Blog Log in Get Started Free

Work Energy And Power MCQs for NEET — Physics Questions with Answers

Practice free Work Energy And Power (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

Which one of the following is not a conservative force

Gravitational force

Electrostatic force between two charges

Magnetic force between two magnetic dipoles

Frictional force

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Friction is a non-conservative force.

● ● ● ● ●

Work done in raising a box depends on

How fast it is raised

The strength of the man

The height by which it is raised

None of the above

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Work in raising a box

= (weight of the box) × (height by which it is raised)

● ● ● ● ●

If the water falls from a dam into a turbine wheel 19.6 m below, then the velocity of water at the turbine is (g = 9.8 m/s²)

9.8 m/s

19.6 m/s

39.2 m/s

98.0 m/s

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Potential energy of water = kinetic energy at turbine

$m g h = \frac{1}{2} m v^{2}$

⇒ $v = \sqrt{2 g h} = \sqrt{2 \times 9 .8 \times 19 .6} = 19 .6 m / s$

● ● ● ● ●

Two identical cylindrical vessels with their bases at the same level each contain a liquid of density; $ρ$ . The height of the liquid in one vessel is h₁ and that in the other vessel is h₂ $(h_{1} > h_{2})$ .The area of either base is A. The work done by gravity in equalizing the levels when the two vessels are connected is :

$(h_{1} - h_{2}) g ρ$

$(h_{1} - h_{2}) g A ρ$

$\frac{1}{2} {(h_{1} - h_{2})}^{2} g A ρ$

$\frac{1}{4} {(h_{1} - h_{2})}^{2} g A ρ$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

If the increase in the kinetic energy of a body is 22%, then the increase in the momentum will be

22%

44%

10%

300%

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$P = \sqrt{2 m E} .$ If m is constant then

$\frac{P_{2}}{P_{1}} = \sqrt{\frac{E_{2}}{E_{1}}} = \sqrt{\frac{1 .22 E}{E}}$ ⇒ $\frac{P_{2}}{P_{1}} = \sqrt{1 .22} = 1 .1$

⇒ $P_{2} = 1 .1 P_{1}$ ⇒ $P_{2} = P_{1} + 0 .1 P_{1} = P_{1} + 10 % of P_{1}$

So the momentum will increase by 10%

● ● ● ● ●

If a body of mass 200 g falls from a height 200 m and its total P.E. is converted into K.E. at the point of contact of the body with earth surface, then what is the decrease in P.E. of the body at the contact (g = 10 m/s²)

200 J

400 J

600 J

900 J

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$Δ U = m g h = 0 .2 \times 10 \times 200$ = 400 J

∴ Gain in K.E. = decrease in P.E. = 400 J.

● ● ● ● ●

If momentum is increased by 20%, then K.E. increases by

44%

55%

66%

77%

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$E = \frac{P^{2}}{2 m} .$ If m is constant then $E \propto P^{2}$

⇒ $\frac{E_{2}}{E_{1}} = {(\frac{P_{2}}{P_{1}})}^{2} = {(\frac{1 .2 P}{P})}^{2} = 1 .44$

⇒ $E_{2} = 1 .44 E_{1} = E_{1} + 0 .44 E_{1}$

$E_{2} = E_{1} + 44 %$ of $E_{1}$

i.e. the kinetic energy will increase by 44%

● ● ● ● ●

The kinetic energy of a body of mass 2 kg and momentum of 2 Ns is

1 J

2 J

3 J

4 J

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$E = \frac{P^{2}}{2 m} = \frac{{(2)}^{2}}{2 \times 2} = 1 J$

● ● ● ● ●

The decrease in the potential energy of a ball of mass 20 kg which falls from a height of 50 cm is

968 J

98 J

1980 J

None of these

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$Δ U = m g h = 20 \times 9 .8 \times 0 .5 = 98 J$

● ● ● ● ●

A 0.5 kg ball is thrown up with an initial speed 14 m/s and reaches a maximum height of 8.0m. How much energy is dissipated by air drag acting on the ball during the ascent

19.6 Joule

4.9 Joule

10 Joule

9.8 Joule

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

If there is no air drag then maximum height

$H = \frac{u^{2}}{2 g} = \frac{14 \times 14}{2 \times 9 .8} = 10 m$

But due to air drag ball reaches up to height 8m only. So loss in energy

$= m g (10 - 8) = 0 .5 \times 9 .8 \times 2 = 9 .8 J$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Work Energy And Power question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.