NEETshala
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
Register free to filter questions
● ● ● ● ●

The gravitational force Fg between two objects does not depend on

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

The gravitational force \( F_g \) between two objects is given by the formula \( F_g = G \frac{m_1 m_2}{r^2} \), where \( G \) is the gravitational constant, \( m_1 \) and \( m_2 \) are the masses of the objects, and \( r \) is the distance between them. The force depends on the product of the masses, not their sum. Therefore, the correct option is 'sum of the masses'.

● ● ● ● ●

Two sphere of mass $m_1$ and $m_2$ are situated in air and the gravitational force between them is F. The space around the masses is now filled with liquid of specific gravity 3. The gravitational force will now be

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

Gravitational force does not depend upon the medium

● ● ● ● ●

A satellite of the earth is revolveing in a circular orbit with a uniform speed v. If the gravitational force suddennly disappears, the satellite will

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

Due to inertia to direction

● ● ● ● ●

Two particle of equal mass go round a circle of radius r. Under the action of their mutual gravitational force. The speed of each particle is =..................

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

cenripetal force provided by the gravitational force of attraction between two particites $ \therefore { m \upsilon^2 \over r } = { G (m) (m) \over (2r)^2 } $ $ \therefore \upsilon = {1 \over 2 } \sqrt { Gm \over r } $

● ● ● ● ●

The distance of the moon and earth is D the mass of earth is 81 times the mass of moon. At what distance from the center of the earth, the gravitational force will be zero

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

For will be zero at the point of zero intensity $ x = { \sqrt {m_1} \over \sqrt {m_1} + \sqrt {m_2} } = { \sqrt m \over \sqrt {81m} + \sqrt m }$ $ D = { 9 \over 10 } D $

● ● ● ● ●

One can easily “ Weight the earth ” by calculating the mass of earth using the formula (in usual notation)

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

$ mg = { G M_e m \over R e^2 } $ where Me and Re is the mass and radius of the earth respectively $ \therefore M_e = { g \over G } Re^2 $

● ● ● ● ●

Three equal masses of m kg each are plced the vertices of an equilateral triangle PQR and a mass of 2m kg is placed at the centroid 0 of the triangle which is at a distance of $\sqrt 2 m$ from each of vertices of triangle. The force in newton. acting on the mass 2m is = ............

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

Here $ F_{OA} = F_{OB} =F_{OC} = { G (m) (2m ) \over r^2 } $ $ \vec F = \vec F_{OA} + \vec F_{OB} + \vec F _ { OC } $

● ● ● ● ●

Which of the following statement about the gravitational constant is true

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

Two point masses A and B having masses in the ratio 4 : 3 are seprated by a distance of lm. When another point mass c of mass M is placed in between A and B the forces A and C is 1/3rd of the force between band C, Then the distance C form A is = m

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

Here$ { m_a \over m_b } = { 4 \over 3 } $ $ FAC = { G (m) (mA) \over x^2 } ....(i) $ $ FAC = { G (m) (mB) \over (1-x)^2 } ....(ii) $ According to given problem $ FAC = { 1 \over 4 } FBC $ with the help of eqn (i) and (ii) $ = { G (m) ( mA) \over x^2 } = { 1 \over 3 } { G (m) ( mB ) \over (1-x) ^2 } $ $ \therefore { m_A \over m_B} = { x^2 \over 3(1-x)^2 } \Rightarrow { 4\over 3} = { x^2 \over (1-x)^2 } \Rightarrow 4 = { x^2 \over (1-x)^2} $ $ \Rightarrow 2 = { x \over 1-x } = 2 - 2x = x $ $ = 3x = 2 \therefore x = { 2 \over 3 } m $

● ● ● ● ●

The gravitational force between two point masses $m_1$ ans $m_2$ at separation r is given by$ F = G { m_1 m_2 \over r^2 } $. The constant k

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

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.