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Physics MCQs for NEET — Practice Questions with Answers

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P, Q and R are three balloons ascending with velocities U, 4U and 8U respectively. If stones of the same mass be dropped from each, when they are at the same height, then

They reach the ground at the same time

Stone from P reaches the ground first

Stone from R reaches the ground first

Stone from Q reaches the ground first

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Explanation

It has lesser initial upward velocity.

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A body is projected up with a speed ‘u’ and the time taken by it is T to reach the maximum height H. Pick out the correct statement

It reaches H/2 in T/2 sec

It acquires velocity u/2 in T/2

Its velocity is u/2 at H/2

Same velocity at 2T

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Explanation

At maximum height velocity $v = 0$

We know that $v = u + a t$ , hence

$0 = u - g T \Rightarrow u = g T$

When $v = \frac{u}{2}$ , then

$\frac{u}{2} = u - g t \Rightarrow g t = \frac{u}{2} \Rightarrow g t = \frac{g T}{2} \Rightarrow t = \frac{T}{2}$

Hence at $t = \frac{T}{2}$ , it acquires velocity $\frac{u}{2}$

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A body falling for 2 seconds covers a distance S equal to that covered in next second. Taking $g = 10 m / s^{2}, S =$

30 m

10 m

60 m

20 m

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Explanation

If u is the initial velocity then distance covered by it in 2 sec

$S = u t + \frac{1}{2} a t^{2} = u \times 2 + \frac{1}{2} \times 10 \times 4 = 2 u + 20$ …(i)

Now distance covered by it in 3^rdsec

$S_{3^{r d}} = u + \frac{g}{2} (2 \times 3 - 1) 10 = u + 25$ …(ii)

From(i) and (ii), $2 u + 20 = u + 25 \Rightarrow u = 5$

∴ $S = 2 \times 5 + 20 = 30 m$

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A body dropped from a height h with an initial speed zero, strikes the ground with a velocity $3 k m / h$ . Another body of same mass is dropped from the same height h with an initial speed $- u^{'} = 4 k m / h$ . Find the final velocity of second body with which it strikes the ground

3 km/h

4 km/h

5 km/h

12 km/h

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Explanation

For first case $v^{2} - 0^{2} = 2 g h$ ⇒ ${(3)}^{2} = 2 g h$

For second case $v^{2} = {(- u)}^{2} + 2 g h$ $= 4^{2} + 3^{2}$

∴ v = 5 km/h

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A ball of mass m₁ and another ball of mass m₂ are dropped from equal height. If time taken by the balls are t₁ and t₂ respectively, then

$t_{1} = \frac{t_{2}}{2}$

$t_{1} = t_{2}$

$t_{1} = 4 t_{2}$

$t_{1} = \frac{t_{2}}{4}$

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Explanation

The time of fall is independent of the mass.

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With what velocity a ball be projected vertically so that the distance covered by it in 5^th second is twice the distance it covers in its 6^th second $(g = 10 m / s^{2})$

58.8 m/s

49 m/s

65 m/s

19.6 m/s

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Explanation

$h_{n^{t h}} = u - \frac{g}{2} (2 n - 1)$

$h_{5^{t h}} = u - \frac{10}{2} (2 \times 5 - 1) = u - 45$

$h_{6^{t h}} = u - \frac{10}{2} (2 \times 6 - 1) = u - 55$

Given $h_{5^{t h}} = 2 \times h_{6^{t h}}$ . By solving we get $u = 65 m / s$

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A body sliding on a smooth inclined plane requires 4 seconds to reach the bottom starting from rest at the top. How much time does it take to cover one-fourth distance starting from rest at the top

1 s

2 s

4 s

16 s

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Explanation

$S = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} a t^{2}$

Hence $t \propto \sqrt{S}$ i.e., if S becomes one-fourth then t will become half i.e., 2 sec

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A ball is dropped downwards. After 1 second another ball is dropped downwards from the same point. What is the distance between them after 3 seconds

25 m

20 m

50 m

9.8 m

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Explanation

Distance between the balls = Distance travelled by first ball in 3 seconds – Distance travelled by second ball in 2 seconds

= $\frac{1}{2} g {(3)}^{2} - \frac{1}{2} g {(2)}^{2} = 45 - 20 = 25 m$

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A stone is thrown with an initial speed of 4.9 m/s from a bridge in vertically upward direction. It falls down in water after 2 sec. The height of the bridge is

4.9 m

9.8 m

19.8 m

24.7 m

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Explanation

Speed of stone in a vertically upward direction is 4.9 m/s. So for vertical downward motion we will consider $u = - 4 .9 m / s$

$h = u t + \frac{1}{2} g t^{2} = - 4 .9 \times 2 + \frac{1}{2} \times 9 .8 \times {(2)}^{2} = 9 .8 m$

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A stone is shot straight upward with a speed of 20 m/sec from a tower 200 m high. The speed with which it strikes the ground is approximately

60 m/sec

65 m/sec

70 m/sec

75 m/sec

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Explanation

Speed of stone in a vertically upward direction is 20m/s. So for vertical downward motion we will consider $u = - 20 m / s$

$v^{2} = u^{2} + 2 g h = {(- 20)}^{2} + 2 \times 9 .8 \times 200 = 4320 m / s$

∴ $v ≃ 65 m / s$ .

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