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Physics MCQs for NEET — Practice Questions with Answers

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N bullets each of mass m kg are fired with a velocity v ms^–1 at the rate of n bullets per second upon a wall. The reaction offered by the wall to the bullets is given by

nmv

$\frac{N m v}{n}$

$n \frac{N m}{v}$

$n \frac{N v}{m}$

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Explanation

Total mass of bullets = Nm, time $t = \frac{N}{n}$

Momentum of the bullets striking the wall = Nmv

Rate of change of momentum (Force) = $\frac{N m v}{t}$ = nmv.

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With what minimum acceleration can a fireman slides down a rope while breaking strength of the rope is $\frac{2}{3}$ of his weight

$\frac{2}{3} g$

$\frac{1}{3} g$

Zero

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Explanation

If man slides down with some acceleration then its apparent weight decreases. For critical condition rope can bear only 2/3 of his weight. If a is the minimum acceleration then,

Tension in the rope $= m (g - a)$ = Breaking strength

⇒ $m (g - a) = \frac{2}{3} m g$ ⇒ $a = g - \frac{2 g}{3} = \frac{g}{3}$

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A ball of mass m moves with speed v and it strikes normally with a wall and reflected back normally, if its time of contact with wall is t then find force exerted by ball on wall

$\frac{2 m v}{t}$

$\frac{m v}{t}$

mvt

$\frac{m v}{2 t}$

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Explanation

For exerted by ball on wall

= rate of change in momentum of ball

= $\frac{m v - (- m v)}{t} = \frac{2 m v}{t}$

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A body of mass 5 kg starts from the origin with an initial velocity $\vec{u} = 30 \hat{i} + 40 \hat{j} m s^{- 1}$ . If a constant force $\vec{F} = - (\hat{i} + 5 \hat{j}) N$ acts on the body, the time in which the y–component of the velocity becomes zero is

5 seconds

20 seconds

40 seconds

80 seconds

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Explanation

$u_{y} = 40 m / s$ , $F_{y} = - 5 N$ , $m = 5 k g$ .

So $a_{y} = \frac{F_{y}}{m} = - 1 m / s^{2}$ (As v = u + at)

∴ $v_{y} = 40 - 1 \times t = 0$ ⇒ $t = 40 \sec$ .

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A ball of mass 0.5 kg moving with a velocity of 2 m/sec strikes a wall normally and bounces back with the same speed. If the time of contact between the ball and the wall is one millisecond, the average force exerted by the wall on the ball is

2000 N

1000 N

5000 N

125 N

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Explanation

$F_{a v} = \frac{Δ p}{Δ t} = \frac{m v - (- m v)}{Δ t} = \frac{2 m v}{Δ t} = \frac{2 \times 0 .5 \times 2}{10^{- 3}}$ = 2000 N

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A particle moves in the xy-plane under the action of a force F such that the components of its linear momentum p at any time t are $p_{x} = 2 \cos t$ , $p_{y} = 2 \sin t$ . The angle between F and p at time t is

90°

0°

180°

30°

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Explanation

Given that $\vec{p} = p_{x} \hat{i} + p_{y} \hat{j} = 2 \cos t \hat{i} + 2 \sin t \hat{j}$

∴ $\vec{F} = \frac{d \vec{p}}{d t} = - 2 \sin t \hat{i} + 2 \cos t \hat{j}$

Now, $\vec{F} . \vec{p} = 0$ i.e. angle between $\vec{F} and \vec{p}$ is 90°.

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A satellite in force-free space sweeps stationary interplanetary dust at a rate $d M / d t = α v$ where M is the mass, v is the velocity of the satellite and $α$ is a constant. What is the deacceleration of the satellite

$- 2 α v^{2} / M$

$- α v^{2} / M$

$+ α v^{2} / M$

$- α v^{2}$

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Explanation

$F = \frac{d p}{d t} = v (\frac{d M}{d t}) = α v^{2}$

∴ $a = \frac{F}{M} = \frac{α v^{2}}{M}$

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Swimming is possible on account of

First law of motion

Second law of motion

Third law of motion

Newton's law of gravitation

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Explanation

Swimming is a result of pushing water in the opposite direction of the motion.

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On a stationary sail-boat, air is blown at the sails from a fan attached to the boat. The boat will

Remain stationary

Spin around

Move in a direction opposite to that in which air is blown

Move in the direction in which the air is blown

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Explanation

The force exerted by the air of fan on the boat is internal and for motion external force is required.

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A man is standing at a spring platform. Reading of spring balance is 60 kg wt. If man jumps outside platform, then reading of spring balance

First increases then decreases to zero

Decreases

Increases

Remains same

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Explanation

For jumping he presses the spring platform, so the reading of spring balance increases first and finally it becomes zero.

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