ICSE Class 10 Physics Chapter Work, Power and Energy PYQs + Solution

The Brainbox Tutorials has especially made the chapter wise PYQs series of all important subjects. Get access to ICSE Class 10 Physics Chapter Work, Power and Energy PYQs + Solution here. All the questions which came in Board exams from the chapter Work, Power and Energy from the year 2014 to 2024 are provided below. Chapter wise PYQs will be very beneficial for the students preparing for the upcoming board exams.

ICSE Class 10 Physics Chapter Work, Power and Energy PYQs + Solution

2024

Q1. When a bell fixed on a cycle rings, then the energy conversion that takes place is:

(a) gravitational potential energy to sound energy
(b) kinetic energy to sound energy
(c) sound energy to electrical energy
(d) sound energy to mechanical energy

Ans: (b) kinetic energy to sound energy

Q2. A force F moves a load from A to C as shown in the figure below. For the calculation
of the work done, which of these lengths would you use as the displacement?

(a) 3m
(b) 4m
(c) 5m
(d) 7m

icse-class-10-physics-chapter-work-energy-and-power-pyqs-solution

Ans: (c) 5m

Q3. Fill in the blanks:
(a) Work done by centripetal force at any instant is __________________ .

Ans: zero

Q4. Sumit does 600 J of work in 10 min and Amit does 300 J of work in 20 min. Calculate
the ratio of the powers delivered by them.

Ans: Power = Work / time

Power developed by Sumit = 600J/(10 x 60)s = 1 W

Power developed by Amit = 300 J/(20 x 60)s = 1/4 W

Sumit’s power : Amit’s power = 1 : 1/4

= 4 : 1

Q5. Two identical marbles A and B arc rolled down along Path l and Path 2 respectively. Path 1 is frictionless and Path 2 is rough.

(a) Which marble will surely reach the next peak?
(b) Along which path/s the mechanical energy will be conserved?
(c) Along which path/s is the law of conservation of energy obeyed?

Ans: (a) Marble A. As Path 1 is frictionless, there will be no energy loss on a frictionless road, marble A will undoubtedly reach the next peak.

(b) Path 1. As there is no friction, Mechanical energy is conserved.

video solution of ICSE Class 10 Physics Chapter Work, Power and Energy PYQs

2023

Q 1. When the speed of a moving object is doubled, then its kinetic energy:
a. remains the same
b. decreases
c. is doubled
d. becomes four times

Ans: d. becomes four times

Q 2. The energy conversion in a washing machine is from ……………
a. magnetic to electrical
b. electrical to mechanical
c. electrical to magnetic
d. magnetic to electrical

Ans: b. electrical to mechanical

Q 3.(a) What should be the angle between the direction of force and the direction of displacement, for work to be negative?
(b) Name the physical quantity obtained using the formula U/h, where U is the potential energy and h is the height.

Ans: (a) 180°

(b) We know, U = mgh

Hence, mg = U/h

Mg = Weight or Force

Therefore, the physical quantity obtained is Force.

Q 4. Calculate the power spent by a crane while lifting a load of mass 2000 kg, at velocity of 1.5 ms-1. (g = 10 ms^-2).

Ans: Power = Force x velocity

= mg x v

= 2000 x 10 x 1.5

=30000 W

Q 5. A car of mass 120 kg is moving at a speed 18 km h^-1 and it accelerates to attain a speed of 54 km h^-1 in 5 seconds. Calculate:
(a) the work done by the engine.
(b) the power of the engine.

Ans: (a) initial velocity (u)= 18 km h^-1 = 18 x 5/18 = 5 m s^-1

final velocity (v)= 54 km h^-1 = 54 x 5/18 = 15 m s^-1

Time(t) = 5 second

We know, a = (v – u)/t

a = (15 – 5)/5 = 2 m ^s-2

Now, v² – u²= 2aS

(15)² – (5)² = 2 x 2 x S

225 – 25 = 4 S

S = 200/4 = 50 m

Now, F = ma

= 120 x 2 = 240 N

(i) Work done = F x S

= 240 N x 50 m

=12000 J

(ii) Power of the engine = work done/time taken

= 12000 J/5 s

=2400 W

2021 Semester-1

Q 1. A coolie raises a load upwards against the force of gravity then the work done by the load is: [1]
(a) zero (b) positive work (c) negative work (d) none of these.

Ans: (c) negative work

Q 2. The energy change during photosynthesis in plants is: [1]
(a) heat to chemical.
(b) light to chemical.
(c) chemical to light.
(d) chemical to heat.

Ans: (b) light to chemical.

Q 3. A body of mass 200 g falls freely from a height of 15 m. [g = 10 ms^-2]
(i) When the body reaches 10 m above the ground, its potential energy will be:
(a) 20000 J (b) 10 J (c) 10000 J (d) 20 J

Ans: (i) (d) 20 J

Potential energy = mgh

= 0.2 kg x 10 m s^-2 x 10 m = 20 J

(ii) The gain in kinetic energy of the body when it reaches 10 m above the ground is:
(a) 20 J (b) 10 J (c) 30 J (d) 25 J

Ans: (ii) (b) 10 J

Initial P.E. at highest point = mgh = 0.2 x 10 x 15 = 30 J

Potential energy at highest point is maximum and Kinetic energy at this point is zero.

Now, Potential energy at 10m height is 20 J

According to the law of conservation of energy,

Kinetic energy at 10 m height = P.E. at 15 m height – P.E. at 10 m height

= 30 J – 20 J = 10 J

(iii) The total mechanical energy it will possess, when it is just about to strike the ground is:
(a) 30000 J (b) 20000 J (c) 30 J (d) 20 J

Ans: (iii) (c) 30 J

According to the law of conservation of energy,

Mechanical energy always remains same during a free fall.

Therefore, total energy when it is just about to strike the ground = 30 J

(iv) The velocity in ms^-1 with which the body will hit the ground is:
(a) 30 (b) 10 (c) 1 0 √3 (d) 1 0 √2

Ans: (c) 1 0 √3

We know, K.E. = 1/2 mv²

30 = 1/2 x 0.2 x v²

v² = 30 x 2/0.2 = 300

v = √300 = 10 √3 m s^-1

Q 4. Which one of the following is the correct mathematical relation?
(a) Power = Force/ Velocity
(b) Power = Force x Acceleration
(c) Power Force/ Acceleration
(d) Power= Force x Velocity

Ans: (d) Power= Force x Velocity

Q 5. Joule = __ erg
(a) 10⁹
(b) 10⁷
(c) 10⁵
(d) 10⁶

Ans: (b) 10⁷

Q 6. A light body A and a heavy body B have the same momentum.
(i) Choose a correct statement from the given options.
(a) kinetic energy of body A and body B will be the same.
(b) kinetic energy of body A is greater than kinetic energy of body B.
(c) kinetic energy of body B is greater than kinetic energy of body A.
(d) unless we know the velocity, we cannot find which body has greater kinetic energy.

Ans: (b) kinetic energy of body A is greater than kinetic energy of body B.

(ii) If the ratio of kinetic energies of A and B is 5 : 2 then which one of the following gives the mass ratio of the bodies respectively?
(a) 5 : 2 (b) 25 : 4 (c) 2 : 5 (d) 4 : 24

Ans: (c) 2 : 5

Given that momentum of A (P_A) = momentum of B (P_B) and mass of A (m_A) < mass of B (m_B)

\[\frac{K.E._A}{K.E._B}=\frac52\\\Rightarrow\frac{{\displaystyle\frac12}m_{A.}v_A^2}{\frac12m_{B.}v_B^2}=\frac52\\\Rightarrow\frac{m_{A.}v_A^2}{m_{B.}v_B^2}=\frac52\\\Rightarrow\frac{\displaystyle\frac{P_A^2}{m_A}}{\frac{P_B^2}{m_B}}=\frac52\\\Rightarrow\frac{\displaystyle\frac1{m_A}}{\displaystyle\frac1{m_B}}=\frac52\;\;\lbrack\;as\;momentums\;of\;A\;and\;B\;are\;same\rbrack\\\Rightarrow\frac{m_B}{m_A}=\frac52\\\Rightarrow\frac{m_A}{m_B}=\frac25\]

2020

Q1. A crane ‘A’ lift a heavy load in 5 seconds whereas another crane ‘B’ does the same work in 2 seconds compare the power of crane ‘A’ to that of crane ‘B’.

Ans: Let the work done by both the cranes be x joule.

Crane A
W_A= x
t_A = 5 s
P_A = W_A/t_A =x/5

Crane B
W_B = x
t_B = 2 s
P_B = W_B/t_B =x/2

Therefore, P_A : P_B=x/5 : x/2

= 2 : 5

Q 2. Define a kilowatt hour. How is it related to Joule?

Ans: kilowatt-hour is the energy spent (or work done) by a source of power 1 kW in 1 hour.

1 kWh = 3.6 × 10⁶ joule.

Q 3. A satellite revolves around a planet in a circular orbit what is the work done by the satellite at any instant? Give reason.

Ans: The work done is zero because the centripetal force on the body at any instant is directed towards the centre of the circular path and the displacement at that instant is along the tangent to the circular path, i.e., normal to the direction of force on the body. Since, θ = 90° , cos 90° = 0. Therefore, work done is zero.

Q 4. Give one example of each when:
(i) Chemical energy changes into electrical energy
(ii) Electrical energy changes into sound energy.

Ans: (i) In a battery when current is drawn from it in the external circuit.

(ii) In a loud speaker, electrical signals vibrate the diaphragm of loud speaker to produce sound.

Q5. The figure alongside shows a simple pendulum of mass 200 g. It is displaced from the mean position A to the extreme position B. The potential energy at the position A is zero. At the position B the pendulum bob is raised by 5 m.

(i) What is the potential energy of the pendulum at the position B?

(ii) What is the total mechanical energy at point C?

(iii) What is the speed of the bob at the position A when released from B?

(Take g = 10 ms^-2 and there is no loss of energy.)

Ans: Given, h = 5 m, m = 200 g = 0.2 kg, g = 10 ms^-2

(i) Potential energy U_Bat B is given by

U_B = m x g x h

Substituting the values we get,

U_B = 0.2 × 10 × 5
⇒ U_B = 10 J

Hence, the potential energy of the pendulum at the position B = 10 J

(ii) Total mechanical energy at point C = 10 J

The total mechanical energy is same at all points of the path due to conservation of mechanical energy.

(iii) At A, bob has only kinetic energy which is equal to potential energy at B,

Therefore,

1/2 x m x (v_A)² = U_B
⇒ 1/2 x 0.2 x (v_A)² = 10
⇒ (v_A)² = √100
⇒ v_A = 10 m s^-1

2019

Q1. Two bodies A and B have masses in the ratio 5 : 1 and their kinetic energies are in the ratio 125 : 9. Find the ratio of their velocities.

Ans: 5 : 3

\[Given,\\\frac{m_A}{m_B}=\frac51\;and\;\frac{KE_A}{KE_B}=\frac{125}9\\Since,\\\frac{KE_A}{KE_B}=\frac{{\displaystyle\frac12}m_Av_A^2}{\frac12m_Bv_B^2}\\\frac{KE_A}{KE_B}=\frac{m_A}{m_B}\times\frac{{(v_A)}^2}{({v_B)}^2}\\substituting\;the\;values\;we\;have,\\\frac{125}9=\frac51\times\frac{{(v_A)}^2}{({v_B)}^2}\\\Rightarrow\frac{{(v_A)}^2}{({v_B)}^2}=\frac{125}9\times\frac15\\\Rightarrow\frac{(v_A)}{({v_B)}}=\sqrt{\frac{25}9}\\\Rightarrow\frac{v_A}{v_B}=\frac53\\Hence,\;ratio\;between\;velocities\;of\;A\;and\;B\;is\;5\;:\;3\\\]

Q 2. A body of mass 10 kg is kept at a height of 5 m, it is allowed to fall and reach the ground.
(i) What is the total mechanical energy possessed by the body at a height of 2 m assuming it is a frictionless medium ?
(ii) What is the kinetic energy possessed by the body just before hitting the ground ( g = 10 m s^-2)

Ans: (i) According to law of conservation of energy, Mechanical energy is always conserved.

Therefore, Mechanical energy at any point during the free fall is equal.

M.E. = Maximum P.E.

= mgh = 10 x 10 x 5 = 500 J

(ii) Kinetic energy of the body is maximum just before hitting the ground.

According to law of conservation of energy, Maximum K.E. = Maximum P.E.

Therefore, K.E. just before hitting the ground = 500 J

2018

Q 1. (i) State and defined SI unit of power.
(ii) How is the unit horse power related to the SI unit of Power?

Ans: (i) The S.I. unit of power is watt

It is defined as the power spent when 1 joule of work is done for 1 second.

(ii) 1 horse power = 746 watt.

Q 2. State the energy change in the following cases while in use:
(i) An electric iron
(ii) A ceiling fan

Ans: (i) In an electric iron, the electrical energy changes into heat energy.

(ii) In a ceiling fan, the electrical energy changes into mechanical energy.

Q 3. (i) Why is the motion of a body moving with a constant speed around circular path said to be accelerated?
(ii) Name the unit of physical quantity obtained by the formula 2K/V² (Where K = kinetic energy, V= Linear velocity)

Ans: (i) The motion of a body moving with a constant speed around a circular path is said to be accelerated because as the direction keeps changing the velocity changes and therefore it’s an accelerated motion.

(ii) kilogram (kg)

\[\frac{2K}{V^2}=\frac{2\times{\displaystyle\frac12}\times m\times V^2}{V^2}=m\\Therefore,\;\frac{2K}{V^2}=m\\\]

Therefore, 2K/V² will have the unit of mass m which is kilogram (kg).

Q 4. (i) Derive a relation between SI and CGS unit of work.

Ans: (i) S.I. unit of work is joule and C.G.S. unit of work is erg.

Since, work = Force x displacement

1 joule = 1 newton x 1 metre

= 10⁵ dyne x 100 cm

= 10⁷dyne x cm

= 10⁷ erg

1 joule = 10⁷ erg

(ii) A force act on a body and displaces it by a distance s in a direction at an angle θ with the direction of force. What should be the value of θ to get the maximum positive work?

Ans: (ii) Work W = F.S. cos θ

For maximum work, cos θ = 1 or θ = 0°

2017

Q 1. If the power of a motor be 100 kW at what speed can it raise a load of 50,000 N?

Ans: P = 100 kW = 100 x 1000 = 100000 W

We know, Power = Force x velocity

100000 = 50000 x v

v = 100000/50000 = 2 m s^-1

2016

Q 1. A boy weighing 40 gf climbs up a stairs of 30 step each 20 cm high in 4 minutes and a girl weighing 30 gf does the same in 3 minutes. Compare:

(i) the work done by them.
(ii) the power developed by them.

Ans: (i) While climbing both the girl and boy have to do work against the force of gravity.

Force of gravity on boy = F₁ = m₁g = 40 x 10 = 400

Force of gravity on girl = F₂ = m₂g = 30 x 10 = 300

Total height climbed up

h = number of steps x height of each step

= 30 x 20 = 600 cm = 6 m

Work done by the boy = W₁ = F₁ x h = 400 x 6 = 2400 J

Work done by the girl = W₂ = F₂ x h = 300 x 6 = 1800 J

Therefore, W₁ : W₂ = 2400 : 1800 = 4 : 3

(ii) The power developed by them = ?

𝑃₁/P₂ = 𝑊₁/t₁ ÷ 𝑊₂/t₂

= 𝑊₁/𝑊₂ x t₂/t₁

= 4/3 x 3/4

=1 : 1

2015

Q 1. (a) How is work done by a force measured when the force :
(i) is in the direction of displacement.
(ii) is at an angle to the direction of displacement.

Ans: (a) (i) Force is in direction of displacement

Work done = Force × Displacement = F x S

(ii) Force is at an angle to the direction of displacement

Work done = Force × Component of displacement in the direction of force = F x S cosθ

where θ is the angle between force and displacement.

(b) State the energy changes in the following while in use :
(i) Burning of a candle.
(ii) A steam engine.

Ans: (b) (i) chemical energy to light energy

(ii) Chemical energy of coal first changes to heat energy of steam and then heat energy charges into mechanical energy.

(c) (i) A scissor is a ………… multiplier.
(ii) 1 kWh = ……….. J.

Ans: (c) (i) speed

(ii) 3.6 x 10⁶

(d) Rajan exerts a force of 150 N in pulling a cart at a constant speed of 10 m s^-1. Calculate the power exerted.

Ans: (d) Power = Force x speed

= 150 N x 10 m s^-1

= 1500 W

2014

Q 1.(i) When does a force do work?
(ii) What is the work done by the moon when it revolves around the earth?

Ans: (i) Work is said to be done only when the force applied on a body makes the body moves (i.e., there is displacement of the body).

(ii) Work done by moon is 0. This is because when it revolves around the earth, it comes back into its initial position. The displacement is 0 hence the work is 0.

Q 2. (a) Calculate the change in the Kinetic energy of a moving body if its
velocity is reduced to 1/3rd of the initial velocity.

Ans: (a) 1/9 th

\[Kinetic\;energy\;KE\;=\;\frac12\;\;mv^2\;\;\;\;\;\\Here,\;m\;is\;mass\;of\;the\;body,\;v\;is\;velocity\;of\;the\;body.\;\\Given\;that\;velocity\;is\;reduced\;to\;\frac13rd\;of\;the\;initial\;velocity,\;\\i.e.,\;V\;=\;\frac v3\\Substituting\;V\;=\;\frac v3\;in\;\;KE\;=\;\frac12\;mv^2\;\\KE\;=\;\frac12\;m{(\frac v3)}^2\;\\=\;\frac12\;m\frac v9\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\\Then,\;\;\;\frac19\left(\frac12mv^2\right)\;\\KE=\frac19\;initial\;KE\;\;\;\;\;\;\;\\Hence,\;Kinetic\;energy\;becomes\;\frac19th\;of\;its\;initial\;kinetic\;energy.\]

(b) State the energy changes in the following devices while in use:
(i) A loud speaker.
(ii) A glowing electric bulb.

Ans: (b) (i) Electric energy to Sound energy

(ii) Electric energy to light and heat energy

Ans: (c) (i) Nuclear energy is the energy that is present in the core of an atom. There is a massive amount of energy in bonds that bind atoms together.

(ii) Nuclear fission

Q 3. (a) A man having a box on his head, climbs up a slope and another man having an identical box walks the same distance on a levelled road. Who does more work against the force of gravity and why?
(b) A body is thrown vertically upwards. Its velocity keeps on decreasing. What happens to its kinetic energy as its velocity becomes zero?

Ans: (a) Man having a box on his head who climbs up a slope does more work against the force of gravity because he has more potential energy by virtue of his position i.e., height.
Work done for man who climbs up slope = F x S cosθ

Where in most cases cosθ becomes 1, so the force is in the same direction as displacement,
So, θ = 0
Therefore, the work done by man climbing up slope = F x S

Now, Work done man travel on a levelled road is zero.
As the walking is along the levelled road so he does no work against gravity.
The displacement of load is normal to the direction of the force of gravity.
Therefore, W= F x S cosθ

Where, θ = 90°

Thus, cos θ = 0
and W = 0

Hence the man having a box on his head climbing up a slope does more work because the work done by the man walking on a leveled road is zero.

(b) The kinetic energy becomes zero.

According to the law of conservation of energy the kinetic energy is converted into potential energy. Potential energy is maximum at this height.

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