Blog Log in Get Started Free

Laws of Motion MCQs for NEET — Physics Questions with Answers

Practice free Laws of Motion (Physics) NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

Chapter Difficulty Show only questions I answered incorrectly Clear Register free for difficulty & keyword filters

● ● ● ● ●

A lift is moving down with acceleration a. A man in the lift drops a ball inside the lift. The acceleration of the ball as observed by the man in the lift and a man standing stationary on the ground are respectively

g, g

g – a, g – a

g – a , g

a, g

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

Explanation

Due to relative motion, acceleration of ball observed by observer in lift = (g – a) and for man on earth the acceleration remains g.

● ● ● ● ●

A monkey of mass 20kg is holding a vertical rope. The rope will not break when a mass of 25 kg is suspended from it but will break if the mass exceeds 25 kg. What is the maximum acceleration with which the monkey can climb up along the rope (g = 10 m/s²)

10 m/s²

25 m/s²

2.5 m/s²

5 m/s²

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

Explanation

Tension the string $= m (g + a) =$ Breaking force

⇒ $20 (g + a) = 25 \times g$ ⇒ $a = g / 4 = 2 .5 m / s^{2}$

● ● ● ● ●

A plumb line is suspended from a ceiling of a car moving with horizontal acceleration of a. What will be the angle of inclination with vertical

tan^–1(a/g)

tan^–1(g/a)

cos^–1(a/g)

cos^–1(g/a)

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

● ● ● ● ●

A block of mass m is placed on a smooth wedge of inclination θ. The whole system is accelerated horizontally so that the block does not slip on the wedge. The force exerted by the wedge on the block (g is acceleration due to gravity) will be

$m g \cos θ$

$m g \sin θ$

$m g / \cos θ$

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

Explanation

When the whole system is accelerated towards left then pseudo force (ma) works on a block towards right.

For the condition of equilibrium

$m g \sin θ = m a \cos θ$ ⇒ $a = \frac{g \sin θ}{\cos θ}$

∴ Force exerted by the wedge on the block

$R = m g \cos θ + m a \sin θ$

R $= m g \cos θ + m (\frac{g \sin θ}{\cos θ}) \sin θ = \frac{m g (\cos^{2} θ + \sin^{2} θ)}{\cos θ}$

R $= \frac{m g}{\cos θ}$

● ● ● ● ●

The linear momentum p of a body moving in one dimension varies with time according to the equation p = a + bt² where a and b are positive constants. The net force acting on the body is

A constant

Proportional to t²

Inversely proportional to t

Proportional to t

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

Explanation

$F = \frac{d p}{d t} \equiv \frac{d}{d t} (a + b t^{2}) = 2 b t$

∴ $F \propto t$

● ● ● ● ●

A man of weight 80 kg is standing in an elevator which is moving with an acceleration of 6 m/s² in upward direction. The apparent weight of the man will be (g = 10 m/s²)

1480 N

1280 N

1380 N

None of these

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

Explanation

The apparent weight of man,

$R = m (g + a) = 80 (10 + 6) = 1280 N$

● ● ● ● ●

N bullets each of mass m kg are fired with a velocity v ms^–1 at the rate of n bullets per second upon a wall. The reaction offered by the wall to the bullets is given by

nmv

$\frac{N m v}{n}$

$n \frac{N m}{v}$

$n \frac{N v}{m}$

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

Explanation

Total mass of bullets = Nm, time $t = \frac{N}{n}$

Momentum of the bullets striking the wall = Nmv

Rate of change of momentum (Force) = $\frac{N m v}{t}$ = nmv.

● ● ● ● ●

With what minimum acceleration can a fireman slides down a rope while breaking strength of the rope is $\frac{2}{3}$ of his weight

$\frac{2}{3} g$

$\frac{1}{3} g$

Zero

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

Explanation

If man slides down with some acceleration then its apparent weight decreases. For critical condition rope can bear only 2/3 of his weight. If a is the minimum acceleration then,

Tension in the rope $= m (g - a)$ = Breaking strength

⇒ $m (g - a) = \frac{2}{3} m g$ ⇒ $a = g - \frac{2 g}{3} = \frac{g}{3}$

● ● ● ● ●

A ball of mass m moves with speed v and it strikes normally with a wall and reflected back normally, if its time of contact with wall is t then find force exerted by ball on wall

$\frac{2 m v}{t}$

$\frac{m v}{t}$

mvt

$\frac{m v}{2 t}$

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

Explanation

For exerted by ball on wall

= rate of change in momentum of ball

= $\frac{m v - (- m v)}{t} = \frac{2 m v}{t}$

● ● ● ● ●

A body of mass 5 kg starts from the origin with an initial velocity $\vec{u} = 30 \hat{i} + 40 \hat{j} m s^{- 1}$ . If a constant force $\vec{F} = - (\hat{i} + 5 \hat{j}) N$ acts on the body, the time in which the y–component of the velocity becomes zero is

5 seconds

20 seconds

40 seconds

80 seconds

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

Explanation

$u_{y} = 40 m / s$ , $F_{y} = - 5 N$ , $m = 5 k g$ .

So $a_{y} = \frac{F_{y}}{m} = - 1 m / s^{2}$ (As v = u + at)

∴ $v_{y} = 40 - 1 \times t = 0$ ⇒ $t = 40 \sec$ .

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

Yes. You can attempt every Laws of Motion 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.