Blog Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

● ● ● ● ●

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

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

Explanation

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

● ● ● ● ●

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

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

Explanation

The separation between the two bodies, two seconds after the release of second body

$= \frac{1}{2} \times 9 .8 [{(3)}^{2} - {(2)}^{2}] = 24 .5 m$

● ● ● ● ●

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

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

Explanation

Time of flight $= \frac{2 u}{g} = \frac{2 \times 100}{10} = 20 s e c$

● ● ● ● ●

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

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

Explanation

$h = \frac{1}{2} g t^{2} = \frac{1}{2} \times 10 \times {(4)}^{2} = 80 m$

● ● ● ● ●

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

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

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 $= h - \frac{h}{4} = \frac{3 h}{4}$

● ● ● ● ●

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

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

$\sqrt{\frac{2 l}{g}}$

$\frac{1}{\sin θ} \sqrt{\frac{2 h}{g}}$

$\sin θ \sqrt{\frac{2 h}{g}}$

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

Explanation

Force down the plane $= m g \sin θ$

∴ Acceleration down the plane $= g \sin θ$

Since $l = 0 + \frac{1}{2} g \sin θ t^{2}$

∴ $t^{2} = \frac{2 l}{g \sin θ} = \frac{2 h}{g \sin^{2} θ} \Rightarrow t = \frac{1}{\sin θ} \sqrt{\frac{2 h}{g}}$

● ● ● ● ●

A particle is projected up with an initial velocity of $80 f t / \sec$ . The ball will be at a height of $96 f t$ from the ground after (Given $g = 32 f t / s^{2}$ )

2.0 and 3.0 sec

Only at 3.0 sec

Only at 2.0 sec

After 1 and 2 sec

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

Explanation

$h = u t - \frac{1}{2} g t^{2} \Rightarrow 96 = 80 t - \frac{32}{2} t^{2}$

$\Rightarrow t^{2} - 5 t + 6 = 0 \Rightarrow t = 2$ sec or 3 sec

● ● ● ● ●

A body falls from rest, its velocity at the end of first second is $(g = 32 f t / \sec^{2})$

$16 f t / \sec$

$32 f t / \sec$

$64 f t / \sec$

$24 f t / \sec$

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

Explanation

$v = g \times t = 32 \times 1 = 32 f t / \sec$

● ● ● ● ●

A stone thrown upward with a speed u from the top of the tower reaches the ground with a velocity 3u. The height of the tower is

$3 u^{2} / g$

$4 u^{2} / g$

$6 u^{2} / g$

$9 u^{2} / g$

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

Explanation

$v^{2} = u^{2} + 2 g h \Rightarrow {(3 u)}^{2} = {(- u)}^{2} + 2 g h \Rightarrow h = \frac{4 u^{2}}{g}$

● ● ● ● ●

Two stones of different masses are dropped simultaneously from the top of a building

Smaller stone hit the ground earlier

Larger stone hit the ground earlier

Both stones reach the ground simultaneously

Which of the stones reach the ground earlier depends on the composition of the stone

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

Explanation

$t = \sqrt{\frac{2 h}{g}}$ and h and g are same.

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every NEET 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.