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

A car turns a corner on a slippery road at a constant speed of 10 m/s. If the coefficient of friction is 0.5, the minimum radius of the arc at which the car turns is ________ meter.

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

$ v = \sqrt { rg } $ $ \therefore r = { v^2 / g } $

A person standing on the floor of a lift drops a coin. The coin reaches the floor of the lift in time to if the lift is stationary and the time $t_2$ if it is accelerated in upward direction. Than

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

$ d = { 1 \over 2 } gt_1^2 and \, d = {1 \over 2 } (g + a ) t_2^2 $ By comparing both eqn $ {1 \over 2 } gt_1^2 = { 1 \over 2 } (g+ a)t_2^2 $ $ \therefore t_1 \gt t_2 $

A lift of mass 1000 kg is moving with an acceleration of $1 ms^{–2}$ in upward direction Tension developed in the rope of lift is__________ N $ (g = 9.8 ms^{-2} )$

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

Tensile force T = m(g+a )

A rope which can withstand a maximum tension of 400 N hangs from a tree. If a monkey of mass 30 kg climbs on the rope in which of the following cases-will the rope break? (take $g =10 ms^{-2} $ and neglect the mass of rope)

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

Calculate F - m(g+a)

Same forces act on two bodies of different mass 2 kg and 5 kg initialy at rest. The ratio of times required to acquire same final velocity is _

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

$ Ft_1 = m_1 ( \nu -\nu_0) = m_1 \nu_1 $ $ Ft_2 = m_2 ( \nu -\nu_0) = m_2 \nu$ $ \therefore { t_1 \over t_2 } = { m_1 \over m_2 } $

A train is moving along a horizontal track. A pendulum suspended from the roof makes an angle of $ 4 ^\circ $ with the vertical, The acceleration of the train is ___ $ms^{-2} (g = 10 ms^{-2} ) $

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

$ ma = mg sin \theta $ $ \therefore a = 10 sin 4 ^\circ $ $ \therefore a = 0.7 ms^{-2} $

A partly hanging uniform chain of length L is resting on a rough horizontal table. l is the maximum possible length that can hang in equilibrium The coefficient of friction between the chain and table is

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

= M/ L Weight of chain of lenght l = lg Weight of a chain of lenght ( L -l ) = (L -l ) g $ f_S = s_ R = s (L- l) g and f_S = l g $ $ \therefore l g = s. ( L - l ) g $ $ \therefore s = { l \over L - l } $

A car of mass 1000 kg travelling at 32 m/s clashes into a rear of a truck of mass 8000 kg moving in the same direction with a velocity of 4 m/s. After the collision the car bounces with a velocity of $8 ms^{–1} $. The velocity of truck after the impact is m/s

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

According to eqn $ m_1 \vec v_1 + m_2 \vec v_2 = m_1 \vec v_1 + m_2 \vec v_2 $ $ \therefore v_2^1 = 9 m/s $

The upper half of an inclined plane of inclination $ \theta $ is perfectly smooth while the lower half is rough A body starting from the rest at top come back to rest at the bottom, then the coefficient of friction for the lower half is given by_ ____

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

For upper half $ v_1^2 - v_o^2 = 2 ad $ $ \nu^2 - 0 = 2 ( g sin \theta )_2 $ $ \therefore v^2 = gl sin \theta $ For lower half $ v_1^2 - v_0^2 = 2 a' d $ $ 0 - gl sin \theta = 2g [ sin \theta - \mu_s cos \theta ] . { l \over 2 } $ $ \therefore 2 = 2 tan \theta $

The motion of a particle of a mass m is describe by $ y = ut + {1 \over 2 } gt^2 $ Find the force acting on the particle.

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

The given equation of motion is $ y = ut + \frac{1}{2}gt^2 $. This equation represents the vertical motion of a particle under the influence of gravity. The force acting on the particle is due to gravity, which is given by $ F = mg $. Hence, the correct option is F = mg.

Ready to ace NEET?

Free access · No credit card required

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.