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 particle moves in the x-y plane with velocity $v_{x} = 8 t - 2$ and $v_{y} = 2$ . If it passes through the point x=14 and y=4 at t=2 sec. The equation of the path is

$x = y^{2} - y + 2$

$x = y + 2$

$x = y^{2} + 2$

$x = y^{2} + y + 2$

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

Explanation

$v_{x} = 8 t - 2 v_{y} = 2 \frac{dx}{dt} = 8 t - 2 \frac{dy}{dt} = 2 \int dx = \int (8 t - 2) dt \int dy = \int 2 dt x = 4 t^{2} - 2 t + c y = 2 t + c_{1} At t = 2 s, x = 14 At t = 2 s, y = 4 14 = 4 {(2)}^{2} - 2 (2) + c 4 = 2 (2) + c_{1} c = 2 c_{1} = 0 x = 4 t^{2} - 2 t + 2 y = 2 t Putting t = \frac{y}{2} in x x = 4 {(\frac{y}{2})}^{2} - 2 (\frac{y}{2}) + 2 x = y^{2} - y + 2$

● ● ● ● ●

A motor boat of mass m moving along a lake with velocity $V_{0}$ . At t=0, the engine of the boat is shut down. Magnitude of resistance force offered to the boat is equal to rV. (V is instantaneous speed). What is the total distance covered till it stops completely? $[Hint : F (x) = mV \frac{dV}{dx} = - rV]$

${mV}_{0} / r$

$3 {mV}_{0} / 2 r$

${mV}_{0} / 2 r$

$2 {mV}_{0} / r$

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

Explanation

$mV \frac{dV}{dx} = - rV \frac{dV}{dx} = - \frac{r}{m} \int_{v_{o}}^{0} d v = \frac{- r}{m} \int_{0}^{s} d x s = \frac{m v_{o}}{r}$

● ● ● ● ●

A particle is moving along positive x-axis. Its position varies as $x = t^{3} - 3 t^{2} + 12 t + 20$ , where x is in meters and t is in seconds.

Initial velocity of the particle is

1 m/s

3 m/s

12 m/s

20 m/s

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

Explanation

$v = \frac{d x}{d t} = 3 t^{2} - 6 t + 12 A t t = 0 v = 12 m / s$

● ● ● ● ●

A particle is moving along positive x-axis. Its position varies as $x = t^{3} - 3 t^{2} + 12 t + 20$ , where x is in meters and t is in seconds.

Initial acceleration of the particle is

Zero

$1 m / s^{2}$

$- 3 m / s^{2}$

$- 6 m / s^{2}$

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

Explanation

$x = t^{3} - 3 t^{2} + 12 t + 20 v = \frac{d x}{d t} = 3 t^{2} - 6 t + 12 a = \frac{d v}{d t} = 6 t - 6 a t t = 0, a = - 6 m / s^{2}$

● ● ● ● ●

A particle is moving along positive x-axis. Its position varies as $x = t^{3} - 3 t^{2} + 12 t + 20$ , where x is in meters and t is in seconds.

Velocity of the particle when its acceleration zero is

1 m/s

3 m/s

6 m/s

9 m/s

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

Explanation

$v = \frac{d x}{d t} = 3 t^{2} - 6 t + 12 a = \frac{d v}{d t} = 6 t - 6 W h e n a = 0, 6 t - 6 = 0 \Rightarrow t = 1 s e c v a t 1 s e c . = 9 m / s$

● ● ● ● ●

Two forces ${\vec{F}}_{1} = 2 \hat{i} + 2 \hat{j} N$ and ${\vec{F}}_{2} = 3 \hat{j} + 4 \hat{k} N$ are acting on a particle.

The resultant force acting on particle is:

$2 \hat{i} + 5 \hat{j} + 4 \hat{k}$

$2 \hat{i} - 5 \hat{j} - 4 \hat{k}$

$\hat{i} - 3 \hat{j} - 2 \hat{k}$

$\hat{i} - \hat{j} - \hat{k}$

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

Explanation

Resultant force = $\vec{F_{1}} + \vec{F_{2}} = 2 \hat{i} + 5 \hat{j} + 4 \hat{k}$

● ● ● ● ●

Two forces ${\vec{F}}_{1} = 2 \hat{i} + 2 \hat{j} N$ and ${\vec{F}}_{2} = 3 \hat{j} + 4 \hat{k} N$ are acting on a particle.

The angle between ${\vec{F}}_{1} & {\vec{F}}_{2}$ is:

$θ = \cos^{- 1} (\frac{3}{2 \sqrt{5}})$

$θ = \cos^{- 1} (\frac{3}{5 \sqrt{2}})$

$θ = \cos^{- 1} (\frac{2}{3 \sqrt{5}})$

$θ = \cos^{- 1} (\frac{\sqrt{3}}{5})$

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

Explanation

angle between ${\vec{F}}_{1} & {\vec{F}}_{2}$ is given by-

$\cos θ = \frac{{\vec{F}}_{1} . {\vec{F}}_{2}}{F_{1} . F_{2}} = \frac{6}{2 \sqrt{2} \times 5} = \frac{3}{5 \sqrt{2}} \Rightarrow θ = \cos^{- 1} (\frac{3}{5 \sqrt{2}})$

● ● ● ● ●

Two forces ${\vec{F}}_{1} = 2 \hat{i} + 2 \hat{j} N$ and ${\vec{F}}_{2} = 3 \hat{j} + 4 \hat{k} N$ are acting on a particle.

The magnitude of the component of force ${\vec{F}}_{1}$ along force ${\vec{F}}_{2}$ is:

$\frac{5}{6}$

$\frac{5}{3}$

$\frac{6}{5}$

$\frac{5}{2}$

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

Explanation

Magnitude of the forcee $\vec{F_{1}} a l o n g \vec{F_{2}} = F_{1} \cos θ$ where $θ$ is the angle between two forces

$\Rightarrow F_{1} \cos θ = F_{1} \times \frac{\vec{F_{1}} . \vec{F_{2}}}{F_{1} F_{2}} = \frac{\vec{F_{1}} . \vec{F_{2}}}{F_{2}} = \frac{6}{5} N$

● ● ● ● ●

$A = 4 i + 4 j - 4 k$ and $B = 3 i + j + 4 k$ , then angle between vectors A and B is:

$180 °$

$90 °$

$45 °$

$0 °$

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

Explanation

The angle between two vector will be given by-

$\cos θ = \frac{\vec{A} . \vec{B}}{A B} = \frac{(4 \hat{i} + 4 \hat{j} - 4 \hat{k}) . (3 \hat{i} + \hat{j} + 4 \hat{k})}{\sqrt{{(4)}^{2} + {(4)}^{2} + {(- 4)}^{2}} \times \sqrt{{(3)}^{2} + {(1)}^{2} + {(4)}^{2}}} = \frac{0}{4 \sqrt{3} \times \sqrt{26}} = 0 \cos θ = \cos 90 ° θ = 90 °$

● ● ● ● ●

The acceleration of a particle starting from rest varies with time according to relation, $a = α t + β$ . Find the velocity of the particle at time instant t.

$α t^{2} + β t$

$α t^{2} + \frac{β t}{2}$

$\frac{α t^{2}}{2} + β t$

$2 α t^{2} + β t$

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

Explanation

Given that, $a = \frac{d v}{d t} = α t + β$

${}_{0}^{v}\int d v = {}_{0}^{t}\int (α t + β) d t = {[\frac{α t^{2}}{2} + β t]}_{0}^{t} \Rightarrow v = \frac{α t^{2}}{2} + β t$

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.