Blog Log in Get Started Free

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

Practice free Motion in a Straight Line (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

A very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is 5m, the number of ball thrown per minute is (take $g = 10 m s^{- 2}$ )

120

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Maximum height of ball = 5 m

So velocity of projection $\Rightarrow u = \sqrt{2 g h} = 10 m / s$

Time interval between two balls (time of ascent)

$= \frac{u}{g} = 1 s e c = \frac{1}{60} m i n$ .

So number of ball thrown per min. = 60

● ● ● ● ●

A body falling from a high Minaret travels 40 meters in the last 2 seconds of its fall to ground. Height of Minaret in meters is (take $g = 10 m s^{- 2}$ )

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Using the equation of motion: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, t is the time, and a is the acceleration (g = -10 m/s^2). Given that the body travels 40 m in the last 2 s, we can substitute the values to find the initial height: 40 = 0 + (1/2)(-10)(2)^2 => h = 45 m.

● ● ● ● ●

A body falls from a height $h = 200 m$ (at New Delhi). The ratio of distance travelled in each 2 sec during t = 0 to t = 6 second of the journey is

1 : 4 : 9

1 : 2 : 4

1 : 3 : 5

1 : 2 : 3

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

Distance travelled from t-0 to t=2 s,

$s_{1} = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} g (2)^{2} = 2 g D i s \tan c e t r a v e l l e d f r o m t = 2 s t o t = 4 s, D i s \tan c e t r a v e l l e d i n 4 s - D i s \tan c e t r a v e l l e d i n 2 s (s_{1}) = \frac{1}{2} g (4)^{2} - \frac{1}{2} g (2)^{2} = 12 \frac{g}{2} = 6 g D i s \tan c e t r a v e l l e d f r o m t = 4 s t o t = 6 s, D i s \tan c e t r a v e l l e d i n 6 s - D i s \tan c e t r a v e l l e d i n 4 s \frac{g (6)^{2}}{2} - \frac{g (4)^{2}}{2} = 10 g$

So distances are in ratio 1:3:5

● ● ● ● ●

A man drops a ball downside from the roof of a tower of height 400 meters. At the same time another ball is thrown upside with a velocity 50 meter/sec. from the surface of the tower, then they will meet at which height from the surface of the tower

100 meters

320 meters

80 meters

240 meters

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

Two balls are dropped from heights h and 2h respectively from the earth surface. The ratio of time of these balls to reach the earth is

$1 : \sqrt{2}$

$\sqrt{2}$ : 1

2 : 1

1 : 4

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$t = \sqrt{\frac{2 h}{g}} \Rightarrow \frac{t_{1}}{t_{2}} = \sqrt{\frac{h_{1}}{h_{2}}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$

● ● ● ● ●

The acceleration due to gravity on the planet A is 9 times the acceleration due to gravity on planet B. A man jumps to a height of 2m on the surface of A. What is the height of jump by the same person on the planet B

18m

$\frac{2}{3} m$

$\frac{2}{9} m$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$H_{\max} = \frac{u^{2}}{2 g} \Rightarrow H_{\max} \propto \frac{1}{g}$

On planet B value of g is $1 / 9$ times to that of A. So value of $H_{\max}$ will become 9 times i.e. $2 \times 9 = 18 m e t r e$

● ● ● ● ●

A parachutist after bailing out falls 50 m without friction. When parachute opens, it decelerates at 2 m/s². He reaches the ground with a speed of 3 m/s. At what height, did he bail out ?

293 m

111 m

91 m

182 m

You've reached today's free limit of 20 questions. Log in to keep practising for free.

● ● ● ● ●

When a ball is thrown up vertically with velocity V₀, it reaches a maximum height of 'h'. If one wishes to triple the maximum height then the ball should be thrown with velocity

$\sqrt{3} V_{o}$

3 V₀

9 V₀

3/2 V₀

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$H_{\max} \propto u^{2}$ ∴ $u \propto \sqrt{H_{\max}}$

i.e. to triple the maximum height, ball should be thrown with velocity $\sqrt{3} u$ .

● ● ● ● ●

A particle moving in a straight line covers half the distance with speed of 3 m/s. The other half of the distance is covered in two equal time intervals with speed of 4.5 m/s and 7.5 m/s respectively. The average speed of the particle during this motion is

4.0 m/s

5.0 m/s

5.5 m/s

4.8 m/s

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

If $t_{1}$ and $2 t_{2}$ are the time taken by particle to cover first and second half distance respectively.

$t_{1} = \frac{x / 2}{3} = \frac{x}{6}$ …(i)

$x_{1} = 4 .5 t_{2}$ and $x_{2} = 7 .5 t_{2}$

So, $x_{1} + x_{2} = \frac{x}{2} \Rightarrow 4 .5 t_{2} + 7 .5 t_{2} = \frac{x}{2}$

$t_{2} = \frac{x}{24}$ …(ii)

Total time $t = t_{1} + 2 t_{2} = \frac{x}{6} + \frac{x}{12} = \frac{x}{4}$

So, average speed $= 4 m / s e c$ .

● ● ● ● ●

The acceleration of a particle is increasing linearly with time t as bt. The particle starts from the origin with an initial velocity v₀ The distance travelled by the particle in time t will be

$v_{0} t + \frac{1}{3} b t^{2}$

$v_{0} t + \frac{1}{3} b t^{3}$

$v_{0} t + \frac{1}{6} b t^{3}$

$v_{0} t + \frac{1}{2} b t^{2}$

You've reached today's free limit of 20 questions. Log in to keep practising for free.

Explanation

$\frac{d v}{d t} = b t \Rightarrow d v = b t d t \Rightarrow v = \frac{b t^{2}}{2} + K_{1}$

At $t = 0, v = v_{0} \Rightarrow K_{1} = v_{0}$

We get $v = \frac{1}{2} b t^{2} + v_{0}$

Again $\frac{d x}{d t} = \frac{1}{2} b t^{2} + v_{0} \Rightarrow x = \frac{1}{2} \frac{b t^{2}}{3} + v_{0} t + K_{2}$

At $t = 0, x = 0 \Rightarrow K_{2} = 0$

$∴ x = \frac{1}{6} b t^{3} + v_{0} t$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

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.