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JEE PHYSICS MCQ: MOTION IN A STRAIGHT LINE

1 Easy

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Application +4 | -1

Two stones of masses

2 \mathrm{~kg}

and

4 \mathrm{~kg}

are dropped from height

30 \mathrm{~m}

and

40 \mathrm{~m}

respectively. The ratio of time taken by the stones to reach the ground is

View answer and explanation

Correct option: D

\sqrt{\frac{3}{4}}

Option D (

\sqrt{\frac{3}{4}}

) is correct because the time taken to reach the ground is proportional to

\sqrt{h}

, so the ratio of times is

\sqrt{\frac{30}{40}} = \sqrt{\frac{3}{4}}

.

More Questions from MOTION IN A STRAIGHT LINE

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Assertion: In a free fall, the initial velocity of a body may not be zero.
Reason: Free fall means the vertical acceleration of the body is equal to

\mathrm{g}

. A ball is thrown vertically upward with speed 15 m/s and it returns to the ground with speed 10 m/s as a constant air resistance acts. The maximum height attained by the ball is Which of the following is not a derived quantites? Assertion: If a body is dropped from the top of a tower of height

\mathrm{h}

and another body is thrown up simultaneously with velocity

\mathrm{u}

from the foot of the tower, then both of them would meet after a time

\frac{\mathrm{h}}{\mathrm{u}}

.
Reason: For a body projected upwards, the distance covered by the body in the last second of its upward journey is always

4.9 \mathrm{~m}

irrespective of velocity of projection.

(\mathrm{g}=9.8 \mathrm{~m} / \mathrm{s}^2)

A particle is having uniform acceleration. If its displacement from

t

(t+1)

120 \mathrm{~m}

and change in velocity is

50 \mathrm{~m} / \mathrm{s}

. Find its displacement (in

\mathrm{m}

(t+2)

second. An aeroplane flies 400m north and 300m south and then flies 1200m upwards, then net displacement is