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Physics NEET Questions — Free MCQ Practice

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A stone is dropped from a certain height which can reach the ground in 5 second. If the stone is stopped after 3 second of its fall and then allowed to fall again, then the time taken by the stone to reach the ground for the remaining distance is

A.

2 sec

B.

3 sec

C.

4 sec

D.

None of these

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Explanation

Total distance $= \frac{1}{2} g t^{2} = \frac{25}{2} g$

Distance moved in 3 sec $= \frac{9}{2} g$

Remaining distance $= \frac{16}{2} g$

If t is the time taken by the stone to reach the ground for the remaining distance then

$\Rightarrow \frac{16}{2} g = \frac{1}{2} g t^{2} \Rightarrow t = 4 \sec$

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A man in a balloon rising vertically with an acceleration of $4 .9 m / \sec^{2}$ releases a ball 2 sec after the balloon is let go from the ground. The greatest height above the ground reached by the ball is $(g = 9 .8 m / \sec^{2})$

A.

14.7 m

B.

19.6 m

C.

9.8 m

D.

24.5 m

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Explanation

Height travelled by ball (with balloon) in 2 sec

$h_{1} = \frac{1}{2} a t^{2} = \frac{1}{2} \times 4 .9 \times 2^{2} = 9 .8 m$

Velocity of the balloon after 2 sec

$v = a t = 4 .9 \times 2 = 9 .8 m / s$

Now if the ball is released from the balloon then it acquire same velocity in upward direction.

Let it move up to maximum height $h_{2}$

$v^{2} = u^{2} - 2 g h_{2}$ ⇒ $0 = {(9 .8)}^{2} - 2 \times (9 .8) \times h_{2}$

∴ $h_{2}$ = 4.9m

Greatest height above the ground reached by the ball $= h_{1} + h_{2} = 9 .8 + 4 .9 = 14 .7 m$

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A balloon is at a height of 81 m and is ascending upwards with a velocity of 12 m/s. A body of 2 kg weight is dropped from it. If $g = 10 m / s^{2}$ , the body will reach the surface of the earth in

A.

1.5 s

B.

4.025 s

C.

5.4 s

D.

6.75 s

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Explanation

$h = u t + \frac{1}{2} g t^{2} - 81 = 12 t - \frac{1}{2} \times 10 \times t^{2} \Rightarrow t = 5 .4 s e c$

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An aeroplane is moving with a velocity u. It drops a packet from a height h. The time t taken by the packet in reaching the ground will be

A.

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

B.

$\sqrt{(\frac{2 u}{g})}$

C.

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

D.

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

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Explanation

The initial velocity of aeroplane is horizontal, then the vertical component of velocity of packet will be zero.

So $t = \sqrt{\frac{2 h}{g}}$

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Water drops fall at regular intervals from a tap which is 5 m above the ground. The third drop is leaving the tap at the instant the first drop touches the ground. How far above the ground is the second drop at that instant

A.

2.50 m

B.

3.75 m

C.

4.00 m

D.

1.25 m

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Explanation

Time taken by first drop to reach the ground $t = \sqrt{\frac{2 h}{g}}$

$\Rightarrow t = \sqrt{\frac{2 \times 5}{10}} = 1 s e c$

As the water drops fall at regular intervals from a tap therefore time difference between any two drops $= \frac{1}{2} \sec$

In this given time, distance of second drop from the tap $= \frac{1}{2} g {(\frac{1}{2})}^{2} = \frac{5}{5} = 1 .25 m$

Its distance from the ground $= 5 - 1 .25 = 3 .75 m$

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A ball is thrown vertically upwards from the top of a tower at $4 .9 m s^{- 1}$ . It strikes the pond near the base of the tower after 3 seconds. The height of the tower is

A.

73.5 m

B.

44.1 m

C.

29.4 m

D.

None of these

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Explanation

$h = u t + \frac{1}{2} g t^{2}, t = 3 \sec, u = - 4 .9 m / s$

$\Rightarrow h = - 4 .9 \times 3 + 4 .9 \times 9 = 29 .4 m$

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An aeroplane is moving with horizontal velocity u at height h. The velocity of a packet dropped from it on the earth's surface will be (g is acceleration due to gravity)

A.

$\sqrt{u^{2} + 2 g h}$

B.

$\sqrt{2 g h}$

C.

2 gh

D.

$\sqrt{u^{2} - 2 g h}$

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Explanation

Horizontal velocity of dropped packet = u

Vertical velocity $= \sqrt{2 g h}$

∴ Resultant velocity at earth $= \sqrt{u^{2} + 2 g h}$

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A rocket is fired upward from the earth's surface such that it creates an acceleration of 19.6 m/sec². If after 5 sec its engine is switched off, the maximum height of the rocket from earth's surface would be [MP PET 1995]

A.

245 m

B.

490 m

C.

980 m

D.

735 m

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Explanation

Given $a = 19 .6 m / s^{2} = 2 g$

Resultant velocity of the rocket after 5 sec

$v = 2 g \times 5 = 10 g m / s$

Height achieved after 5 sec, $h_{1} = \frac{1}{2} \times 2 g \times 25 = 245 m$

On switching off the engine it goes up to height $h_{2}$ where its velocity becomes zero.

$0 = {(10 g)}^{2} - 2 g h_{2} \Rightarrow h_{2} = 490 m$

∴ Total height of rocket $= 245 + 490 = 735 m$

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A bullet is fired with a speed of $1000 m / \sec$ in order to hit a target 100 m away. If $g = 10 m / s^{2}$ , the gun should be aimed

A.

Directly towards the target

B.

5 cm above the target

C.

10 cm above the target

D.

15 cm above the target

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Explanation

Bullet will take $\frac{100}{1000} = 0 .1$ sec to reach target.

During this period vertical distance (downward)

travelled by the bullet $= \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times {(0 .1)}^{2} m = 5 c m$

So the gun should be aimed 5 cm above the target.

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A body starts to fall freely under gravity. The distances covered by it in first, second and third second are in ratio

A.

1 : 3 : 5

B.

1 : 2 : 3

C.

1 : 4 : 9

D.

1 : 5 : 6

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Explanation

$S_{n} = u + \frac{g}{2} (2 n - 1)$ ; when $u = 0$ , $S_{1} : S_{2} : S_{3} = 1 : 3 : 5$

Physics MCQs for NEET — Practice Questions with Answers

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