force which is directly proportional to
product of magnitude of charges and inversely proportional to square of distance b/w
them. 1
→ 1 q1q2
F12 = 2
rˆ21  1
4 p ε0 r21
25. (1) Resistance is the opposition offered to both alternating current and direct current
while impedance is the opposition offered to alternating current only. 1
(2) Resistance is independent of frequency of source while impedance depends on
frequency. 1

section-C

26.
Circuit diagram 1
Working 1
Output waveform 1
Full-wave rectifier
D1 Input
P1 S1

t
RL

output
Output

t
P2 S2
     D2        1
During +ve half cycle diode D1 is forward biased and diode D2 is reverse biased. The
forward current flows due to D1. During –ve half cycle, diode D1 is reverse biased and
Page-3

diode D2 is forward biased. The forward current flows due to D2. The output waveforms is
shown in figure.  1
Or
V-I characteristics 1
Explanation 2
Forward
current

Reverse
Vk
voltage

Forward
voltage

Reverse
current

It is found that beyond forward voltage V = VK called knee voltage, the conductivity is very
high. Potential barrier is overcome and the current increase rapidly.
But reverse current is due to flow of minority carriers, which is very small. 1
It shows that the diode conducts when forward biased and does not conduct when reverse
biased. This characteristics makes it suitable for use for rectification. 1
27. Def. of closest approach 1
Formula 1
Effect 1
The minimum distance up to which an alpha particle travel along the central line of the
nucleus before it rebounds is called distance of closest approach. 1
Ze ( 2 e )
r0 =  1
1 
4 p ε0  mv 2 
2 
1

i.e., r0 a
K. E
As K.E doubled, r0 is halved. 1
28. Calculation of energy of radiation 1½
Calculation of K.E of photoelectron 1½
C
(i) E = hn = h  1½
l
6.63 × 10 −34 × 3 × 108
=  1½
330 × 10 −9
= 6.027 × 10–19 J 1½
(ii) K.E. of photoelectron
K.E = E – f0 = hn – f0 1½

Page-4

= (6.02 × 10–19 – 3.5 × 10–19) 1½
= 2.527 × 10–19 J 1½
29. Expression of torque 2
Effect of non-uniform field 1

(+q) B  
F = +qE

 
F = –qE
(–q)
A C

→ →
(i) Force on +q,   F = q E
→ →
Force on –q,   F = −q E
Total force = 0 1
t = qE × BC
\ t = qE × 2a sin q
→ → →
t = p×E 1
(ii) If electric field is non-uniform, then dipole experiences a translatory force as well as
torque. 1
Or
Eq. Capacitance 1
Energy 1
Charge 1
12 pf 12 pf

50 V
1 1 1
= +
CS 12 12
CS = 6 pf = 6 × 10–12 f 1
1 1 −12
U = CV2 = × 6 × 10 × 50 × 50
2 2
= 75 × 10–10 J 1
q = CV
= 6 × 50 = 300 × 10–12 C
= 3 × 10–10 C 1
Page-5

30. Statement of Kirchhoff’s laws 1+1
Justification ½+½
(i) Junction Rule: At any junction, the sum of currents entering the junction is equal to
the sum of currents leaving the junction. 1
(ii) Loop Rule: The algebraic sum of all the potential differences across the elements of
circuit (i.e., emf of current sources and resistances) in a closed circuit is zero. 1
Justification: The first law is in accordance with law of conservation of charge. ½
The second law is in accordance with law of conservation of energy. ½
section-D (Case Study)
31. (i) (d) both a and c. 1
(ii) (b) 4 mA 1
Or
(s) = [M L–3 T3 A2]–1

(iii) It is defined as the change in the resistivity per unit original resistivity per unit rise in
temperature. 1
(iv) Current density (J) is vector quantity. 1
32. (i) The sources of light which continuously emit light of same wavelength, same
frequency and of same phase are called coherent sources. 1
(ii) x = nl (Constructive interference) ½
l
x = ( 2n + 1 ) (Destructive interference) ½
2
(iii) Zero. 1
Or
A light in which vibration of light vectors are restricted in a one particular plane.
(iv) When d is very large, fringe width will decrease or cannot be seen separately. 1

section-E
33. D1 D2 D1 D2 

B1

F1
P Q
F2

I1 I2 I1 B2 I2

C1 C2 C1 C2 1
Consider C1 D1 and C2 D2 two infinite long straight conductors carrying currents I1 and I2
in same direction, at a distance r apart held ||el to each other.
Mag. field Induction at pt. P on C2 D2 due to current I1 in C1 D1.
Page-6

m 2 I1 ar
B1 = 0 ⊥ to plane of paper acting inwards given by right hand rule. 1
4p r
\ The unit length of C2 D2 experience a force F2.
F2 = B1 I2 × 1 = B1 I2
m 2 I1 I 2
F2 = 0 ...(1) 1
4p r
According to Fleming’s left hand rule force on C2 D2 acts in the plane of paper ⊥ to C2 D2,
directed towards C1 D1.
||ly C1 D1 also experience force given by equation (1), which acts in the plane of paper ⊥ to
C1 D1 directed towards C2 D2.
Hence C1 D1 and C2 D2 attract each other carrying current in same direction. 1
el
One Ampere—is that much current which when flowing through each of two || uniform
long linear conductors placed in free space at a distance of 1m from each other will attract
or repel each other with a force of 2 × 10–7 N/m of their length. 1
Or
Diagram 1
Principle ½
Construction 1½
Working 2
T

T1 T2
M

P Q

L
N S

S R

 1
Principle: When a current carrying coil placed in magnetic field, it experiences a torque. ½
Construction: It consists of a rectangular coil PQRS of large no. of turns of insulated copper
wire wound over a non-magnetic material frame. A soft iron cylindrical core is placed such
that coil can rotate without touching it. Coil is suspended b/w two cylindrical magnets by
a phosphor bronze wire. Upper end of the coil is connected to movable torsion head and
lower end is connected to hair spring. 1½
Working: Function of cylindrical core and magnet is to provide radial magnetic field
t = n I AB
If k is the restoring torque per unit twist and Q be the twist in the wire.
In equilibrium t = tR (Restoring torque)

Page-7

n IAB = Kq
Kq
I=
n AB
= G q
K
where G = galvanometer constant
n AB
I a q i.e., linear scale deflection 2
34. It is the phenomenon of reflection of light into a denser medium from an interface of this
denser medium and a rarer medium. 1
Two essential conditions of TIR:
1. Light should travel from denser to rarer medium.
2. Angle of incidence in denser medium should be greater than critical angle for the pair
of media in contact. 2
B
Rarer Medium B1
Air (a)
To
A1 A2 90° B2 A3R t al
A ef In
X le te Y
ct rn
io al
i>c n
i=c at
B3 A3

Denser Medium
Water (b)

O
Optical fibres are the threads of glass or quartz of ref. index 1.5 coated with a thin layer of
material having low ref. index nearly 1.48.
When light falls at one end of the optical fibre. The refracted ray falls with angle greater
than critical angle TIR takes place and finally ray come out of other end without any loss.
Low 

High 
 2
Or
Huygen’s Principle: According to Huygen’s Principle:
(i) Every point on primary wavefront acts as fresh source of disturbance which travel in
all direction with velocity of light and called as secondary wavelets. 1
(ii) Surface obtained by joining secondary wavelets tangentially in forward direction
called secondary wavefront. 1

Page-8

Refraction of plane wavefront
It C1 is the speed of light in rarer medium and C2 is speed of light in denser medium then
C
m= 1  ...(1) 1
C2
AB is a plane wavefront incident on XY. According to Huygen’s principle, every pt. on
AB is a source of secondary wavelets.
Let secondary wavelets from B strike XY at A′ in t-seconds.
\ BA′ = C1 × t...(2)
Taking C2 × t as radius draw an arc at B′ with A as a centre.
A′B′ is secondary wavefront.
\ AB′ = C2 × t...(3) 1
2
B
1 Rarer

i
i A'
X r Y
A

r

B Denser

1' 2' 3'

BA′ C1 × t
In D AA′ B sin i = =
AA′ AA′
AB′ C ×t
In D AA′ B′ sin r = = 2
AA ′ AA′
sin i C
Divide = 1 =m
sin r C2
sin i
or  m= 1
sin r
It is clear that incident rays, normal and refracted rays all lie in the same plane.
E = E0 sin wt be the alternating emf.
35. (i) Let ...(1)
q
V= = E0 sin wt
C
q = CE0 sin wt
dq d
I= = (CE0 sin wt) 1
dt dt
E0
= sin ( wt + p / 2 )
1 / wC
The current will be maximum if sin (wt + p/2) = 1
E0
I = I0 =
1 / wC
\ I = I0 sin (wt + p/2)...(2) 1

Page-9

It shows alternating current leads by p/2 to the alternating voltage.

A E0

I0 B
t
O

OA = E = E0 sin wt
OB = I = I0 sin (wt + p/2)
(ii) Faraday’s 1st law: Whenever there is a change in the magnetic flux linked with a coil,
an emf is induced in it. It lasts so long as change in flux continuous.
Faraday’s 2nd law: Rate of change of magnetic flux linked with a coil is directly
proportional to emf induced in it. 1
df
e = − 1
dt
Or
A.C. Generator: It is a device used to convert mechanical energy into electrical energy.  1
Principle: It is based on principle of electromagnetic induction. Whenever mag. flux linked
with a coil change, induced emf. produces in coil.  1

C B

B C

N S N S
I I
D A

A D

B1 B2 B1 B2
R1 R2 R2 R1

V V
         1 
Working: As the armature coil is rotated in the mag. field angle q b/w field and normal
to the coil changes continuously. An emf is induced in the coil. The direction of induced
current is shown in figure.
Let N = no. of turns in the coil
A = area of each turn
→

B = strength of mag field
→ →
f = N( B . A) = NBA cos q

= NBA cos wt

Page-10

df −d
e= (NBA cos wt) = NBAw sin wt
dt dt
e
e will be may if sin wt = 1 
2
\ emax = e0 = NBAw

e
\ e = e0 = sin wt
T/2 T
O
t
 qqq

Page-11

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