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Laws of Motion MCQs for NEET — Physics Questions with Answers

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An object with a mass 10 kg moves at a constant velocity of 10 m/sec. A constant force then acts for 4 second on the object and gives it a speed of 2 m/sec in opposite direction. The acceleration produced in it, is

3 m/sec²

–3 m/sec²

0.3 m/sec²

–0.3 m/sec²

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Explanation

$\vec{a} = \frac{\vec{v_{2}} - \vec{v_{1}}}{t} = \frac{(- 2) - (+ 10)}{4} = \frac{- 12}{4} = - 3 m / s^{2}$

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The force acting on the object of mass 1 kg moving with acceleration $3 m / s^{2}$ is -

30 N

– 30 N

3 N

– 3 N

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Explanation

$F = m a = 1 \times (3) = 3 N$

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A machine gun is mounted on a 2000 kg car on a horizontal frictionless surface. At some instant the gun fires bullets of mass 10 gm with a velocity of 500 m/sec with respect to the car. The number of bullets fired per second is ten. The average thrust on the system is

550 N

50 N

250 N

250 dyne

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Explanation

u = velocity of bullet

$\frac{d m}{d t} =$ Mass thrown per second by the machine gun

= Mass of bullet × Number of bullet fired per second

$= 10 g \times 10 b u l l e t / \sec = 100 g / \sec = 0 .1 k g / \sec$

∴ Thrust $= \frac{u d m}{d t} = 500 \times 0 .1 = 50 N$

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A particle of mass 0.3 kg is subjected to a force F = –kx with k = 15 N/m. What will be its initial acceleration if it is released from a point 20 cm away from the origin

5 m/s²

10 m/s²

3 m/s²

15 m/s²

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Explanation

Force on particle at 20 cm away $F = k x$

F = 15 × 0.2 = 3 N $[As k = 15 N / m]$

∴ Acceleration = $\frac{Force}{Mass} = \frac{3}{0 .3} = 10 m / s^{2}$

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An elevator weighing 6000 kg is pulled upward by a cable with an acceleration of 5 ms^–2. Taking g to be 10 ms^–2, then the tension in the cable is

6000 N

9000 N

60000 N

90000 N

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Explanation

$T = m (g + a) = 6000 (10 + 5) = 90000 N$

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The ratio of the weight of a man in a stationary lift and when it is moving downward with uniform acceleration ‘a’ is 3 : 2. The value of ‘a’ is (g-Acceleration due to gravity of the earth)

$\frac{3}{2} g$

$\frac{g}{3}$

$\frac{2}{3} g$

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If force on a rocket having exhaust velocity of 300 m/sec is 210 N, then rate of combustion of the fuel is

0.7 kg/s

1.4 kg/s

0.07 kg/s

10.7 kg/s

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Explanation

$F = u (\frac{d m}{d t})$ ⇒ $\frac{d m}{d t} = \frac{F}{u} = \frac{210}{300} = 0 .7 k g / s$

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In an elevator moving vertically up with an acceleration g, the force exerted on the floor by a passenger of mass M is

$\frac{1}{2} M g$

Zero

2 Mg

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Explanation

$R = m (g + a) = m (g + g) = 2 m g$

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A 5000 kg rocket is set for vertical firing. The exhaust speed is 800 ms^–1. To give an initial upward acceleration of 20 ms^–2, the amount of gas ejected per second to supply the needed thrust will be (g = 10 ms^–2)

127.5 kg s^–1

187.5 kg s^–1

185.5 kg s^–1

137.5 kg s^–1

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Explanation

$F = \frac{u d m}{d t} = m (g + a)$

⇒ $\frac{d m}{d t} = \frac{m (g + a)}{u} = \frac{5000 \times (10 + 20)}{800} = 187 .5 k g / s$

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If a person with a spring balance and a body hanging from it goes up and up in an aeroplane, then the reading of the weight of the body as indicated by the spring balance will

Go on increasing

Go on decreasing

First increase and then decrease

Remain the same

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Explanation

Initially due to upward acceleration apparent weight of the body increases but then it decreases due to decrease in gravity.

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