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
● ● ● ● ●

A ball of mass 0.2 kg is thrown vertically upwards by applying a force by hand. If the hand moves 0.2 m while applying the force and the ball goes upto 2 m height further, find the magnitude of the force.$ (Consider g = 10 ms^{–2}) $

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

$ H_{max} = { u^2 \over 2g } $ $ \therefore u = \sqrt { 2gH_{max} }$ This velocity is supplied to the ball by hand and initially the hand was at rest. It acquires this velocity in distance of 0.2 meter. $ \therefore a = { u^2 \over 2S } $ So upword force F = m (g+a)

● ● ● ● ●

Formula for true force is

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

According to newton's second law force = rate of change of linear momentum.

● ● ● ● ●

A particle moves in the X–Y plane under the influence of a force such that its linear momentum is $ \vec P (t) = A[ \hat i cos (kt) -\hat j sin(kt)] $ where A and k are constants .The angle betweenn the force and momentum is

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

$ \vec P (t) = A[ \hat i cos (kt) -\hat j sin(kt)] $ $ \vec F = { d \over dt} \left ( \vec P (t) \right) $ $ = Ak \left( - \hat i sin kt - \hat j cos ky \right) $ $ \vec F . \vec P = A^2 k ( -cos kt .sin kt + sin kt . cos kt ) $ $ \vec F . \vec P = 0 $

● ● ● ● ●

Force of 5 N acts on a body of weight 9.8 N. what is the acceleration produced in $ms^{-2}$

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

As weight W = mg = 9.8 N Therefore m = 1 Kg a = F / m

● ● ● ● ●

Same force acts on two bodies of different masses 2 kg and 4 kg initially 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

$ t = {V \over a } \Rightarrow t \alpha { 1 \over a } ( v is constant ) $ $ \therefore {t_1 \over t_2 } = {a_2 \over a_1} = { m_1 \over m_2 } = {2 \over 4 } = { 1 \over 2 } $ $ a \alpha 1/m as F is constant $

● ● ● ● ●

Which of the following quantities measured from different inertial reference frames are same

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

Force is a quantity that is invariant under Galilean transformations between different inertial reference frames. This means that when observed from different inertial frames, the force remains the same. In contrast, quantities like velocity, displacement, and kinetic energy can vary depending on the reference frame.

● ● ● ● ●

When the speed of a moving body is doubled

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

A particle moves in the XY Plane under the action of a force F such that the components of its linear momentum P at any time t are Px = 2 cost, Py = 2 sint. The angle between F and P at time t is

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

$ \vec P = Px \hat i + Py \hat j , \vec F = { \vec {dp} \over dt} $ Now $ \vec F . \vec P = 0 $ $ \therefore \theta = 90 ^\circ $

● ● ● ● ●

A player caught a cricket ball of mass 150 g moving at the rate of $20 ms^{-1} $ . If the catching process be completed in 0.1 s the force of the blow exerted by the ball on the hands of player is

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

Force exerted by ball, F = m (dv /dt)

● ● ● ● ●

A body of mass 5 kg starts from the origin with an initial velocity $ \vec u = 30 \hat i + 40 \hat j ms^{-1} $. If a constant Force
$ \vec F = - \left( \hat i + 5 \hat j \right) N $ acts on the body, the time in which the y-component of the velocity becomes

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

$ \nu_y = 40ms^{-1} Fy = -5N m = 5kg $ $ so a_y = { F_y \over m } = -1 ms^{-1} $ $ as V_y = u_y + at $ $ 0 = 40 -lt \Rightarrow t = 40 s $

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.