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Motion in a Straight Line MCQs for NEET — Physics Questions with Answers

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Two bodies of different masses m_a and m_b are dropped from two different heights a and b. The ratio of the time taken by the two to cover these distances are

a : b

b : a

$\sqrt{a} : \sqrt{b}$

$a^{2} : b^{2}$

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Explanation

$h = \frac{1}{2} g t^{2} \Rightarrow t = \sqrt{2 h / g}$

$t_{a} = \sqrt{\frac{2 a}{g}} and t_{b} = \sqrt{\frac{2 b}{g}} \Rightarrow \frac{t_{a}}{t_{b}} = \sqrt{\frac{a}{b}}$

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A body falls freely from rest. It covers as much distance in the last second of its motion as covered in the first three seconds. The body has fallen for a time of

3 s

5 s

7 s

9 s

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Explanation

$\frac{1}{2} g {(3)}^{2} = \frac{g}{2} (2 n - 1) \Rightarrow n = 5 s$

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A stone is dropped into water from a bridge 44.1 m above the water. Another stone is thrown vertically downward 1 sec later. Both strike the water simultaneously. What was the initial speed of the second stone ?

12.25 m/s

14.75 m/s

16.23 m/s

17.15 m/s

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Explanation

Time taken by first stone to reach the water surface from the bridge be t, then

$h = u t + \frac{1}{2} g t^{2} \Rightarrow 44 .1 = 0 \times t + \frac{1}{2} \times 9 .8 t^{2}$

$t = \sqrt{\frac{2 \times 44 .1}{9 .8}} = 3 s e c$

Second stone is thrown 1 sec later and both strikes simultaneously. This means that the time left for second stone $= 3 - 1 = 2 s e c$

Hence $44 .1 = u \times 2 + \frac{1}{2} 9 .8 {(2)}^{2}$

$\Rightarrow 44 .1 - 19 .6 = 2 u \Rightarrow u = 12 .25 m / s$

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A body is thrown vertically upwards. If air resistance is to be taken into account, then the time during which the body rises is

Equal to the time of fall

Less than the time of fall

Greater than the time of fall

Twice the time of fall

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Explanation

Let the initial velocity of ball be u

Time of rise $t_{1} = \frac{u}{g + a}$ and height reached $= \frac{u^{2}}{2 (g + a)}$

Time of fall $t_{2}$ is given by

$\frac{1}{2} (g - a) t_{2}^{2} = \frac{u^{2}}{2 (g + a)}$

$\Rightarrow t_{2} = \frac{u}{\sqrt{(g + a) (g - a)}} = \frac{u}{(g + a)} \sqrt{\frac{g + a}{g - a}}$

∴ $t_{2} > t_{1}$ because $\frac{1}{g + a} < \frac{1}{g - a}$

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A ball P is dropped vertically and another ball Q is thrown horizontally from the same height and at the same time. If air resistance is neglected, then

Ball P reaches the ground first

Ball Q reaches the ground first

Both reach the ground at the same time

The respective masses of the two balls will decide the time

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Explanation

Vertical component of velocities of both the balls are same and equal to zero. So $t = \sqrt{\frac{2 h}{g}}$

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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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Yes. You can attempt every Motion in a Straight Line question on this page for free without logging in, and check the correct answer with a detailed explanation instantly.

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.