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

● ● ● ● ●

What average horsepower is developed by an 80 kg man while climbing in 10 s a flight of stairs that rises 6 m vertically

A.

0.63 HP

B.

1.26 HP

C.

1.8 HP

D.

2.1 HP

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

Explanation

$p = \frac{m g h}{t} = \frac{80 \times 9 .8 \times 6}{10} W = \frac{470}{746} H P = 0 .63 H P$

● ● ● ● ●

A quarter horse power motor runs at a speed of 600 r.p.m. Assuming 40% efficiency, the work done by the motor in one rotation will be

A.

7.46 J

B.

7400 J

C.

7.46 ergs

D.

74.6 J

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

Explanation

Motor makes 600 revolution per minute

∴ n = $600 \frac{revolution}{minute} = 10 \frac{r e v}{\sec}$

∴ Time required for one revolution $= \frac{1}{10}$ sec

Energy required for one revolution = power × time

= $\frac{1}{4} \times 746 \times \frac{1}{10} = \frac{746}{40} J$

But work done = 40% of input

$= 40 % \times \frac{746}{40} = \frac{40}{100} \times \frac{746}{40} = 7 .46 J$

● ● ● ● ●

An engine pumps up 100 kg of water through a height of 10 m in 5 s. Given that the efficiency of the engine is 60% . If g = 10 ms^–2, the power of the engine is

A.

3.3 kW

B.

0.33 kW

C.

0.033 kW

D.

33 kW

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

Explanation

Work output of engine = mgh = $100 \times 10 \times 10 = 10^{4} J$

Efficiency ( $η$ ) = $\frac{output}{input}$

∴ Input energy = $\frac{outupt}{η}$

$= \frac{10^{4}}{60} \times 100 = \frac{10^{5}}{6} J$

∴ Power = $\frac{input energy}{time}$ = $\frac{10^{5} / 6}{5} = \frac{10^{5}}{30} = 3 .3 k W$

● ● ● ● ●

A force of $2 \hat{i} + 3 \hat{j} + 4 \hat{k} N$ acts on a body for 4 second, produces a displacement of $(3 \hat{i} + 4 \hat{j} + 5 \hat{k}) m .$ The power used is-

A.

9.5 W

B.

7.5 W

C.

6.5 W

D.

4.5 W

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

Explanation

$P = \frac{\vec{F} . \vec{s}}{t} = \frac{(2 \hat{i} + 3 \hat{j} + 4 \hat{k}) . (3 \hat{i} + 4 \hat{j} + 5 \hat{k})}{4} = \frac{38}{4} = 9 .5 W$

● ● ● ● ●

An engine pump is used to pump a liquid of density ρ continuously through a pipe of cross-sectional area A. If the speed of flow of the liquid in the pipe is v, then the rate at which kinetic energy is being imparted to the liquid is

A.

$\frac{1}{2} A ρ v^{3}$

B.

$\frac{1}{2} A ρ v^{2}$

C.

$\frac{1}{2} A ρ v$

D.

$A ρ v$

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

Explanation

Energy supplied to liquid per second by the pump

= $\frac{1}{2} \frac{m v^{2}}{t}$ = $\frac{1}{2} \frac{V ρ v^{2}}{t}$ = $\frac{1}{2} A \times (\frac{l}{t}) \times ρ \times v^{2}$ $[\frac{l}{t} = v]$

$= \frac{1}{2} A \times v \times ρ \times v^{2}$ = $\frac{1}{2} A ρ v^{3}$

● ● ● ● ●

A uniform chain of length L and mass M is lying on a smooth table and one third of its length is hanging vertically down over the edge of the table. If g is acceleration due to gravity, the work required to pull the hanging part on to the table is

A.

MgL

B.

MgL/3

C.

MgL/9

D.

MgL/18

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

Explanation

The hanging part of the chain has a length L/3 and mass M/3. The work required to lift it vertically through a height L/3 is (M/3)g(L/3) = MgL/9. However, since the chain is pulled horizontally, the work done is half of this value, which is MgL/18.

● ● ● ● ●

A particle of mass m is moving in a horizontal circle of radius r under a centripetal force equal to –K/r², where K is a constant. The total energy of the particle is

A.

$\frac{K}{2 r}$

B.

$- \frac{K}{2 r}$

C.

$- \frac{K}{r}$

D.

$\frac{K}{r}$

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

Explanation

Here $\frac{m v^{2}}{r} = \frac{K}{r^{2}}$

∴ K.E. $= \frac{1}{2} m v^{2} = \frac{K}{2 r}$

$U = - \int_{\infty}^{r} F . d r = - \int_{\infty}^{r} (- \frac{K}{r^{2}}) d r = - \frac{K}{r}$

Total energy $E = K.E. + P.E. = \frac{K}{2 r} - \frac{K}{r} = - \frac{K}{2 r}$

● ● ● ● ●

The displacement x of a particle moving in one dimension under the action of a constant force is related to the time t by the equation $t = \sqrt{x} + 3$ , where x is in meters and t is in seconds. The work done by the force in the first 6 seconds is

A.

9 J

B.

6 J

C.

0 J

D.

3 J

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

Explanation

$x = {(t - 3)}^{2}$ ⇒ $v = \frac{d x}{d t} = 2 (t - 3)$

at $t = 0$ ; $v_{1} = - 6 m / s$ and at $t = 6 \sec$ , $v_{2} = 6 m / s$

so, change in kinetic energy $= W = \frac{1}{2} m v_{2}^{2} - \frac{1}{2} m v_{1}^{2} = 0$

● ● ● ● ●

A force $F = - K (y i + x j)$ (where K is a positive constant) acts on a particle moving in the xy-plane. Starting from the origin, the particle is taken along the positive x-axis to the point (a, 0) and then parallel to the y-axis to the point (a, a). The total work done by the force F on the particles is

A.

$- 2 K a^{2}$

B.

$2 K a^{2}$

C.

$- K a^{2}$

D.

$K a^{2}$

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

Explanation

While moving from (0,0) to (a,0)

Along positive x-axis, y = 0

$∴ \vec{F} = - k x \hat{j}$

i.e. force is in negative y-direction while displacement is in positive x-direction.

∴ W₁ = 0

Because force is perpendicular to displacement

Then particle moves from (a, 0) to (a, a) along a line parallel to y-axis (x = +a) during this $\vec{F} = - k (y \hat{i} + a \hat{J})$

The first component of force, $- k y \hat{i}$ will not contribute any work because this component is along negative x-direction $(- \hat{i})$ while displacement is in positive y-direction (a,0) to (a,a). The second component of force i.e. $- k a \hat{j}$ will perform negative work

∴ $W_{2} = (- k a \hat{j}) (a \hat{j})$ = $(- k a) (a) = - k a^{2}$

So net work done on the particle W = W₁ + W₂

= $0 + (- k a^{2}) = - k a^{2}$

● ● ● ● ●

If g is the acceleration due to gravity on the earth's surface, the gain in the potential energy of an object of mass m raised from the surface of earth to a height equal to the radius of the earth R, is

A.

$\frac{1}{2} m g R$

B.

2 mgR

C.

mgR

D.

$\frac{1}{4} m g R$

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

Explanation

Gain in potential energy $Δ U = \frac{m g h}{1 + \frac{h}{R}}$

If h = R then $Δ U = \frac{m g R}{1 + \frac{R}{R}} = \frac{1}{2} m g R$

Physics MCQs for NEET — Practice Questions with Answers

Frequently Asked Questions