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Physics MCQs for NEET — Practice Questions with Answers

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A player caught a cricket ball of mass 150 gm moving at a rate of 20 m/s. If the catching process be completed in 0.1 s, then the force of the blow exerted by the ball on the hands of the player is

0.3 N

30 N

300 N

3000 N

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Explanation

Force exerted by the ball on hands of the player

$= \frac{m d v}{d t} = \frac{0 .15 \times 20}{0 .1} = 30 N$

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A rocket has a mass of 100 kg. 90% of this is fuel. It ejects fuel vapours at the rate of 1 kg/sec with a velocity of 500 m/sec relative to the rocket. It is supposed that the rocket is outside the gravitational field. The initial upthrust on the rocket when it just starts moving upwards is

Zero

500 N

1000 N

2000 N

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Explanation

$F = u (\frac{d m}{d t}) = 500 \times 1 = 500 N$

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In which of the following cases forces may not be required to keep the

Particle going in a circle

Particle going along a straight line

The momentum of the particle constant

Acceleration of the particle constant

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Explanation

If momentum remains constant then force will be zero because $F = \frac{d P}{d t}$

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A ball of mass 150g starts moving with an acceleration of 20 m/s². When hit by a force, which acts on it for 0.1 sec. The impulsive force is

0.5 N-s

0.1 N-s

0.3 N-s

1.2 N-s

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Explanation

Impulse = Force × time = m a t

$= 0 .15 \times 20 \times 0 .1 = 0 .3 N - s$

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A body, whose momentum is constant, must have constant

Force

Velocity

Acceleration

All of these

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Explanation

For a given mass $P \propto v$ . If the momentum is constant then it's velocity must have constant.

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Rocket engines lift a rocket from the earth surface because hot gas with high velocity

Push against the earth

Push against the air

React against the rocket and push it up

Heat up the air which lifts the rocket

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Explanation

It works on the principle of conservation of momentum.

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A force of 50 dynes is acted on a body of mass 5 g which is at rest for an interval of 3 seconds, then impulse is

$0 .15 \times 10^{- 3} N s$

$0 .98 \times 10^{- 3} N s$

$1 .5 \times 10^{- 3} N s$

$2 .5 \times 10^{- 3} N s$

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Explanation

Impulse, $I = F \times Δ t = 50 \times 10^{- 5} \times 3$ = $1 .5 \times 10^{- 3} N - s$

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A bullet of mass 0.1 kg is fired with a speed of 100 m/sec, the mass of gun is 50 kg. The velocity of recoil is

0.2 m/sec

0.1 m/sec

0.5 m/sec

0.05 m/sec

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Explanation

According to principle of conservation of linear momentum, $m_{G} v_{G} = m_{B} v_{B}$

$\Rightarrow v_{G} = \frac{m_{B} v_{B}}{m_{G}} = \frac{0 .1 \times 10^{2}}{50} = 0 .2 m / s$

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Two forces of magnitude F have a resultant of the same magnitude F. The angle between the two forces is

45°

120°

150°

60°

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Explanation

$F = \sqrt{{(F)}^{2} + {(F)}^{2} + 2 F . F \cos θ}$ ⇒ $θ = 120 °$

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Two forces with equal magnitudes F act on a body and the magnitude of the resultant force is F/3. The angle between the two forces is

$\cos^{- 1} (- \frac{17}{18})$

$\cos^{- 1} (- \frac{1}{3})$

$\cos^{- 1} (\frac{2}{3})$

$\cos^{- 1} (\frac{8}{9})$

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Explanation

$F_{_{n e t}}^{2} = F_{1}^{2} + F_{2}^{2} + 2 F_{1} F_{2} \cos θ$

⇒ ${(\frac{F}{3})}^{2} = F^{2} + F^{2} + 2 F^{2} \cos θ$

⇒ $\cos θ = (- \frac{17}{18})$

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