HBSE Class 12 Sample Paper 2026 Answers Physics

MARKING SCHEME (PHYSICS) (2025- 26)
1
1. c) 1
K
2. c) 31.4 wb 1
3. b) Ultraviolet Rays 1
4. d) Magnetic Flux and Power Both 1
I0
5. b) 1
2
6. d) 2.14 eV 1
R
7. b) 1
2
8. c) O8
17
1
9. c) 25 1
10. Zero 1
Vd
11. 1
2
12. X3 due to more neutrons
7
1
13. 0K 1
C
14. 1
4
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

Production of infrared waves.
19. Reason of heat waves. 1+1

Infrared waves are produced by hot bodies and vibrations of molecules.
They are referred as heat waves because they are readily absorbed by water molecules and
increases their thermal energy and heat them.
OR

Production of X-rays
Two uses
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
g = 1.5 l
=?
1 1
fa = 0.2 m 20cm ½
P 5

1 1 1
(a g – 1) –
fa R1 R2

1 1 1
(1.5 – 1) – ...(1) ½
20 R1 R2

1 g 1 1
–1 –
fl l R1 R2

1 1.5 1 1
–1 –
1000 l R1 R2
From (1) and (2), on solving
5
l 1.67
3

Meaning of ionization energy
21. Value for H-atom 1+1=2

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
Relation with stability 1+1=2

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

Formula
22. Calculation 1+1=2

Change in Voltage
dynamic Resistance = 1
Change in Current
.1V
= = 10 Ohm 1
.01A

Def. of magnetic susceptibility
23. Identification of A and B 1+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

Statement of Gauss's Law
24. Proof 1+1

The total flux associated with a sealed surface equals 1/ 0 times the charge encompassed by
the closed surface 1

Let’s say the charge is equal to q.
Let’s make a Gaussian sphere with radius = r.
Now imagine surface A or area ds has a ds vector
At ds, the flux is:
 
d = E ds cos
But , = 0
Hence , Total flux: 1
=E4 r 2

Hence, E = 1 / 4 0 q / r2 × 4 r2
=q/ 0
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

Circuit diagram
26. Working 1+1+1
Output waveform

During +ve half cycle diode D 1 is forward biased and diode D 2 is reverse biased. The
forward current flows due to D1. During –ve half cycle, diode D1 is reverse biased and diode
D2 is forward biased. The forward current flows due to D2. The output waveforms is shown
in figure. 1
OR

V-I characteristics
Explanation 1+2

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.
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.

Part - I
27. Part - II (Formula + Calculation) 1+2

i) When an electron undergoes a transition from 2nd excited state (n = 3) to the 1st
excited state (n = 2) in hydrogen atom, first spectral line of Balmer series is emitted. 1
ii) the ratio of the wavelengths of the Balmer series to the Paschen series: 2
Ratio= b/ P =4R/9R=4/9
The ratio between the wavelengths of the most energetic spectral lines in the Balmer
and Paschen series is: 4/9

Calculation of energy of radiation
28. Calculation of K.E of photoelectron 1½ + 1½

i) 1½

ii) K.E. of photoelectron
K.E. = E – 0 = h – 0
= (6.02 × 1019 – 3.5 × 10–19) 1½
= 2.527 × 1019J

Expression of torque
29. Effect of non-uniform field 2+1

(i) 2

(ii) If electric field is non-uniform, then dipole experiences a translatory force as well
as torque. 1
OR

Eq. Capacitance
Energy 1+1+1
Charge

1+1+1

30. 1. Let I1 be the current through MN, I2 be the current through PO, and I3 be the current
through SP.
2. Apply KCL at junction P: I1 = I2 + I3. 1
3. Apply KVL in loop MNP: E1 - I1R1 - I2R2 = 0, where E1 is the voltage source in branch
MNP, R1 is the resistance in branch MN, and R2 is the resistance in branch NP. ½
4. Apply KVL in loop POQ : E2 - I2R3 - I3R4 = 0, where E2 is the voltage source in branch
POQ, R3 is the resistance in branch OP, and R4 is the resistance in branch PQ. ½
5. Solve the system of equations : You now have three equations with three unknowns
(I1, I2, I3). Solve these equations to find the values of I1, I2, and I3. 1

I1 = MN = 4A,
I2 = TO = 0A,
I3 = SP = 4A
SECTION - D (CASE STUDY)
31. (i) (D) 1
(ii) (B) 1
(iii) (C) 1
(iv) (D) OR (A) 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=n (Constructive interference)
λ
x = (2n + 1) (Destructive interference) 1
2

(iii) 1

OR
Two light sources are considered coherent if they maintain a constant phase difference
and have the same frequency or wavelength. 1
(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.

m0 2 I1 ar
B1 = ⊥ 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

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)

n IAB = Kq
Kq
I=
n AB
=Gq
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
A A1 A2 90° B2 A3R tal I
ef
X le nter Y
ct
io nal
i>c n
i=c at
3 B A
3

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

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'

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.
35. (i) Let E = E0 sin wt be the alternating emf. ...(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

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 1

V V

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

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