Class 11 Question Paper 2025 Physics

ANNUAL EXAMINATION

Question Paper
2025
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FOR CBSE AND STATE BOARD
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Annual Exam 2025 Question Paper

SESSION ENDING EXAMINATION 2024-25
CLASS: XI ​ ​ ​ ​ ​ ​ ​ ​ ​ ​ SUBJECT: PHYSICS
TIME: 3 HOURS​ ​ ​ ​ ​ ​ ​ ​ ​ M. MARKS: 70
GENERAL INSTRUCTIONS:
(I)​ All the questions are compulsory
(II)​ There are 33 questions in the paper. Section A contains 12 MCQs and 4 Assertion-Reason
questions of 1 mark each, Section B contains 5 Short answer questions of 2 marks each,
Section C contains 7 short answer questions of 3 marks each, Section D contains 2 Case-study
Based questions of 4 marks, Section E contains 3 long answer questions of 5 marks each.
(III)​ There is no overall choice in the paper however an internal choice has been given in 1 question
of 2 marks, 1 question of 3 marks, 1 question of 4 marks and all the questions of 5 marks.
SECTION A
MULTIPLE CHOICE QUESTIONS (1 X 12 = 12 MARKS)
1. Out of the following pairs, which one does not have identical dimensions? (1)
(a)​ Moment of Inertia and moment of force
(b)​ Work and torque
(c)​ Angular momentum and Planck’s constant
(d)​ Impulse and momentum
2. Speeds of two identical cars are u and 4u at a specific instant. The ratio of the respective distances at (1)
which the two cars are stopped from that instant is
(a)1:1 (b) 1:4 (c) 1:8 (d) 1: 16
3. If the velocity-time graph has the slope AMB, what would be the shape of the corresponding (1)
acceleration-time graph?

→ →
4. ^ ^ ^
The angle between vectors 𝐴 = 10𝑖 + 10𝑗 − 5𝑘 𝑎𝑛𝑑 𝐵 = 10𝑖 − 5𝑗 + 10𝑘 is
^ ^ ^ (1)
(a)​ 30 ͦ (b) 45 ͦ (c) 60 ͦ (d) 90 ͦ
5. If normal force is doubled, then coefficient of friction is (1)
(a)​ Halved (b) tripled (c) doubled (d) not changed
6. Instantaneous power can be expressed as (1)
→ → → →2 → → →
(a)​ 𝐹. 𝑣 (b)1/2𝐹. 𝑣 (c)𝐹. 𝑡 (d)𝐹 𝑥 𝑣

Annual Exam 2025 Question Paper
7. A system consists of 3 particles each of mass m and located at (1,1), (2,2), (3,3). The co-ordinates of (1)
the centre of mass are
(a)​ (6,6) (b) (3,3) (c) (2,2) (d) (1,1)
8. The breaking force for a wire of diameter D of a material is F. The breaking force for a wire of the (1)
same material of radius D is
(a)​ F (b) 2F (c) F/4 (d) 4F
9. If the liquid neither rises nor falls in a capillary tube, the angle of contact is (1)
(a)​ 0ͦ (b) 45 ͦ (c) 90 ͦ (d) 180 ͦ
10. The temperature gradient of 0.5 m long rod is 80 ͦ C/m. If the temperature of hotter end of the rod is (1)
30 ͦC, then the temperature of the colder end is
(a)40 ͦC (b) - 10 ͦC (c) 10 ͦC (d) 0 ͦ ͦC
11. If the gas has f degrees of freedom, ratio of specific heats of the gas is (1)
1+𝑓 1 𝑓 2
(a)​ 2
(b) 1 + 𝑓 (c) 1 + 2 (d) 1 + 𝑓
12. If a spring has time period T and is cut into 2 equal parts, then the time period of oscillation of (1)
each part will be
(a)​ √2 𝑇 (b) T/ √2 (c) 2T (d) T/2
ASSERTION REASON QUESTIONS (QUES NO. 13-16)
Each of these questions contain two statements, Assertion and Reason. Each of these questions also
have four alternative choices, only one of which is the correct answer. You have to select one of the
codes (a), (b), (c) & (d) given below.
(a)​ Both Assertion and Reason are true and Reason is the correct explanation of Assertion
(b)​Both Assertion and Reason are true but Reason is not the correct explanation of Assertion
(c)​ Assertion is true but Reason is false
(d)​Both Assertion and Reason are false
13. ASSERTION: The speed of a body in uniform circular motion is constant (1)
REASON: In uniform circular motion, the acceleration of the body is constant.
14. ASSERTION: Impulsive force is large and acts for a short time. (1)
REASON: Finite change in momentum should be produced by the force.
15. ASSERTION: Surface tension decreases with increase in temperature. (1)
REASON: On increasing temperature, kinetic energy increases and inter-molecular forces decrease.
16. ASSERTION: In an adiabatic compression, the internal energy and temperature of the system gets (1)
decreased.
REASON: Adiabatic compression is a slow process.
SECTION B (TWO MARKS QUESTIONS) 2 X 5 = 10 MARKS
17. Write the expressions for the acceleration due to gravity ‘g’ (2)
(i)​ at a height h (h >> R), where R is the radius of the Earth and
(ii)​ depth d from the Earth’s surface
Also, show the variation graphically.
( Assume Earth to be a sphere of uniform density).
18. If momentum[P], area [A] and time [T] are taken as fundamental quantities, then find the (2)
dimensional formula for coefficient of viscosity in terms of P,A & T.
19. The displacement x of a particle varies with time t as x = 4t2 - 15t + 25. Find the instantaneous (2)
velocity and acceleration of the particle at t = 0 s. What can you infer about the acceleration of the
particle?
OR
Prove s = ut + 1/2 at2 graphically
20. Will two spheres of equal masses and radii, one solid and the other hollow have equal moments of (2)
inertia? Give reason
21. At what temperature is the rms velocity of hydrogen molecule equal to that of an oxygen molecule at (2)
47 ͦC?

Annual Exam 2025 Question Paper
SECTION C (THREE MARKS QUESTIONS) 3 X 7 = 21 MARKS
22. What is meant by banking of roads? What is the need for banking of a road? Derive the expression (3)
for the angle of banking ignoring friction between the tyres and the road.
23. Derive a relation between torque and angular momentum (3)
24. Define ‘Escape velocity’. Prove that moon would depart forever if its speed were increased by 42% (3)
OR
(i)​ State Kepler’s third law of periods.
(ii) Two satellites are at different heights (H1 > H2) from the surface of the Earth. Which would have
greater velocity?
25. Explain why: (3)
(a) a brass tumbler feels much colder than a wooden tray on a chilly day.
(b) the earth without its atmosphere would be inhospitably cold.
(c) heating systems based on circulation of steam are more efficient in warming a building than those
based on circulation of hot water.
26. A spring balance has a scale that reads from 0 to 50 kg. The length of the scale is 20 cm. A body is (3)
suspended from this balance, when displaced and released, oscillates with a certain period. What is
the value of the spring constant?
27. The equation of a plane progressive wave is y (x,t) = 10 sin 2π(t - 0.005x), where y and x are in cm (3)
and t in seconds. Calculate
(a)​ The wavelength
(b)​ Frequency
(c)​ Velocity of the wave.
28. Discuss various modes of vibration in open organ pipe. Hence, show that the ratio of frequencies of (3)
different harmonics with first harmonic in open organ pipe is 1:2:3
SECTION D CASE STUDY BASED QUESTION (4 x 2 = 8 MARKS)
29. Conservation of Momentum
This principle is a consequence of Newton’s second and third laws of motion. In an isolated system, (4)
internal forces between pairs of particles in the system cause momentum change in individual
particles. Let a bomb be at rest, then its momentum will be zero. If the bomb explodes into two equal
parts, then the parts fly off in exactly opposite directions with same speed, so that the total
momentum is still zero. Here, no external force is applied on the system of particles.
(i)​ The momentum of a system is conserved
(a)​ Always (b) never
(b)​ in the absence of an external force (d) none of the above
(ii)​ A shell of mass 10 kg is moving with a velocity of 10 ms-1, then it blasts and forms two parts
of masses 9 kg and 1 kg respectively. If the first mass is stationary, the velocity of the second is
(a)​ 1m/s (b) 10 m/s (c) 100 m/s (d) 1000 m/s
(iii)​ Two masses of M and 4M are moving with equal kinetic energy. The ratio of their linear
momenta is
(a)​ 1:8 (b) 1:4 (c) 1:2 (d) 4:1
(iv)​ A bullet of mass 0.1 kg is fired with a speed of 100 m/s. The mass of gun being 50 kg, then
the recoil velocity of the gun is
(a)​ 0.2 m/s (b) 0.5 m/s (c) 0.1 m/s (d) 0.05 m/s
OR
A bullet is fired from a rifle. If the rifle recoils freely, then the kinetic energy of the rifle is
(a)​ Less than that of the bullet
(b)​ more than that of the bullet
(c)​ Same as that of the bullet
(d)​ Equal to or less than that of the bullet

Annual Exam 2025 Question Paper
30. First law of thermodynamics (4)
Heat and work are two modes of energy transfer to a system. Heat is the energy transfer arising due
to temperature difference between the system and the surroundings. Work is energy transfer brought
about by other means, such as moving the piston of a cylinder containing the gas, by raising or
lowering some weight connected to it. The first law of thermodynamics is the general law of
conservation of energy applied to any system in which energy transfer from a system to the
surroundings occurs through heat and work. According to the first law of thermodynamics, if some
heat is supplied to a system which is capable of doing work, then the quantity of heat ∆𝑄 absorbed
by the system will be equal to the sum of the increase in its internal energy ∆U and the external
work ∆W done by the system on the surroundings.
∆𝑄 = ∆U + ∆W
As ∆W = P∆V, so ∆𝑄 = ∆U + 𝑃∆V
(i)​ In a cyclic process, work done by the system is
(a)​ Zero
(b)​ More than the heat given to the system
(c)​ Equal to heat given to the system
(d)​ Independent of heat given to the system
(ii)​ The increase in internal energy of a system is equal to the work done on the system. Which
process does the system undergo?
(a)​ Isochoric (c) Isobaric
(b)​ Isothermal (d) Adiabatic
(iii)​ The heat of 110 J is added to a gaseous system whose internal energy is 40 J, then the
amount of work done will be
(a)​ 70 J (b) 140 J (c) -70 J (d) -140 J
(iv)​ A thermodynamic process is carried out from an original state D to an intermediate state E
by the linear process shown in the figure

The work done by the gas from D to E to F is
(a)​ 100 J (b) 800 J (c) 300 J (d) 250 J
OR
In an Isothermal change of an ideal gas, ∆U = 0. The change in heat energy ∆Q is equal to
(a)​ 0.5 ∆W (b) ∆W (c) 1.5 ∆W (d) 2 ∆W
SECTION E (FIVE MARKS QUESTIONS) 5 X 3 = 15 MARKS
31. Show that the total mechanical energy of a freely falling body under gravity is conserved. Draw a (5)
graph to show the variation of Potential energy, kinetic energy and total energy of a freely falling
body with height from the ground.
OR
Identify the type of collision in which
(i)​ Linear momentum and kinetic energy is conserved
(ii)​ Linear momentum is conserved but kinetic energy is not conserved
(iii)​ Bodies stick together after the collision
What is the relation between velocity of approach and velocity of separation in (i) and (ii) cases.

Annual Exam 2025 Question Paper
For an elastic collision in one dimension of two bodies with equal masses, prove that the particles
interchange their velocities after the collision.
32. (i)​ What is a ‘Projectile’? (5)
(ii)​ A projectile is fired upward at an angle θ with the horizontal with velocity u, show that the
path of projectile is a parabola.
(iii)​ Find the angle of projection at which the horizontal range and maximum height of the
projectile are equal.
OR
(i)​ State the ‘Parallelogram law of vector addition’.
(ii)​ Derive an expression for the magnitude of the resultant of two vectors inclined at angle θ.
(iii)​ Two equal forces have the square of their resultant equal to three times their product. Find the
angle between them.
33. (i)​ State and prove Bernoulli’s theorem. (5)
(ii)​ Why it is dangerous to stand near the platform when a train is passing on the railway track?
OR
(i)​ Derive an expression for excess pressure inside a soap bubble.
(ii)​ Eight drops of equal radius coalesce to form a bigger drop. What is the ratio of surface energy
of bigger drop to smaller one? Surface tension of liquid is σ.

SESSION ENEDING EXAM 2024-25
CLASS XI PHYSICS
MODEL ANSWER SHEET
SECTION A (MULTIPLE CHOICE QUESTIONS) MARKS
1. (a)​ Moment of inertia and moment of force 1
2. (d) 1:16 1
3. 1

4. (d) 90 ͦ 1
5. (d) not changed 1
→ →
6. (a)𝐹. 𝑣 1
7. (c)(2,2) 1
8. (d) 4F 1
9. (c)90 ͦ ͦ 1
10. (b)​ - 10 ℃
ͦ 1
11. (d)1 + 𝑓
2 1
12. (b)T/√2 1
ASSERTION-REASON QUESTIONS
13. (c)​ Assertion is true but Reason is false 1
14. (a)Both Assertion and Reason are true and Reason is the correct explanation of 1
Assertion
15. (a)Both Assertion and Reason are true and Reason is the correct explanation of 1
Assertion
16. (d)​ Both Assertion and Reason are false 1
SECTION B (2 MARKS QUESTIONS)
17. ℎ −2 1/2
(i)​ (
At a height h >> R, 𝑔' = 𝑔 1 + 𝑅)
(ii)​ (
At a depth d, 𝑔' = 𝑔 1 −
𝑑
𝑅 ) 1/2

1

18. Using dimensions,
𝑎 𝑏 𝑐
η∝𝑃 𝐴 𝑇
𝑎 𝑏 𝑐
η= 𝑘𝑃 𝐴 𝑇
Putting dimensions
[ML-1T-1] = k [ MLT-1]a [L2]b [T]c 1
[ML-1T-1] = k [M]a [L]a + 2b [T]-a + c
On comparing the powers of M, L and T

a = 1, a + 2b = -1, -a + c = -1 1/2
On solving a = 1, b = -1, c = 0
1 −1 0
Therefore, η = 𝑘 𝑃 𝐴 𝑇 1/2

19. Given, x = 4t2 - 15t + 25
𝑑𝑥
Instantaneous velocity, v = 𝑑𝑡 = 8t - 15 1/2
1/2
At t = 0 s, v = -15 m/s
𝑑𝑣
Instantaneous acceleration, a = 𝑑𝑡 = 8 1/2
2
At t = 0 s, a = 8 m/s
Acceleration is constant throughout the motion. 1/2
OR

½(graph)

The v-t graph for accelerated motion is shown
Area of v-t graph = area of triangle + area of rectangle 1/2
= 1/2 x t x (v-u) + u x t
= 1/2t (at) + ut ( since v = u + at) 1/2
Therefore, area of v-t graph = ut + 1/2 at2
But area under v-t graph = displacement s 1/2
Therefore, s = ut + 1/2 at2
Hence proved.
20. No, 1/2
Hollow sphere will have greater moment of inertia as mass is distributed away from
the axis of rotation. 1 1/2
21. 3𝑘𝑇𝐻 1/2
RMS velocity of hydrogen molecule, vH2 = 𝑚𝐻
1/2
3𝑘𝑇𝑂
RMS velocity of oxygen molecule, vO2 = 𝑚𝑂 1/2
Given that vH2 = vO2, 𝑇𝑂 = 47 ͦC + 273 = 320 K, 𝑚𝐻 = 2, 𝑚𝑂 = 32
𝑇𝐻 𝑇𝑂 1/2
Therefore, 𝑚 = 𝑚𝑂
𝐻
𝑇𝑂 320 𝑥 2
𝑇𝐻 = 𝑚𝑂
𝑚𝐻 = 32
= 20 K
SECTION C (3 MARKS QUESTIONS)

22. Banking of roads: Raising of outer edge of the road a little above the inner edge in 1
hilly areas.
It is done to provide the necessary centripetal force for a vehicle to move with a 1/2
reasonable speed without skidding.

1/2 (fig.)

1/2
From fig,
N cos θ = Mg
2
𝑀𝑣
N sin θ = 𝑅
2
𝑣
1/2
Tan θ = 𝑅𝑔
2
𝑣
Angle of banking, θ = tan -1 ( 𝑅𝑔 )

→ → →
23. We have, 𝐿 = 𝑟 𝑥 𝑝 1/2
Differentiating wrt t
→ →
→ → → 1
𝑑𝐿 𝑑𝑟 𝑑𝑝
𝑑𝑡
= 𝑑𝑡 x 𝑝 + 𝑟 𝑥 𝑑𝑡
→ → → → 1
= 𝑣 𝑥 𝑚𝑣 + 𝑟 𝑥 𝐹
→
=0+τ
→ →
𝑑𝐿 1/2
τ= 𝑑𝑡
24. Escape velocity: The velocity required by a body to be projected from the surface of 1
the Earth so as to overcome the gravitational pull of the Earth.
The percentage increase in the orbital velocity of moon required to escape the
gravitational pull of the Earth 1
𝑣𝑒 − 𝑣𝑜 2𝑔𝑅 − 𝑔𝑅
𝑣𝑜
x 100 = x 100
𝑔𝑅

= ( 2 − 1)𝑥 100 = (1.414 - 1) x 100 = 41.4 % ≈42% 1
OR
(i)​ Kepler’s third law of periods: The square of time period of revolution of a
planet is directly proportional to the cube of semi-major axis of the elliptical orbit. 1
T2 ∝ R3
𝐺𝑀
(ii)​ The orbital velocity of satellite v0 = 𝑅+𝐻
, 1
Hence, lesser the height, greater will be the velocity of the satellite. So, satellite at a
height H1 will have more velocity. 1

25. (a) brass is a good conductor of heat, it readily transfers heat away from your body
when you touch it, while wood is a poor conductor, so very little heat is transferred
from your hand to the wood. 1
(b) the lower layers of the earth’s atmosphere reflect infrared radiations(heat) from
earth back to the surface of the earth; thus the heat radiation received by the earth 1
from the sun during the day are kept trapped by the atmosphere.
(c) because steam contains more heat compared to hot water.
1

26.​ F = mg = 50 kg x 9.8 ms-2 = 490 N 1
Use Hooke’s law F = kx 1
490 = k x 0.20
K = 2450 N/m 1

27.​ y (x,t) = 10 sin 2π(t - 0.005x)
Comparing with y = A sin (ωt - kx)
(i)​
2π
k = λ = 2π x 0.005, λ = 1/0.005 = 200 cm 1
1
(ii)​ w = 2πν = 2π, ν = 1 Hz
1
(iii)​ V = ν λ = 1 x 200 = 200 cm/s

28.​ Open organ pipe:
2𝐿
We have λ = 𝑛

1
First mode of vibration: n = 1,λ = λ1, λ1 = 2L, L = λ1/2
𝑣 1 γ𝑃
Frequency v1 = λ = 2𝐿 ρ
= fundamental frequency or
1

first harmonic
Second mode of vibration: n = 2,λ = λ2, λ2 = L, L = λ2/2 1
𝑣 1 γ𝑃
Frequency v2 = λ2
= 𝐿 ρ
= 2v1 second harmonic
Third mode of vibration: n = 3,λ = λ3, λ3 = 2L/3, L = 3λ
3/2
𝑣 31 γ𝑃
Frequency v3 = λ = 2𝐿 ρ
= 3v1 third harmonic 1
1

Hence, v1:v2:v3 = 1:2:3
Hence proved
29. SECTION D ( CASE-STUDY BASED QUESTION 4 MARKS)
(i) (b) in the absence of an external force 1

(ii) (c) 100 m/s 1
(iii) (c) 1:2 1
(iv) (a)​ 0.2 m/s 1
OR
(a)less than that of the bullet
30.
(i) (c)equal to heat given to the system 1
(ii) (d)adiabatic process 1
(iii) (a)70J 1
(iv) (b)800 J 1
OR
(b)∆𝑊
SECTION E (LONG ANSWER QUESTION 5 MARKS)
30. (i)​ Elastic collision 1/2
(ii)​ Inelastic collision 1/2
(iii)​ Perfectly inelastic collision 1/2
In an elastic collision, velocity of approach = velocity of separation 1/2
In an inelastic collision, velocity of approach > velocity of separation

Elastic collision in one dimension:
Linear momentum of the system remains conserved
m1u1 + m2u2 = m1v1 + m2v2 ………………………..(1) ½
Kinetic energy of the system is also conserved
½ m1u12 + ½ m2u22 = ½ m1v12 + ½ m2v22……………(2) ½
From (1) and (2)
m1 (u1 – v1) = m2 (v2 – u2)…………..(3)
m1 (u12 – v12) = m2 (v22- u22)………..(4)
Dividing (4) by (3)
u1 + v1 = v2 + u2 ½
v2 = u1 + v1 - u2
Putting the value of v2 in eq (1)
m1u1 + m2u2 = m1v1 + m2(u1 + v1 - u2)
(𝑚1− 𝑚2)+2𝑚2𝑢2
Therefore v1 = 𝑚1+𝑚2 1
(𝑚2− 𝑚1)+2𝑚1𝑢1
Similarly, v2 = 𝑚1+𝑚2

If masses are equal, m1 = m2 = m
So, v1 = u2 and v2 = u1 ½
Hence, in an elastic collision in one dimension of equal masses, velocities are
interchanged.
OR

For a freely-falling body of mass m falling from a height h,

1 (fig.)

1

At point A, 1
P.E. = mgh, K.E. = 0
T.E. = P.E. + K.E. = mgh ………………….(1)
At point B,
P.E. = mg (h-x), K.E = ½ m v2
Using v2 = u2 + 2gx
v2 = 0 + 2gx
K.E. = ½ m (2gx) = mgx 1
T.E. = mg(h – x) + mgx = mgh ………….(2)
At point C
P.E. = 0
K.E. = ½ m V2
Using V2 = u2 + 2gh
V2 = 2gH 1
K.E. = 1/2m (2gh) = mgh
T.E. = mgh………………………………………..(3)
Hence, mechanical energy of a freely falling body remains conserved.

31. Projectile: A body which when thrown moves under the effect of gravity alone. 1/2

½
For motion along X-axis (fig.)
ux = u cos θ, ax = 0
using Sx = uxt + ½ ax t2 ½
x = (u cos θ)t ……………………(1)

For motion along Y-axis
uy = u sin θ, ay = -g
Using Sy = uyt + ½ ayt2 ½
y = u sin θ t– ½ gt2 ……………….(2)

Putting the value of t from (1) in (2)
𝑥 𝑥 2 ½
y = u sin θ 𝑢 𝑐𝑜𝑠θ – ½ g ( 𝑢 𝑐𝑜𝑠θ )
2
𝑥
y = x tanθ – ½ g 2 2
𝑢 𝑐𝑜𝑠 θ 1/2
2
Since, y α x
hence, the motion of projectile is parabolic.
1/2
2
Range R =
𝑢 𝑠𝑖𝑛2θ
, 1/2
𝑔
2 2
𝑢 𝑠𝑖𝑛 θ
Maximum height attained Hmax = 2𝑔
For equal range and maximum height, R = Hmax
2 2
𝑢 𝑠𝑖𝑛 θ
2
𝑢 𝑠𝑖𝑛2θ
𝑔
= 2𝑔
2 sinθ cosθ = sin2θ/2 1
Tan θ = 4
Θ = tan-1 (4) approx 75.96 degrees

OR 1/2(fig.)
Parallelogram law of vector addition- “If two co-initial vectors can be represented
in magnitude and direction by the two adjacent sides of a parallelogram, then their
resultant will be represented completely in magnitude and direction by the diagonal of
that parallelogram”.

½

½
Construction- Draw a perpendicular from S to OP produced.
From right angled triangle SNP,
𝑆𝑁
𝑃𝑆
= sinθ or SN = PS sinθ = B sinθ
𝑃𝑁 ½
𝑃𝑆
= cosθ or PN = PS cosθ = B cosθ
Using Pythagoras theorem in right angles triangle ONS, 1/2
OS2 = ON2 + SN2 = (OP + PN)2 + SN2
Or R2 = (A + B cosθ)2 + (B sinθ)2
= A2 + B2 (cos2θ + sin2θ) + 2AB cosθ
= A2 + B2 + 2AB cosθ 1/2
2 2
Or R = 𝐴 + 𝐵 + 2𝐴𝐵 𝑐𝑜𝑠θ
1/2
R2 = A2 + B2 + 2AB cosθ 1/2
Given A = B = x, R2 = 3AB = 3X2
R2 = x2 + x2 + 2x2 cosθ = 2x2 (1 + cosθ) 1/2
3x2 = 2x2 (1 + cosθ) 1/2
1 + cosθ = 3/2
Cosθ = 1/2, θ = 60 ͦ

33. (i)​ Bernoulli’s theorem: 1
“ It states that for an ideal liquid (non-viscous, incompressible) in a streamlined flow,
the sum of pressure energy, kinetic energy, potential energy per unit volume remains
constant provided there is no source or sink of fluid throughout its flow”.

1/2 (fig.)

Work done per sec by the pressure energy at A = P1A1v1 = P1∆𝑉
Work done per sec by the pressure energy at B = P2A2v2 = P2∆𝑉
= (P1 - P2)∆𝑉……(1) 1/2
Change in potential energy from A to B = mg (h2 - h1)……………………,(2) 1/2
2 2
Change in kinetic energy from A to B = 1/2 m (𝑣2 − 𝑣1)…………………(3) 1/2
Using work-energy conservation,
Work done per sec by pressure energy = Change in P.E. + Change in K.E.
2 2
(P1 - P2)∆𝑉 = mg (h2 - h1) + 1/2 m (𝑣2 − 𝑣1)
Dividing by ∆𝑉
2 2 1
(P1 - P2) = ρg (h2 - h1) + 1/2 ρ(𝑣2 − 𝑣1)
𝑃 1 2 𝑃 1 2
Or ρ1 + 2
ρ𝑣1 + ρ𝑔h1 = ρ2 + 2
ρ𝑣2 + ρ𝑔h2

𝑃 1 2
ρ
+ 2
ρ𝑣 + ρ𝑔h = constant

Hence proved. 1
(ii)​ When a train is passing on the railway track, the velocity of air between the
train and the person increases and pressure decreases, while the atmospheric pressure
is greater behind him. The pressure difference will pull him towards the train.

OR
(i)​ Excess pressure inside a soap bubble:

1

Work done in increasing the area of the bubble
W = (pi - po) x 4πR2 x δ𝑅………….(1) 1

Increase in surface energy = ∆𝑈 = Surface tension x Increase in surface area

= σ 𝑥 2 [ 4π ( R + δ𝑅)2 - 4πR2] (since a soap
bubble has two surfaces)
∆𝑈 = 2σ x [4π ( R2 + δ𝑅2 + 2Rδ𝑅 - R2]
Since δ𝑅<<R, neglecting δ𝑅2
∆𝑈
= 16πRδ𝑅σ ……………………….(2) 1
From (1) and (2)
W = ∆𝑈
(pi - po) x 4πR2 x δ𝑅 = 16πRδ𝑅σ
4σ
(pi - po) = 𝑅
1

(ii)​ Volume of 8 small drops = Volume of 1 big drop
4 3 4 3
8 x 3 π𝑟 = 3 π𝑅
R = 2r
1
Surface energy of big drop U1 = 4πR2σ
Surface energy of small drop U2 = 4πr2σ
𝑈1 𝑅
2

𝑈2
= 2
𝑟
U1 : U2 = 4 : 1

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