HBSE Class 11 Sample Paper 2026 Answers Physics

BsEH Sample PAPER PHYSICS (2025-26)
Marking Scheme
Physics Class-XI

section-A
1. (c) 663.8. 1
2. (d) [P1A1/2 T–1]. 1
3. (d) both Newton’s second and third law. 1
4. (d) at first greater than mg, and later becomes equal to mg. 1
5. (c) C. 1
6. (c) least in (b). 1
7. (b) energy. 1
8. (a) P1 > P2. 1
9. (b) –2P0V0. 1
10. scaler. 1
1
11. MR 2 .  1
2
12. Decreases. 1
13. uniform motion. 1
14. zero. 1
15. dw = t (dQ). 1
16. (d). 1
17. (c). 1
18. (d).1
section-B
19. n1 = 10, W = [ML2T–2] ½
\ a = 1, b = 2 c = –2
a b c
 M
L
T

n2 = n1
1 
1 
1   ½

M2 
L2 
T2 
1 1 2
 1 kg
 1 m
 1 s

= 10




1 g 
1 cm 
1 s 
2
 10 3 g
 10 2 cm

= 10

 1

g 
cm 
= 10 × 103 × 104 1
= 108 1
\ 10 J = 108 ergs
OR
A

[P] =
2   ½

V 
Page-1

[ML–1 T–2] × [L6] = [A]
\ [A] = [ML5 T–2] ½
[b] = [V] ½
3
\ [b] = [L ]. ½
20. When bullet is fired from gun, then gun recoils back with some velocity which is known as
recoil velocity of gun. ½
According to law of conservation of liner momentum
m1u1 + m2u2 = m1 v1 + m2v2 ½
0 = m1v1 + m2v2
−m1v1
v2 =  1
m2
m1 m2 m1 m2
B G B G
u1 = 0 u2 = 0 v1 v2
Before firing After firing
OR
–1 –1
m = 3 kg,   u = 2 ms ,  v = 3.5 ms
t = 25 sec., F = ?
m
v  u

F = ma =  ½
t
3
3.5  2

=  ½
25
= 0.18 N ½
The direction is along the direction of motion. ½
21. Positive work done
 
W= F
S
= FS cos q
when q < 90° then W is +ve. ½
For example when a body falls freely under the action of gravity, q = 0 and work done is
positive.
 ½
Negative work done
when q > 90° then W is –ve. ½
For example when body is thrown up against gravity then work done is negative. ½
Or
Work done is stopping the body = force × distance
  = K.E of body, which is same for two bodies.
As retarding force applied is the same, therefore distance moved by both the bodies before
coming to rest must be same. 2
22. Moment of inertia (I) plays the same role in rotational motion as mass (m) plays in linear
motion. 1

Page-2

Moment of Inertia: of a body about a given axis as the sum of the products of masses of
all the particles and square of their respective perpendicular distances from the axis of
rotation. 1
23. Law of orbits:– Every planet revolves around the sun in an elliptical orbit. The sun is
situated at one foci of the ellipse.
Law of areas:– The areal velocity of planet around the sun is constant. 1
24. Any two differences 1+1
Isothermal Process Adiabatic Process
1. Temperature of system remains constant 1. Heat of the system remains constant
        dT = 0         dQ = 0
2. Isothermal is slow process. 2. Adiabatic process is fast.
y = a sin wt
25. ½
dy d
V= =
a sin t

dt dt
= aw cos wt
dV d
A= =
aw cos t
 ½
dt dt
= –w2 a sin wt
or A = –w2y ½
A
2
+a
T/4
O t
T 3T T
2 4
2
–a

*
section-C
26. Let the gas be heated at constant volume
dT be the rise in temperature.
\ dQ = CvdT
dV = 0
\ PdV = dw = 0
Acc. to first law of thermodynamics
dQ = dU + dW
CvdT = dU + 0
\ dU = CvdT 1
Now heat the gas at constant pressure.
dQ′ = CpdT
dW′ = PdV
dQ′ = dU′ + dW′ 1

Page-3

CpdT = CvdT + PdV
[ dU′ = dU because dT is the same rise in temp.]
(Cp – Cv) dT = PdV
(Cp – Cv) dT = R dT
Cp – Cv = R
which is Mayer’s formula.
OR
(i) Yes, this happens when the gas undergoes adiabatic compression.
As dQ = dU + dW
As dQ = 0 in adiabatic process
\ dU + dW = 0
or dU = –dW
In compression, work is done on the gas, so dW is negative.
Hence dU is +ve i.e., internal energy of the gas increase. Hence, temperature of gas
increases. 2
(ii) Concept of internal energy is given by first law of thermodynamics and concept of
temperature is given by zeroth law of thermodynamics. ½+½
27. Let us consider a small spherical ball of radius r, density r is dropped in a liquid of density
r′ and coefficient of viscosity η. U
viscous force, F = 6p η rv F

4
wt. of ball W = mg = pr3 rg
3
4
wt. of the liquid displaced U = m1g = pr3 r′g 1
3
At equilibrium
W = U + F 1
4 3 4 3
pr rg = pr r′g + 6pηr v
3 3 Motion

4 3
6p ηr v = pr (r – r′)g
3 W

2r 2  

 g
v= . 1
9

OR

Ductile— The breaking point is widely separated from the point of elastic limit on the
stress-strain graph. These materials show large plastic region.

Brittle— These are the materials which show very small plastic range beyond elastic limit.

Elastomers— With in elastic limit, stress is not proportional to strain. 1+1+1

Page-4

Small plastic
Large plastic range
range

stress
stress

strain
strain
(Ductile) (Brittle)
elastic
stress limit

strain

28. No. of DOF = 3 ½
1
Av. energy per molecule per DOF =
K T ½
2 B
3
Av. energy per molecule with 3 DOF = K T
2 B
3
Total energy of 1g mole of gas = K T × N ½
2 B
3
U = RT [ KB × N = R]
2
dU d 3

Cv = =
RT   ½
dT dT
2 
3
Cv = R
2
Cp – Cv = R
3 5
Cp = R + R = R
2 2
Cp 5
g= = = 1.67.  1
Cv 3
29. The potential energy of a spring is the energy associated with the state of compression or
expansion of an elastic spring. 1

O

A
Page-5

When the spring is compressed or elongated it tends to recover its original length. So
Restoring force ∝ extension or compression.
–F ∝ x
F = – kx 1
where k is spring constant.
Small amount of work done is displacing by distance dx is
dW = –Fdx = kxdx
x x
 x2

W= ∫ dw = ∫
0
kx dx = k


2 0
1 2
W= kx  1
2
This work done is stored in the form of potential energy of spring.
30. It states that if two vectors are represented by the two sides of a triangle taken in the same
order then resultant is represented by third side of a triangle taken in opposite order. 1
Q

R
B

 Q
O P N
A
In DOQN,
(OQ)2 = (ON)2 + (QN)2
= (OP + PN)2 + (QN)2...(1) 1
QN
In DPNQ sin q = ⇒ QN = B sin q
PQ
PN
cos q = ⇒ PN = B cos q
PQ
Put in (1)
R2 = (A + B cos q)2 + (B sin q)2
= A2 + B2 cos2 q + 2AB cos q + B2 sin2 q ... .
R= A 2  B2  2 AB cos
.  1

section-D
31. A simple pendulum consists of a massless string of length l whose one end is connected to
a spherical body of mas m known as bob and other end is connected to rigid support.
When the bob is displaced to position P, through a small angle q from vertical.
Various forces acting on bob are:
(i) Weight mg to bob vertically downward.
(ii) Tension T along PS. 1

Page-6

S



T

 P
Sin
mg 
mg Cos 
O
mg

Resolve mg into two components as shown in fig.
T = mg cos q...(1)
mg sin q will provide restoring torque.
t = –(mg sin q)l = –mg l sin q 1
t ≅ mg l q[ q is very small]
or t *∝ q
t = –k q
where k = spring factor = mg l
Inertia factor = ml2
Inertia factor
T = 2p
Spring factor

ml 2
= 2 p
mgl
l
T = 2p . 2
g
OR
(i) Newton gave an empirical relation to calculate velocity of sound in gas.
B B- bulk modulus of gas
v=  1
r r- density of gas
Newton assumed that the change in pressure and volume of gas when sound wave
propagated through it, are isothermal.
\ PV = constant
Differentiating it 1
PdV + VdP = 0
−dP
P= =B
dV / V
P
\ v=
r
P = hrg = 0.76 × 13.6 ×103 × 9.8 of Hg – Column

Page-7

r = 1.293 k/m3 for air
so we get v = 280 ms–1 1
(ii) h = 300 m,    g = 9.8 ms ,  v = 340 ms–1
–2

If t1 = time taken to strike the surface of water
1 2
S = ut + at
2
1
300 = 0 + × 9.8 t12
2
300
t1 = = 7.82 s  1
4.9
Time taken by sound to reach the top of tower
h 300
t2 = = = 0.88 sec. ½
v 340
Total time after which splash of sound is heared
= t1 + t2
= 7.82 + 0.88
= 8.70 sec. ½
32. Let R = radius of curvature of liquid meniscus
P = atmospheric pressure
S = Surface tension of liquid
r = radius of capillary tube
A

B

h

A C

B D E

Fig. 1
The pressure of point A = P
2S
The pressure of point B = P−
R
Pressure at point C and D is also P.
In order to attain equilibrium, the liquid level rise in the capillary tube upto height h.
Now, pressure at E = Pressure at B + Pressure due to height, h

2S 
=
P    hg  2

R
As there is equilibrium,
Pressure at E = Pressure at D
Page-8

2S
P
h
g = P
R
2S
=hrg O
R
2S
or h=
Rrg

In Fig. (2) G
H

GH r 
Cos q = =
OG R
r
or R=
cos q
2S cos 
So, we get h=
r
g
OR Fig. 2
Newton’s law of cooling states that the rate of loss of heat of a body is directly proportional
to the difference in temperature of the body and the surroundings, provided he difference
in temperature not more than 40°C. 1
Let a body of mass m, specific heat s at the temperature T. Let T0 is the temperature of
surroundings.  T > T0
dQ
Rate of loss of heat = −
dt
−dQ
∝ (T – T0)
dt
−dQ
= k (T – T0)
dt
d

msT
= k (T – T0)[ Q = msT]
dt
−dT k
=
T  T0

dt ms
−dT loge (T–To)
= K (T – T0)
dt
dT
= Kdt

T  T0

on Integrating both sides
loge (T – T0) = –Kt + C t

This equation is similar to a straight line; y = mx + C 3
33. ux = u cos q
uy = u sin q
ax = 0
ay = –g

Page-9

A

u
uy

H


O ux B C

(a) Path of Projectile
(1) Motion along x-axis
1
x = uxt + a t2
2 x
1
x = u cos q t + (0) t2
2
x = (u cos q) t
x
t= (1)
ucos q
  (2) Motion along y-axis
1
y = uyt + a t2
2 y
1
= (u sin q)t + (–g)t2
2
x g x2
= u sin q ·

u cos  2 u2 cos 2 
1 gx 2
y = x tan q –
2 u2 cos 2 q
This is the equation of parabola. Hence path of projectile is parabolic. 2
(b) Time of flight
Time of flight, T = time of ascent + time of descent
T=t+t
T
or t=
2
use vy = uy + ayt
At highest point, vy = 0
T
\ 0 = u sin q + (–g)
2
2u sin q
or T=  2
g
T
Maximum height attained y = H, t=
2

Page-10

1
y = uyt + a t2
2 y

u sin   1 u2 sin 2 
H = u sin q
  g 

g  2 g2

u2 sin 2  1 u2 sin 2 
=

g 2 g
u2 sin 2 q
H=  1
2g
OR
(i) Muzzle speed of bullet vB = 150 ms–1
= 540 km h–1
speed of police van, vp = 30 km h–1
speed of thief car, vT = 192 kmh–1 1
since the bullet is sharing the velocity of the police man.
So effective velocity VB = vB + vP
= 540 + 30 = 570 kmh–1 1
The speed of bullet w.r.t. the thief’s car moving in the same direction
VBT = VB – VT = 570 – 192 = 378 kmh–1 1
dv
(ii) we know a=
dt
dv = a dt 1
Integrating on both sides
v t
∫ dv = a ∫ dt
u 0



v
vu = a t
0
t

[v – u] = a [t – 0]
v = u + at 1

section-E  (Case Study)
34. (i) (c) (ii) (B) 1×4=4
(iii) rolling (iv) m = tan q.
Or
By using lubricants, we can reduce friction.
35. (i) (B) (ii) (B) 1×4=4
(iii) (A) (iv) w = mg
w 49
\ m= =
g 9.8
= 5 kg
Or
Decreases. qqq

Page-11