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

● ● ● ● ●

To a person, going eastward in a car with a velocity of 25 km/hr, a train appears to move towards north with a velocity of $25 \sqrt{3}$ km/hr. The actual velocity of the train will be

A.

25 km/hr

B.

50 km/hr

C.

5 km/hr

D.

$5 \sqrt{3}$ km/hr

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

Explanation

$v_{T} = \sqrt{v_{T C}^{2} + v_{C}^{2}}$ = $\sqrt{(25 {\sqrt{3)}}^{2} + {(25)}^{2}}$

= $\sqrt{1875 + 625}$ = $\sqrt{2500}$ = 50 km/hr

● ● ● ● ●

A bus is moving with a velocity 10 m/s on a straight road. A scooterist wishes to overtake the bus in 100 s. If the bus is at a distance of 1 km from the scooterist, with what velocity should the scooterist chase the bus

A.

50 m/s

B.

40 m/s

C.

30 m/s

D.

20 m/s

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

Explanation

Let the velocity of the scooterist =v

Relative velocity of scooterist with respect to bus = (v – 10)

$\Rightarrow S = (v - 10) \times 100 \Rightarrow 1000 = (v - 10) \times 100$

∴ $v = 10 + 10 = 20 m / s$

● ● ● ● ●

The x and y coordinates of the particle at any time are x=5t-2 $t^{2}$ and y=10t respectively, where x and y are in the metres and t is in seconds. The acceleration of the particle at t=2s is :

A.

0

B.

5 $m / s^{2}$

C.

-4 $m / s^{2}$

D.

-8 $m / s^{2}$

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

Explanation

(c) Given, x=5t-2 $t^{2}$

The velocity of the particle in the x-direction,

$v_{x} = \frac{d x}{d t} = \frac{d}{d t} (5 t - 2 t^{2}) = 5 - 4 t$

Acceleration, $a_{x} = \frac{d v_{x}}{d t} = - 4 m s^{- 2}$

The velocity of the particle in the y-direction,

$v_{y} = \frac{d y}{d t} = 10 ∴ A c c e l e r a t i o n a_{y} = \frac{d v_{y}}{d t} = 0 ∴ N e t a c c e l e r a t i o n o f t h e p a r t i c l e, {\vec{a}}_{n e t} = a_{x} \hat{i} + a_{y} \hat{j} = (- 4 m s^{2}) \hat{i}$

or $a_{n e t} = - 4 m s^{- 2}$

● ● ● ● ●

Preeti reached the metro station and found that the escalator was not working. She walked up the stationary escalator in time $t_{1}$ . On other days, if she remains stationary on the moving escalator, then the escalator takes her up in time $t_{2}$ . The time taken by her to walk up on the moving escalator will be

A.

$\frac{t_{1} - t_{2}}{2}$

B.

$\frac{t_{1} t_{2}}{t_{2} - t_{1}}$

C.

$\frac{t_{1} t_{2}}{t_{2} + t_{1}}$

D.

$t_{1} - t_{2}$

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

Explanation

(c)

$S p e e d o f P r e e t i, v_{1} = \frac{h}{t_{1}} S p e e d o f e s c a l a t o r, v_{2} = \frac{h}{t_{2}} S p e e d o f P r e e t i w i t h e s c a l a t o r = v_{1} + v_{2} (V e l o c i t y o f b o t h w i l l c o m b i n e t o g i v e h i g h v e l o c i t y) T i m e = \frac{h}{v_{1} + v_{2}} = \frac{h}{\frac{h}{t_{1}} + \frac{h}{t_{2}}} = \frac{t_{1} t_{2}}{t_{1} + t_{2}}$

● ● ● ● ●

A particle moves so that its position vector is given by $r = c o s ω t \hat{x} + s i n ω t \hat{y} w h e r e ω$ is a constant. Which of the following is true?

A.

Velocity and acceleration both are parallel to r.

B.

Velocity is perpendicular to r and acceleration is directed towards to origin.

C.

Velocity is not perpendicular to r and acceleration is directed away from the origin.

D.

Velocity and acceleration both are perpendicular to r.

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

● ● ● ● ●

A body is moving with velocity 30 m/s towards east. After 10 s its velocity becomes 40 m/s towards north. The average acceleration of the body is

A.

$7 m / s^{2}$

B.

$\sqrt{7} m / s^{2}$

C.

$5 m / s^{2}$

D.

$1 m / s^{2}$

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

Explanation

Average acceleration $= \frac{Change in velocity}{Total time}$

$a = \frac{| v_{f} - v_{i} |}{∆ t} = \frac{| 40 j - 30 i |}{∆ t} = \frac{\sqrt{40^{2} + 30^{2}}}{10} = \frac{\sqrt{1600 + 900}}{10} = 5 {ms}^{- 2}$

● ● ● ● ●

A particle moves in the x-y plane according to rule $x = a \sin ωt$ and $y = a \cos ωt$ . The particle follows

A.

an elliptical path

B.

a circular path

C.

a parabolic path

D.

a straight line path inclined equally to x and y-axis

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

Explanation

$x = a \sin ωt \Rightarrow \frac{x}{a} = \sin ωt x = a \cos ωt \Rightarrow \frac{y}{a} = \cos ωt \to x^{2} + y^{2} = a^{2}$

This is an equation of a circle, so the particle follows a circular path.

● ● ● ● ●

The coordinates of a moving particle at a time t, are given by, x=5 sin 10t, y=5 cos 10t. The speed of the particle is:

A.

25

B.

50

C.

10

D.

$50 \sqrt{2}$

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

Explanation

(B)

$x = 5 \sin 10 t y = 5 \cos 10 t v_{x} = \frac{dx}{dt} v_{y} = \frac{dy}{dt} = 5 \times 10 \cos l 0 t = - 50 \sin l 0 t = 50 \cos l 0 t Velocity \vec{v} = v_{x} \hat{i} + v_{y} \hat{j} = 50 \cos l 0 t \hat{i} - 50 \sin l 0 t \hat{j} Speed = Magnitude of velocity = \sqrt{v_{x}^{2} + v_{y}^{2}} = \sqrt{{(50 \cos l 0 t)}^{2} + {(- 50 \sin l 0 t)}^{2}} = 50 \sqrt{\cos^{2} 10 t + \sin^{2} 10 t} units = 50 units$

● ● ● ● ●

For a rocket moving in free space, the fraction of mass to be disposed of Off to attain a speed equal to two times the exhaust speed is given by (given $e^{2}$ = 7.4)

A.

0.40

B.

0.37

C.

0.50

D.

0.86

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

Explanation

$\begin{aligned} v & = v_{r} \ln (\frac{m_{0}}{m}) \end{aligned} o r \begin{aligned} 2 v_{r} & = v_{r} \ln (\frac{m_{0}}{m}) \end{aligned} ∴ \begin{aligned} 2 & = \ln (\frac{m_{0}}{m}) \end{aligned} ∴ \begin{aligned} \frac{m_{0}}{m} & = e^{2} = 7.4 \end{aligned} o r \begin{aligned} \frac{m}{m_{0}} & = 0.14 \end{aligned}$

This is a fraction of mass remaining. Hence fraction of mass disposed of = 0.86.

● ● ● ● ●

A simple pendulum hangs from the roof of a train moving on horizontal rails. If the string is inclined towards the front of the train, then train is

A.

Moving with constant velocity

B.

In accelerated motion

C.

In retarded motion

D.

At rest

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

Explanation

Using concept of inertia

Physics MCQs for NEET — Practice Questions with Answers

Frequently Asked Questions