UP Board Class 9 Science Chapter 10 Work, Energy And Power Multiple Choice Questions

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Work, Energy And Power Multiple Choice Question And Answers

Question 1. When a horse-pulls a cart, who does the work ?

  1. Cart
  2. Wheels
  3. Road
  4. Horse

Answer: 4. Horse

When a horse-pulls a cart, work is being done by horse.

Question 2. If 10 N of force is applied to an object, but the object does not move, then how much work being done by the force ?

  1. Zero
  2. 10 J
  3. 10N
  4. 20 J

Answer: 1. Zero

If there is no displacement due to application of force, then net work done will be zero.

Question 3. A 4 N of force displaces a body by 2m, the work done will be

  1. 5 J
  2. 8 J
  3. 8 N
  4. 5 N

Answer: 2. 8 J

Work done by given force =force x displacement = 4 x 2 =8J

Question 4. A car is moving uniformly in a circular racing track of radius 1 km. If the car makes two complete round. How much work is done by the car, if car has a force of 20 kN?

  1. Zero
  2. 20 kJ
  3. 200 kJ
  4. 2000 kJ

Answer: 1. Zero

During a circular motion, force is perpendicular to the displacement, so net work done is zero.

Question 5. What kind of energy is possessed by a running horse?

  1. Kinetic energy
  2. Potential energy
  3. No energy
  4. Heat energy

Answer: 1. Kinetic energy

A running horse has kinetic energy.

UP Board Class 9 Science Chapter 10 Work, Energy And Power Multiple Choice Questions

Question 6. If two stones A and B are dropped from a tower, then which one has maximum kinetic energy ?

  1. Lighter stone
  2. Heavier stone
  3. Both have equal
  4. None of the above

Answer: 2. Heavier stone

If two stones are dropped from some height, then the heavier stone has greater kinetic energy w.r.t. lighter one.

Question 7. A monkey weighing 50 kg climbs up a vertical tree of height 300 m. How much potential energy does it gain? [Take, g=9.8 m / \(s^2\) ]

  1. 147 \(\times 10^4 \mathrm{~J}\)
  2. 14.7 \(\times 10^3 \mathrm{~J}\)
  3. 14.7 \(\times 10^2 \mathrm{~J}\)
  4. 14.7 \(\times 10^4 \mathrm{~J}\)

Answer: 4. 14.7 \(\times 10^4 \mathrm{~J}\)

Given, m=50 kg, g=9.8 \(ms^{-2}\), b=300 m

Work done by the body = mgh

=50 \(\times 9.8300=14.7 \times 10^4 \mathrm{~J}\)

Gain in PE = Work done =14.7 \(\times 10^4\) J

Question 8. A scooter of mass 150 kg is travelling at 10 m/s. If its speed increases to 15 m/s, then by how much amount does its kinetic energy increases ?

  1. 9500 J
  2. 9425 J
  3. 9375 J
  4. 9753 J

Answer: 3. 9375 J

Given, m=150 kg, \(v_1\)=10 m / s and \(v_2\)=15 m/ s

The change in kinetic energy

=\(\mathrm{KE}_2-\mathrm{KE}_1=\frac{1}{2} m v_2^2-\frac{1}{2} m v_1^2\)

=\(\frac{1}{2} \times 150 \times 15^2-\frac{1}{2} \times 150 \times 10^2\)

= 16875-7500= 9375 J

Question 9. A ball is allowed to fall freely from a tower. Which energy is possessed at the middle point during the fall ?

  1. Kinetic only
  2. Potential only
  3. Both (1) and (2)
  4. Heat only

Answer: 3. Both (1) and (2)

At middle point of fall, the ball has both kinetic and potential energy.

Question 10. In head phone, the electrical energy is converted to what ?

  1. Light energy
  2. Sound energy
  3. Chemical energy
  4. Heat energy

Answer: 2. Sound energy

Electrical energy changes into sound energy.

Question 11. Which mathematical relation of energy of a stone of mass m fading freely from height h remains conserved at every point in its downward motion?

  1. E = mgh
  2. E=\(\frac{1}{2} m v^2\)
  3. E=\(m g h+\frac{1}{2} m v^2\)
  4. None of these

Answer: 3. E=\(m g h+\frac{1}{2} m v^2\)

Total energy ot the free hilling stone is conserved.

So, total energy = kinetic energy + potential energy.

Or E=\(m g+\frac{1}{2} m v^2[latex]

Question 12. If a stone of mass 5 kg drops from height 20 m, then what will be the velocity at the surface of earth ? (hike, g = 10 m/s²)

  1. 20 m/s
  2. 30 m/s
  3. 35 m/s
  4. 40 m/s

Answer: 1. 20 m/s

Given, mass of stone, m = 5 kg

Height, h = 20 m and g =10 m/s²

So, potendal energy at highest point = mgh

= 5 x 10 x 20 = 1000 J

Now, according to conservation of energy,

kinetic energy at surface of the earth = potential energy at height h

So, 1/2 m [latex]v^2\) =\(m g h^h\)

⇒ \(v^2 =\frac{2 m g^b}{m}\)

v =\(\sqrt{2 g^b}=\sqrt{2 \times 10 \times 20}=20 \mathrm{~m} / \mathrm{s}\)

Question 13. Mukesh drops a ball of 200 g from a tower of height 20 m. What will be its kinetic energy at the height of 5 m ?

  1. 30 J
  2. 50 J
  3. 60 J
  4. 65 J

Answer: 1. 30 J

Given, m = 200 g = 0.2 kg, height, h = 20 m

When ball reaches at height of 5 m, its velocity is given by

⇒ \(v^2=v^2+2 g\)

⇒ \(v^2=0^2+2 \times 10 \times 15\)

v=\(\sqrt{300}=\sqrt{100 \times 3}=10 \sqrt{3}\)

So, kinetic energy \(\left.=\frac{1}{2} m v^2=\frac{1}{2} \times 0.2 \times 300=30\right]\)

Question 14. An object of mass 50 kg is raised to a height of 7 m above the ground. What is its potential energy? If object is allowed to fall, then its kinetic energy when it is half way down.

  1. 3500 J, 1650 J
  2. 3800 J, 1750 J
  3. 3800 J, 1650 J
  4. 3500 J, 1750 J

Answer: 4. 3500 J, 1750 J

Given, mass, m = 50 kg, height, h = 7 m

As, potential energy is given by PR = mgh =50x 10x 7 = 3500J

When it is allowed to fall its potential energy gets converted into kinetic energy.

So, when it reaches the half-way, half of its potential energy gets converted to kinetic energy.

Question 15. A horse does 5000 J of work in 100 s. What is its power ?

  1. 50 W
  2. 50 J
  3. 10W
  4. 10J

Answer: 1. 50 W

Given, work done (W) = 5000 J

Time taken (r) =100 s

We know that,

Power,P=\(\frac{W}{t}=\frac{5000}{100}=50 \mathrm{~J} / \mathrm{s}\)

Thus, power, P=50 W

Question 16. The power uses of household water pump is 200 W. What is the mass of water, the pump can lift per minute from a 7 m deep tank? [Take, g =10 m/s²]

  1. 172 kg
  2. 272 kg
  3. 150 kg
  4. 250 kg

Answer: 1. 172 kg

Given,power,.P = 200 W^ = 60 s,height,h = 7m

Energy supplied to the pump = Power x Time = 200 Wx 60s =12000J

So, energy =mgh

12000 = m x 10 x 7

m = 172 kg

So, mass of the water is 172 kg.

Question 17. A certain household has consumed 200 units of energy during a month. How much energy is then in joules?

  1. 72 \(\times 10^5 \mathrm{~J}\)
  2. 72 \(\times 10^7 \mathrm{~J}\)
  3. 7.2 \(\times 10^8 \mathrm{~J}\)
  4. 72 \(\times 10^{\mathrm{B}} \mathrm{J}\)

Answer: 3. 7.2 \(\times 10^8 \mathrm{~J}\)

Given, energy consumed = 200 units

= 200 kWh = 200x 1000×3600

= 7.2 x 108 J

Question 18. Ankit does 500 J of work in 15 min and Bablu does 700 J of work in 30 min. Who expends more power, Ankit or Bablu ?

  1. Ankit
  2. Bablu
  3. Both same
  4. Insufficient information

Answer: 1. Ankit

Power expended by Ankit =\(\frac{W}{t}=\frac{500}{15 \times 60}\)=0.55J

and power expended by Bablu =\(\frac{700}{30 \times 60}\)=0.38 J

Ankit expends more power.

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