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

● ● ● ● ●

If M the mass of the earth and R its radius, the ratio of the gravitational acceleration and the gravitational constant is

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

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

$M R^{2}$

$\frac{M}{R}$

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

Explanation

(b) Acceleration due to gravity g= $\frac{G M}{R^{2}}$

$S o, \frac{g}{G} = \frac{M}{R^{2}}$

● ● ● ● ●

Two sphere of mass m and M are situated in air and the gravitational force between them is F. The space around the masses is now filled with a liquid of specific gravity 3. The gravitational force will now be

$\frac{F}{3}$

$\frac{F}{9}$

3 F

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

Explanation

(a) Gravitational force does not depend on the medium.

● ● ● ● ●

Earth binds the atmosphere because of:

Gravity

Oxygen between earth and atmosphere

Both (a) and (b)

None of these

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

Explanation

(a)

Various atom of gases present in the earth atmosphere has masses and a gravitational force acts between theses atoms and earth.

also this is the force that holds the various gases near the surface and the atmosphere is present.

● ● ● ● ●

A body of mass m rises to height h = R/5 from the earth's surface, where R is earth's radius. If g is acceleration due to gravity at earth's surface, the increase in potential energy is

mgh

$\frac{4}{5} m g h$

$\frac{5}{6} m g h$

$\frac{6}{7} m g h$

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

Explanation

Substituting $R = 5 h w e g e t ∆ U = \frac{m g h}{1 + 1 / 5} = \frac{5}{6} m g h$

● ● ● ● ●

Spot the wrong statement :

The acceleration due to gravity ‘g’ decreases if

We go down from the surface of the earth towards its centre

We go up from the surface of the earth

We go from the equator towards the poles on the surface of the earth

The rotational velocity of the earth is increased

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

Explanation

● ● ● ● ●

Which of the following statements is true?

g is less at the earth's surface than at a height above it or a depth below it

g is the same at all places on the surface of the earth

g has its maximum value at the equator

g is greater at the poles than at the equator

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

Explanation

(d)

The value of 'g' that is gravity is greater at the poles because the gravitational pull is maximum at the poles and decreases as it comes down toward the equator.

● ● ● ● ●

A spring balance is graduated on sea level. If a body is weighed with this balance at consecutively increasing heights from earth's surface, the weight indicated by the balance

Will go on increasing continuously

Will go on decreasing continuously

Will remain same

Will first increase and then decrease

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

Explanation

(b) Because value of g decreases with increasing height

● ● ● ● ●

The gravitational field due to a mass distribution is $E = K / x^{3}$ in the x-direction. (K is a constant). Taking the gravitational potential to be zero at infinity, its value at a distance x is

K/x

K/2x

$K / x^{2}$

K/2 $x^{2}$

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

Explanation

(d) Gravitational potential $= \int I d x = \int_{x}^{\infty} \frac{K}{x^{3}} d x = K {(\frac{x^{- 3 + 1}}{- 3 + 1})}_{x}^{\infty} = |\frac{- K}{2 x^{2}}| = \frac{K}{2 x^{2}}$

● ● ● ● ●

The change in potential energy, when a body of mass m is raised to a height nR from the earth's surface is (R = Radius of earth)

$m g R \frac{n}{n - 1}$

nmgR

C.
mgR $\frac{n^{2}}{n^{2} + 1}$
D.

$m g R \frac{n}{n + 1}$

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

Explanation

(d) $∆ U = \frac{m g h}{1 + \frac{h}{R}} = \frac{m g n R}{1 + \frac{n R}{R}} = \frac{n m g R}{n + 1}$

● ● ● ● ●

If the earth suddenly shrinks (without changing mass) to half of its present radius, the acceleration due to gravity will be

g/2

g/4

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

Explanation

(b) $g = \frac{G M}{R^{2}}$ . If radius shrinks to half of its present value then g will becomes four time

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.