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

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The average power required to lift a 100 kg mass through a height of 50 metres in approximately 50 seconds would be

50 J/s

5000 J/s

100 J/s

980 J/s

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Explanation

$P = \frac{m g h}{t}$ = $\frac{100 \times 9 .8 \times 50}{50} = 980 J / s$

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The power of a pump, which can pump 200kg of water to a height of 200m in 10sec is (g = 10 m/s²)

40 kW

80 kW

400 kW

960 kW

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Explanation

$p = \frac{m g h}{t} = \frac{200 \times 10 \times 200}{10} = 40 k W$

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A 60 kg man runs up a staircase in 12 seconds while a 50 kg man runs up the same staircase in 11, seconds, the ratio of the rate of doing their work is

6 : 5

12 : 11

11 : 10

10 : 11

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Explanation

$P = \frac{m g h}{t}$

⇒ $\frac{P_{1}}{P_{2}} = \frac{m_{1}}{m_{2}} \times \frac{t_{2}}{t_{1}}$ (As h = constant)

∴ $\frac{P_{1}}{P_{2}} = \frac{60}{50} \times \frac{11}{12} = \frac{11}{10}$

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What average horsepower is developed by an 80 kg man while climbing in 10 s a flight of stairs that rises 6 m vertically

0.63 HP

1.26 HP

1.8 HP

2.1 HP

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Explanation

$p = \frac{m g h}{t} = \frac{80 \times 9 .8 \times 6}{10} W = \frac{470}{746} H P = 0 .63 H P$

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A quarter horse power motor runs at a speed of 600 r.p.m. Assuming 40% efficiency, the work done by the motor in one rotation will be

7.46 J

7400 J

7.46 ergs

74.6 J

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Explanation

Motor makes 600 revolution per minute

∴ n = $600 \frac{revolution}{minute} = 10 \frac{r e v}{\sec}$

∴ Time required for one revolution $= \frac{1}{10}$ sec

Energy required for one revolution = power × time

= $\frac{1}{4} \times 746 \times \frac{1}{10} = \frac{746}{40} J$

But work done = 40% of input

$= 40 % \times \frac{746}{40} = \frac{40}{100} \times \frac{746}{40} = 7 .46 J$

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An engine pumps up 100 kg of water through a height of 10 m in 5 s. Given that the efficiency of the engine is 60% . If g = 10 ms^–2, the power of the engine is

3.3 kW

0.33 kW

0.033 kW

33 kW

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Explanation

Work output of engine = mgh = $100 \times 10 \times 10 = 10^{4} J$

Efficiency ( $η$ ) = $\frac{output}{input}$

∴ Input energy = $\frac{outupt}{η}$

$= \frac{10^{4}}{60} \times 100 = \frac{10^{5}}{6} J$

∴ Power = $\frac{input energy}{time}$ = $\frac{10^{5} / 6}{5} = \frac{10^{5}}{30} = 3 .3 k W$

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A force of $2 \hat{i} + 3 \hat{j} + 4 \hat{k} N$ acts on a body for 4 second, produces a displacement of $(3 \hat{i} + 4 \hat{j} + 5 \hat{k}) m .$ The power used is-

9.5 W

7.5 W

6.5 W

4.5 W

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Explanation

$P = \frac{\vec{F} . \vec{s}}{t} = \frac{(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) . (3 \hat{i} + 4 \hat{j} + 5 \hat{k})}{4} = \frac{38}{4} = 9 .5 W$

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An engine pump is used to pump a liquid of density ρ continuously through a pipe of cross-sectional area A. If the speed of flow of the liquid in the pipe is v, then the rate at which kinetic energy is being imparted to the liquid is

$\frac{1}{2} A ρ v^{3}$

$\frac{1}{2} A ρ v^{2}$

$\frac{1}{2} A ρ v$

$A ρ v$

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Explanation

Energy supplied to liquid per second by the pump

= $\frac{1}{2} \frac{m v^{2}}{t}$ = $\frac{1}{2} \frac{V ρ v^{2}}{t}$ = $\frac{1}{2} A \times (\frac{l}{t}) \times ρ \times v^{2}$ $[\frac{l}{t} = v]$

$= \frac{1}{2} A \times v \times ρ \times v^{2}$ = $\frac{1}{2} A ρ v^{3}$

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A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is

MgL

MgL/3

MgL/9

MgL/18

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Explanation

The hanging part of the chain has a length L/3 and mass M/3. The work required to lift it vertically through a height L/3 is (M/3)g(L/3) = MgL/9. However, since the chain is pulled horizontally, the work done is half of this value, which is MgL/18.

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A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to –K/r², where K is a constant. The total energy of the particle is

$\frac{K}{2 r}$

$- \frac{K}{2 r}$

$- \frac{K}{r}$

$\frac{K}{r}$

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Explanation

Here $\frac{m v^{2}}{r} = \frac{K}{r^{2}}$

∴ K.E. $= \frac{1}{2} m v^{2} = \frac{K}{2 r}$

$U = - \int_{\infty}^{r} F . d r = - \int_{\infty}^{r} (- \frac{K}{r^{2}}) d r = - \frac{K}{r}$

Total energy $E = K.E. + P.E. = \frac{K}{2 r} - \frac{K}{r} = - \frac{K}{2 r}$

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