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

● ● ● ● ●

A river is flowing from W to E with a speed of 5 m/min. A man can swim in still water with a velocity 10 m/min. In which direction should the man swim so as to take the shortest possible path to go to the south.

30° with downstream

60° with downstream

120° with downstream

South

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

● ● ● ● ●

A train is moving towards east and a car is along north, both with same speed. The observed direction of car to the passenger in the train is

East-north direction

West-north direction

South-east direction

None of these

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

Explanation

If the train is moving towards the east and the car is moving towards the north with the same speed, then from the perspective of the passenger in the train, the car will appear to be moving in the west-north direction due to the relative motion between the train and the car.

● ● ● ● ●

If the body is moving in a circle of radius r with a constant speed v, its angular velocity is

v²/r

v/r

r/v

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

Explanation

$v = r ω$ ⇒ $ω = \frac{v}{r}$ = constant [As v and r are constant]

● ● ● ● ●

Two racing cars of masses m₁ and m₂ are moving in circles of radii r₁ and r₂ respectively. Their speeds are such that each makes a complete circle in the same duration of time t. The ratio of the angular speed of the first to the second car is

m₁ : m₂

r₁ : r₂

1 : 1

m₁ r₁ : m₂ r₂

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

Explanation

As time periods are equal therefore the ratio of angular speeds will be one. $ω = \frac{2 π}{T}$

● ● ● ● ●

If a particle moves in a circle describing equal angles in equal times, its velocity vector

Remains constant

Changes in magnitude

Changes in direction

Changes both in magnitude and direction

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

Explanation

It is always directed in a direction of tangent to circle.

● ● ● ● ●

A motor cyclist going round in a circular track at constant speed has

Constant linear velocity

Constant acceleration

Constant angular velocity

Constant force

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

Explanation

Linear velocity, acceleration and force varies in direction.

● ● ● ● ●

Two bodies of mass 10 kg and 5 kg moving in concentric orbits of radii R and r such that their periods are the same. Then the ratio between their centripetal acceleration is

R/r

r/R

R²/r²

r²/R²

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

Explanation

$\frac{a_{R}}{a_{r}} = \frac{ω_{R}^{2} \times R}{ω_{r}^{2} \times r} = \frac{T_{r}^{2}}{T_{R}^{2}} \times \frac{R}{r} = \frac{R}{r}$ [As T_r = T_R]

● ● ● ● ●

A particle is moving in a horizontal circle with constant speed. It has constant

Velocity

Acceleration

Kinetic energy

Displacement

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

Explanation

K.E. $= \frac{1}{2} m v^{2}$ . Which is scalar, so it remains constant.

● ● ● ● ●

The angular speed of a fly wheel making 120 revolutions/minute is

$2 π r a d / s$

$4 π^{2} r a d / s$

$π r a d / s$

$4 π r a d / s$

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

Explanation

$120 r e v / \min = 120 \times \frac{2 π}{60} r a d / \sec = 4 π r a d / \sec$

● ● ● ● ●

Certain neutron stars are believed to be rotating at about 1 rev/sec. If such a star has a radius of 20 km, the acceleration of an object on the equator of the star will be

$20 \times 10^{8} m / \sec^{2}$

$8 \times 10^{5} m / \sec^{2}$

$120 \times 10^{5} m / \sec^{2}$

$4 \times 10^{8} m / \sec^{2}$

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

Explanation

$a = ω^{2} r = 4 π^{2} n^{2} r = 4 π^{2} \times 1^{2} \times 20 \times 10^{3}$

∴ a = ${8×10}^{5} m / s e c^{2}$

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.