Are these Physics questions free to attempt?

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.

Do I need an account to practise Physics MCQs?

No account is required to attempt questions and view answers. A free account adds bookmarks, personal notes, and progress tracking.

Are these previous year NEET questions?

The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.

Physics NEET Questions — Free MCQ Practice

● ● ● ● ●

A cricketer can throw a ball to a maximum horizontal distance of 100 m. With the same effort, he throws the ball vertically upwards. The maximum height attained by the ball is

A.

100 m

B.

80 m

C.

60 m

D.

50 m

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

Explanation

Maximum range $= \frac{u^{2}}{g} = 100 m$

Maximum height $= \frac{u^{2}}{2 g} = \frac{100}{2} = 50 m$

● ● ● ● ●

A ball is projected with velocity V₀ at an angle of elevation 30°. Mark the correct statement

A.

Kinetic energy will be zero at the highest point of the trajectory

B.

Vertical component of momentum will be conserved

C.

Horizontal component of momentum will be conserved

D.

Gravitational potential energy will be minimum at the highest point of the trajectory

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

Explanation

Since horizontal component of velocity is constant, hence momentum is constant.

● ● ● ● ●

Neglecting the air resistance, the time of flight of a projectile is determined by

A.

U_vertical

B.

U_horizontal

C.

U = U²_vertical+ U²_horizontal

D.

U = U(U²_vertical+ U²_horizontal )^1/2

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

Explanation

Time of flight = $\frac{2 u \sin θ}{g} = \frac{2 u_{y}}{g} = \frac{2 \times u_{vertical}}{g}$

● ● ● ● ●

A ball is thrown from a point with a speed v₀ at an angle of projection θ. From the same point and at the same instant a person starts running with a constant speed v₀/2 to catch the ball. Will the person be able to catch the ball? If yes, what should be the angle of projection?

A.

Yes, 60°

B.

Yes, 30°

C.

No

D.

Yes, 45°

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

Explanation

Person will catch the ball if its velocity will be equal to horizontal component of velocity of the ball.

$\frac{v_{0}}{2} = v_{0} \cos θ$ ⇒ $\cos θ = \frac{1}{2}$ ⇒ $θ = 60 °$

● ● ● ● ●

A stone is thrown at an angle θ to the horizontal reaches a maximum height H. Then the time of flight of stone will be

A.

$\sqrt{\frac{2 H}{g}}$

B.

$2 \sqrt{\frac{2 H}{g}}$

C.

$\frac{2 \sqrt{2 H \sin θ}}{g}$

D.

$\frac{\sqrt{2 H \sin θ}}{g}$

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

Explanation

$H = \frac{u^{2} \sin^{2} θ}{2 g}$ and $T = \frac{2 u \sin θ}{g}$ ⇒ $T^{2} = \frac{4 u^{2} \sin^{2} θ}{g^{2}}$

∴ $\frac{T^{2}}{H} = \frac{8}{g}$ ⇒ $T = \sqrt{\frac{8 H}{g}} = 2 \sqrt{\frac{2 H}{g}}$

● ● ● ● ●

The horizontal range of a projectile is $4 \sqrt{3}$ times its maximum height. Its angle of projection will be

A.

45°

B.

60°

C.

90°

D.

30°

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

Explanation

$R = 4 H \cot θ,$ if R = $4 \sqrt{3} H$ then $\cot θ = \sqrt{3}$ ⇒ $θ = 30 °$

● ● ● ● ●

The maximum horizontal range of a projectile is 400 m. The maximum value of height attained by it will be

A.

100 m

B.

200 m

C.

400 m

D.

800 m

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

Explanation

$R_{\max} = \frac{u^{2}}{g}$ = 400 m $(For θ = 45 °)$

$H_{\max} = \frac{u^{2}}{2 g} = \frac{400}{2} = 200 m$ $(For θ = 90 °)$

● ● ● ● ●

A particle is moving eastwards with velocity of 5 m/s. In 10 seconds the velocity changes to 5 m/s northwards. The average acceleration in this time is

A.

Zero

B.

$\frac{1}{\sqrt{2}} m / s^{2}$ toward north-west

C.

$\frac{1}{\sqrt{2}} m / s^{2}$ toward north-east

D.

$\frac{1}{2} m / s^{2}$ toward north-west

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

● ● ● ● ●

Two cars are moving in the same direction with the same speed 30 km/hr. They are separated by a distance of 5 km, the speed of a car moving in the opposite direction if it meets these two cars at an interval of 4 minutes, will be

A.

40 km/hr

B.

45 km/hr

C.

30 km/hr

D.

15 km/hr

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

Explanation

The two car (say A and B) are moving with same velocity, the relative velocity of one (say B) with respect to the other $A, {\vec{v}}_{B A} = {\vec{v}}_{B} - {\vec{v}}_{A} = v - v = 0$

So the relative separation between them (= 5 km) always remains the same.

Now if the velocity of car (say C) moving in opposite direction to A and B, is ${\vec{v}}_{C}$ relative to ground then the velocity of car C relative to A and B will be ${\vec{v}}_{r e l .} = {\vec{v}}_{C} - \vec{v}$

But as $\vec{v}$ is opposite to v_C

So $v_{r e l} = v_{c} - (- 30) = (v_{C} + 30) k m / h r .$

So, the time taken by it to cross the cars A and B $t = \frac{d}{v_{r e l}} \Rightarrow \frac{4}{60} = \frac{5}{v_{C} + 30}$

$\Rightarrow v_{C} = 45 k m / h r .$

● ● ● ● ●

A man standing on a road hold his umbrella at 30° with the vertical to keep the rain away. He throws the umbrella and starts running at 10 km/hr. He finds that raindrops are hitting his head vertically, the speed of raindrops with respect to the road will be

A.

10 km/hr

B.

20 km/hr

C.

30 km/hr

D.

40 km/hr

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

Explanation

When the man is at rest w.r.t. the ground, the rain comes to him at an angle 30° with the vertical. This is the direction of the velocity of raindrops with respect to the ground.

Here ${\vec{v}}_{r g} =$ velocity of rain with respect to the ground

${\vec{v}}_{m g} =$ velocity of the man with respect to the ground.

and ${\vec{v}}_{r m} =$ velocity of the rain with respect to the man,

We have ${\vec{v}}_{r g} = {\vec{v}}_{r m} + {\vec{v}}_{m g}$ ......(i)

Taking horizontal components equation (i) gives

$v_{r g} \sin 30 ° = v_{m g} = 10 k m / h r$

or $v_{r g} = \frac{10}{\sin 30 °} = 20 k m / h r$

Physics MCQs for NEET — Practice Questions with Answers

Frequently Asked Questions