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Physics MCQs for NEET — Practice Questions with Answers

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A body is released from a great height and falls freely towards the earth. Another body is released from the same height exactly one second later. The separation between the two bodies, two seconds after the release of the second body is

4.9 m

9.8 m

19.6 m

24.5 m

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Explanation

The separation between the two bodies, two seconds after the release of second body

$= \frac{1}{2} \times 9 .8 [{(3)}^{2} - {(2)}^{2}] = 24 .5 m$

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An object is projected upwards with a velocity of 100 m/s. It will strike the ground after (approximately)

10 sec

20 sec

15 sec

5 sec

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Explanation

Time of flight $= \frac{2 u}{g} = \frac{2 \times 100}{10} = 20 s e c$

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A stone dropped from the top of the tower touches the ground in 4 sec. The height of the tower is about

80 m

40 m

20 m

160 m

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Explanation

$h = \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times {(4)}^{2} = 80 m$

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A body is released from the top of a tower of height h. It takes t sec to reach the ground. Where will be the ball after time t/2 sec

At h/2 from the ground

At h/4 from the ground

Depends upon mass and volume of the body

At 3h/4 from the ground

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Explanation

Let the body after time $t / 2$ be at x from the top, then

$x = \frac{1}{2} g \frac{t^{2}}{4} = \frac{g t^{2}}{8}$ …(i)

$h = \frac{1}{2} g t^{2}$ …(ii)

Eliminate t from (i) and (ii), we get $x = \frac{h}{4}$

∴ Height of the body from the ground $= h - \frac{h}{4} = \frac{3 h}{4}$

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A body is slipping from an inclined plane of height h and length l. If the angle of inclination is θ, the time taken by the body to come from the top to the bottom of this inclined plane is

$\sqrt{\frac{2 h}{g}}$

$\sqrt{\frac{2 l}{g}}$

$\frac{1}{\sin θ} \sqrt{\frac{2 h}{g}}$

$\sin θ \sqrt{\frac{2 h}{g}}$

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Explanation

Force down the plane $= m g \sin θ$

∴ Acceleration down the plane $= g \sin θ$

Since $l = 0 + \frac{1}{2} g \sin θ t^{2}$

∴ $t^{2} = \frac{2 l}{g \sin θ} = \frac{2 h}{g \sin^{2} θ} \Rightarrow t = \frac{1}{\sin θ} \sqrt{\frac{2 h}{g}}$

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A particle is projected up with an initial velocity of $80 f t / \sec$ . The ball will be at a height of $96 f t$ from the ground after (Given $g = 32 f t / s^{2}$ )

2.0 and 3.0 sec

Only at 3.0 sec

Only at 2.0 sec

After 1 and 2 sec

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Explanation

$h = u t - \frac{1}{2} g t^{2} \Rightarrow 96 = 80 t - \frac{32}{2} t^{2}$

$\Rightarrow t^{2} - 5 t + 6 = 0 \Rightarrow t = 2$ sec or 3 sec

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A body falls from rest, its velocity at the end of first second is $(g = 32 f t / \sec^{2})$

$16 f t / \sec$

$32 f t / \sec$

$64 f t / \sec$

$24 f t / \sec$

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Explanation

$v = g \times t = 32 \times 1 = 32 f t / \sec$

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A stone thrown upward with a speed u from the top of the tower reaches the ground with a velocity 3u. The height of the tower is

$3 u^{2} / g$

$4 u^{2} / g$

$6 u^{2} / g$

$9 u^{2} / g$

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Explanation

$v^{2} = u^{2} + 2 g h \Rightarrow {(3 u)}^{2} = {(- u)}^{2} + 2 g h \Rightarrow h = \frac{4 u^{2}}{g}$

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Two stones of different masses are dropped simultaneously from the top of a building

Smaller stone hit the ground earlier

Larger stone hit the ground earlier

Both stones reach the ground simultaneously

Which of the stones reach the ground earlier depends on the composition of the stone

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Explanation

$t = \sqrt{\frac{2 h}{g}}$ and h and g are same.

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A body thrown up with an initial speed of $96 f t / \sec$ reaches the ground after $(g = 32 f t / \sec^{2})$

3 sec

6 sec

12 sec

8 sec

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Explanation

Time of flight $= \frac{2 u}{g} = \frac{2 \times 96}{32} = 6 s e c$

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