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

● ● ● ● ●

Which of the following is the evidence to show that there must be a force acting on earth and directed towards the sun?

Deviation of the falling bodies towards east

Revolution of the earth around the sun

Phenomenon of day and night

The apparent motion of the sun around the earth

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

Explanation

(b)

The earth revolves around the sun due to the gravitation pull of the sun. Due to this gravitational attraction between this celestial body, the centripetal force is generated which binds the solar system together. Hence revolution of the earth around the sun is the evidence to show that there must be a force acting on earth and directed towards the sun.

● ● ● ● ●

A mass of $6 \times 10^{24} k g$ is to be compressed in a sphere in such a way that the escape velocity from the sphere is $3 \times 10^{8} m / s$ . Radius of the sphere should be $(G = 6.67 \times 10^{- 11} N - m^{2} / k g^{2})$

9 km

9 m

9 cm

9 mm

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

Explanation

$(d) v_{e} = \sqrt{\frac{2 G M}{R}} = \sqrt{\frac{2 \times 6.67 \times 10^{- 11} \times 6 \times 10^{24}}{R}} = 3 \times 10^{8} B y s o l v i n g R = 9 m m$

● ● ● ● ●

The mass and diameter of a planet have twice the value of the corresponding parameters of earth. Acceleration due to gravity on the surface of the planet is

$9.8 m / s e c^{2}$

$4.9 m / s e c^{2}$

$980 m / s e c^{2}$

$19.6 m / s e c^{2}$

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

Explanation

(b) $\frac{g'}{g} = \frac{M'}{M_{2}} {[\frac{R}{R'}]}^{2} = [\frac{2 M}{M}] {[\frac{R}{2 R}]}^{2} = \frac{1}{2}$

$g^{1} = \frac{g}{2} = \frac{9.8}{2} = 4.9 m / s^{2}$

● ● ● ● ●

Force of gravity is least at

The equator

The poles

A point in between equator and any pole

None of these

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

Explanation

(a)

The equator has the least gravitational pull, that is there will be the least weight for an object there.

● ● ● ● ●

The velocity with which a projectile must be fired so that it escapes earth’s gravitation does not depend on:

Mass of the earth

Mass of the projectile

Radius of the projectile’s orbit

Gravitational constant

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

Explanation

(b) We know that, $V_{e} = \sqrt{\frac{2 G M}{r}}$

From above expression we can see, mass of projectile does not matter.

● ● ● ● ●

The gravitational force between two stones of mass 1 kg each separated by a distance of 1 metre in vacuum is

Zero

6.675 $\times 10^{‐ 5} n e w t o n$

$6.675 \times 10^{‐ 11} n e w t o n$

$6.675 \times 10^{‐ 8} n e w t o n$

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

Explanation

(c) $F = G \frac{m_{1} m_{2}}{r^{2}} = 6.675 \times \frac{1 \times 1}{1^{2}} \times 10^{‐ 11} = 6.675 \times 10^{‐ 11} N$

● ● ● ● ●

The radius of a planet is $\frac{1}{4}$ of earth’s radius and its acceleration due to gravity is double that of earth’s acceleration due to gravity. How many times will the escape velocity at the planet’s surface be as compared to its value on earth’s surface ?

$\frac{1}{\sqrt{2}}$

$\sqrt{2}$

$2 \sqrt{2}$

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

Explanation

$(a) v = \sqrt{2 g R} \Rightarrow \frac{v_{p}}{v_{e}} = \sqrt{\frac{g_{p}}{g_{e}} \times \frac{R_{e}}{R_{p}}} = \sqrt{2 \times \frac{1}{4}} = \frac{1}{\sqrt{2}} s o, v_{p} = \frac{v_{e}}{\sqrt{2}}$

● ● ● ● ●

The escape velocity for the earth is $V_{d}$ . The escape velocity for a planet whose radius is four times and density is nine times that of the earth, is

$36 v_{d}$

$12 v_{d}$

$6 v_{d}$

$20 v_{d}$

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

Explanation

(b) $v \propto R \sqrt{ρ} so, \frac{v_{p}}{v_{e}} = \frac{R_{p}}{R_{e}} \times \sqrt{\frac{ρ_{p}}{ρ_{e}}} = 4 \sqrt{9} = 12 \Rightarrow v_{p} = 12 v_{e}$

● ● ● ● ●

Two particles of equal mass go round a circle of radius R under the action of their mutual gravitational attraction. The speed of each particle is

$ν = \frac{1}{2 R} \sqrt{\frac{1}{G m}}$

$ν = \sqrt{\frac{G m}{2 R}}$

$ν = \frac{1}{2} \sqrt{\frac{G m}{R}}$

$ν = \sqrt{\frac{4 G m}{R}}$

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

Explanation

i.e. $\frac{m v^{2}}{R} = \frac{G m \times m}{{(2 R)}^{2}}$

$\Rightarrow ν = \frac{1}{2} \sqrt{\frac{G m}{R}}$

● ● ● ● ●

The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from the surface of the earth is (g = acceleration due to gravity at the surface of the earth)

$\frac{g}{9}$

$\frac{g}{3}$

$\frac{g}{4}$

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

Explanation

(a) $g' = g {(\frac{R}{R + h})}^{2} = g {(\frac{R}{R + 2 R})}^{2} = \frac{g}{9}$

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.