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 man throws a ball vertically upward and it rises through 20 m and returns to his hands. What was the initial velocity (u) of the ball and for how much time (T) it remained in the air $[g = 10 m / s^{2}]$

A.

u = 10 m/s, T = 2s

B.

u = 10 m/s, T = 4s

C.

u = 20 m/s, T = 2s

D.

u = 20 m/s, T = 4s

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

Explanation

$u = \sqrt{2 g h} = \sqrt{2 \times 10 \times 20} = 20 m / s$

and $T = \frac{2 u}{g} = \frac{2 \times 20}{10} = 4$ sec

● ● ● ● ●

A particle when thrown, moves such that it passes from same height at 2s and 10s, the height is

A.

g

B.

2 g

C.

5 g

D.

10 g

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

Explanation

let initial velocity of the particle= u

after time $t_{1}$ :

$h = u t_{1} - \frac{1}{2} g {t_{1}}^{2} . . . . . . (1)$

after time $t_{2}$ :

$h = u t_{2} - \frac{1}{2} g {t_{2}}^{2} . . . . . . (2)$

From (1) & (2) :

$u (t_{1} - t_{2}) = \frac{1}{2} g ({t_{1}}^{2} - {t_{2}}^{2}) u = \frac{1}{2} g (t_{1} + t_{2})$

Put it in eqution (1) -

$h = \frac{1}{2} g t_{1} (t_{1} + t_{2}) - \frac{1}{2} g {t_{1}}^{2} h = \frac{1}{2} g t_{1} t_{2}$

If $t_{1}$ and $t_{2}$ are the time, when body is at the same height then, $h = \frac{1}{2} g t_{1} t_{2} = \frac{1}{2} \times g \times 2 \times 10 = 10 g$

● ● ● ● ●

Two balls A and B of same masses are thrown from the top of the building. A, thrown upward with velocity V and B, thrown downward with velocity V, then

A.

Velocity of A is more than B at the ground

B.

Velocity of B is more than A at the ground

C.

Both A & B strike the ground with same velocity

D.

None of these

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

Explanation

$v^{2} = u^{2} + 2 g h \Rightarrow v = \sqrt{u^{2} + 2 g h}$

so for both the cases velocity will be equal.

● ● ● ● ●

A ball is dropped from top of a tower of 100m height. Simultaneously another ball was thrown upward from bottom of the tower with a speed of 50 m/s ( $g = 10 m / s^{2}$ ). They will cross each other after

A.

1 s

B.

2 s

C.

3 s

D.

4 s

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

● ● ● ● ●

A cricket ball is thrown up with a speed of 19.6 ms^–1. The maximum height it can reach is

A.

9.8 m

B.

19.6 m

C.

29.4 m

D.

39.2 m

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

Explanation

$H_{\max} = \frac{u^{2}}{2 g} = \frac{19 .6 \times 19 .6}{2 \times 9 .8} = 19 .6 m$

● ● ● ● ●

A very large number of balls are thrown vertically upwards in quick succession in such a way that the next ball is thrown when the previous one is at the maximum height. If the maximum height is 5m, the number of ball thrown per minute is (take $g = 10 m s^{- 2}$ )

A.

120

B.

80

C.

60

D.

40

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

Explanation

Maximum height of ball = 5 m

So velocity of projection $\Rightarrow u = \sqrt{2 g h} = 10 m / s$

Time interval between two balls (time of ascent)

$= \frac{u}{g} = 1 s e c = \frac{1}{60} m i n$ .

So number of ball thrown per min. = 60

● ● ● ● ●

A body falling from a high Minaret travels 40 meters in the last 2 seconds of its fall to ground. Height of Minaret in meters is (take $g = 10 m s^{- 2}$ )

A.

60

B.

45

C.

80

D.

50

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

Explanation

Using the equation of motion: s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, t is the time, and a is the acceleration (g = -10 m/s^2). Given that the body travels 40 m in the last 2 s, we can substitute the values to find the initial height: 40 = 0 + (1/2)(-10)(2)^2 => h = 45 m.

● ● ● ● ●

A body falls from a height $h = 200 m$ (at New Delhi). The ratio of distance travelled in each 2 sec during t = 0 to t = 6 second of the journey is

A.

1 : 4 : 9

B.

1 : 2 : 4

C.

1 : 3 : 5

D.

1 : 2 : 3

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

Explanation

Distance travelled from t-0 to t=2 s,

$s_{1} = u t + \frac{1}{2} a t^{2} = 0 + \frac{1}{2} g (2)^{2} = 2 g D i s \tan c e t r a v e l l e d f r o m t = 2 s t o t = 4 s, D i s \tan c e t r a v e l l e d i n 4 s - D i s \tan c e t r a v e l l e d i n 2 s (s_{1}) = \frac{1}{2} g (4)^{2} - \frac{1}{2} g (2)^{2} = 12 \frac{g}{2} = 6 g D i s \tan c e t r a v e l l e d f r o m t = 4 s t o t = 6 s, D i s \tan c e t r a v e l l e d i n 6 s - D i s \tan c e t r a v e l l e d i n 4 s \frac{g (6)^{2}}{2} - \frac{g (4)^{2}}{2} = 10 g$

So distances are in ratio 1:3:5

● ● ● ● ●

A man drops a ball downside from the roof of a tower of height 400 meters. At the same time another ball is thrown upside with a velocity 50 meter/sec. from the surface of the tower, then they will meet at which height from the surface of the tower

A.

100 meters

B.

320 meters

C.

80 meters

D.

240 meters

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

● ● ● ● ●

Two balls are dropped from heights h and 2h respectively from the earth surface. The ratio of time of these balls to reach the earth is

A.

$1 : \sqrt{2}$

B.

$\sqrt{2}$ : 1

C.

2 : 1

D.

1 : 4

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

Explanation

$t = \sqrt{\frac{2 h}{g}} \Rightarrow \frac{t_{1}}{t_{2}} = \sqrt{\frac{h_{1}}{h_{2}}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$

Physics MCQs for NEET — Practice Questions with Answers

Frequently Asked Questions