Blog Log in Get Started Free

Physics MCQs for NEET — Practice Questions with Answers

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

All Physics Chemistry Botany Zoology

Chapter Language Difficulty Show only questions I answered incorrectly Register free to filter questions

● ● ● ● ●

The masses and radii of the earth and moon are $M_{1,} R_{1}$ and $M_{2}, R_{2}$ respectively. Their centres are distance d apart. The minimum velocity with which a particle of mass m should be projected from a point midway between their centres so that it escapes to infinity is

$2 \sqrt{\frac{G}{D} (M_{1} + M_{2})}$

$2 \sqrt{\frac{2 G}{D} (M_{1} + M_{2})}$

$2 \sqrt{\frac{G M}{D} (M_{1} + M_{2})}$

$2 \sqrt{\frac{G M (M_{1} + M_{2})}{D (R_{1} + R_{2})}}$

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

Explanation

(a) Gravitational potential at mid point

$V = \frac{- G M_{1}}{d / 2} + \frac{- G M_{2}}{d / 2} N o w, P E = m \times V = \frac{- 2 G M}{d} (M_{1} + M_{2}) [m = m a s s o f p a r t i c l e] S o, f o r p r o j e c t i n g p a r t i c l e f r o m m i d p o i n t t o i n f i n i t y$

$K E = |P E| \Rightarrow \frac{1}{2} m v^{2} = \frac{2 G m}{d} (M_{1} + M_{2}) \Rightarrow v = 2 \sqrt{\frac{G (M_{1} + M_{2})}{d}}$

● ● ● ● ●

If mass of earth is M, radius is R and gravitational constant is G, then work done to take 1 kg mass from earth surface to infinity will be

$\sqrt{\frac{G M}{2 R}}$

$\frac{G M}{R}$

$\sqrt{\frac{2 G M}{R}}$

$\frac{G M}{2 R}$

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

Explanation

(b) Potential energy of the 1 kg mass which is placed at the earth surface= $- \frac{G M}{R}$

its potential energy at infinite = 0

Work done = change in potential energy = $\frac{G M}{R}$

● ● ● ● ●

If $v_{e}$ and $v_{o}$ represent the escape velocity and orbital velocity of a satellite corresponding to a circular orbit of radius R, then

$v_{e} = v_{o}$

$\sqrt{2} v_{o} = v_{e}$

$v_{e} = v_{o} / \sqrt{2}$

$v_{e}$ and $v_{o}$ are not related

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

Explanation

(b) $v_{e} = \sqrt{2 g R} a n d v_{o} = \sqrt{g R} ∴ \sqrt{2} v_{o} = v_{e}$

● ● ● ● ●

If r represents the radius of the orbit of a satellite of mass m moving around a planet of mass M, the velocity of the satellite is given by:

$v^{2} = g \frac{M}{r}$

$v^{2} = \frac{G M m}{r}$

$v = \frac{G M}{r}$

$v^{2} = \frac{G M}{r}$

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

Explanation

$F_{e} = \frac{m v^{2}}{r} a n d F_{g} = \frac{G M m}{r^{2}}$

For the satellite to revolve in the orbit the centripetol force must be blamed by gravitational force i.e.

$F_{C} = F_{g} \Rightarrow \frac{m v^{2}}{r} = \frac{G M m}{r^{2}} \Rightarrow v^{2} = \frac{G M}{r}$

● ● ● ● ●

An earth satellite of mass m revolves in a circular orbit at a height h from the surface of the earth. R is the radius of the earth and g is acceleration due to gravity at the surface of the earth. The velocity of the satellite in the orbit is given by

$\frac{g R^{2}}{R + h}$

$\frac{g R}{R + h}$

$\sqrt{\frac{g R^{2}}{R + h}}$

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

Explanation

(d) $v_{o} = \sqrt{\frac{G M}{r}} = \sqrt{\frac{g R^{2}}{R + h}}$

● ● ● ● ●

A satellite which is geostationary in a particular orbit is taken to another orbit. Its distance from the centre of earth in new orbit is 2 times that of the earlier orbit. The time period in the second orbit is

4.8 hours

$48 \sqrt{2}$ hours

24 hours

$24 \sqrt{2}$ hours

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

Explanation

(b) $T \propto r^{3 / 2}$ . If r becomes double then time period will becomes ${(2)}^{\frac{3}{2}}$ times.
So new time period will be $24 \times 2 \sqrt{2} h r$ i.e. $T = 48 \sqrt{2}$

● ● ● ● ●

The ratio of the K.E. required to be given to the satellite to escape earth's gravitational field to the K.E. required to be given so that the satellite moves in a circular orbit just above earth atmosphere is

One

Two

Half

Infinity

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

Explanation

(b) K.E. required for satellite to escape from earth's gravitational field

$\frac{1}{2} m v_{e}^{2} = \frac{1}{2} m {(\sqrt{\frac{2 G M}{R}})}^{2} = \frac{G M m}{R}$

K.E. required for satellite to move in circular orbit

$\frac{1}{2} m v_{o}^{2} = \frac{1}{2} m {(\sqrt{\frac{G M}{R}})}^{2} = \frac{G M m}{2 R}$

The ratio between these two energies = 2

● ● ● ● ●

An astronaut orbiting the earth in a circular orbit 120 km above the surface of earth, gently drops a spoon out of space-ship. The spoon will

Fall vertically down to the earth

Move towards the moon

Will move along with space-ship

Will move in an irregular way then fall down

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

Explanation

out of the space-ship.

● ● ● ● ●

The period of a satellite in a circular orbit around a planet is independent of

The mass of the planet

The orbital radius of satellite around the planet

The mass of the satellite

All the three parameters (a), (b) and (c)

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

Explanation

$T = 2 π \sqrt{\frac{R^{3}}{GM}}$

so, it is independent of mass of satellite.

● ● ● ● ●

Two identical satellites A and B go round a planet P in circular orbits having radii 4R and R respectively. If the speed of the satellite A is 3V, the speed of the satellite B will be ?

12 V

6 V

$\frac{4}{3} v$

$\frac{3}{2} v$

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

Explanation

(b) $v = \sqrt{\frac{G M}{R}} \Rightarrow \frac{v_{A}}{v_{B}} = \sqrt{\frac{R_{B}}{R_{A}}} = \sqrt{\frac{R}{4 R}} = \frac{1}{2} ∴ \frac{v_{A}}{v_{B}} = \frac{3 V}{v_{B}} = \frac{1}{2} ∴ v_{B} = 6 V$

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Physics 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.