Are these NEET questions free to attempt?

Yes. You can attempt every NEET 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 NEET 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.

NEET NEET Questions — Free MCQ Practice

● ● ● ● ●

A moving body of mass m and velocity 3 km/h collides with a rest body of mass 2m and sticks to it. Now the combined mass starts to move. What will be the combined velocity

A.

3 km/h

B.

2 km/h

C.

1 km/h

D.

4 km/h

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

Explanation

Converting the given velocities to SI units, the initial velocity of the moving body is (3 km/h) = (3/3.6) m/s = 5/6 m/s. Using conservation of momentum, (m*(5/6) + 2m0) = 3mv', where v' is the final velocity of the combined mass. Solving this, we get v' = (5/18) m/s = 1 km/h.

● ● ● ● ●

If a skater of weight 3 kg has initial speed 32 m/s and second one of weight 4 kg has 5 m/s. After collision, they have speed (couple) 5 m/s. Then the loss in K.E. is

A.

48 J

B.

96 J

C.

Zero

D.

None of these

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

Explanation

Loss in K.E. = (initial K.E. – Final K.E.) of system

$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} u_{2}^{2} - \frac{1}{2} (m_{1} + m_{2}) V^{2}$

$= \frac{1}{2} 3 \times {(32)}^{2} + \frac{1}{2} \times 4 \times {(5)}^{2} - \frac{1}{2} \times (3 + 4) \times {(5)}^{2}$

= 986.5 J

● ● ● ● ●

A metal ball of mass 2 kg moving with a velocity of 36 km/h has an head on collision with a stationary ball of mass 3 kg. If after the collision, the two balls move together, the loss in kinetic energy due to collision is

A.

40 J

B.

60 J

C.

100 J

D.

140 J

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

Explanation

$v = 36 k m / h = 10 m / s$

By law of conservation of momentum

$2 \times 10 = (2 + 3) V$ ⇒ V = 4 m/s

Loss in K.E. $= \frac{1}{2} \times 2 \times {(10)}^{2} - \frac{1}{2} \times 5 \times {(4)}^{2} = 60 J$

● ● ● ● ●

A body of mass 2kg is moving with velocity 10 m/s towards east. Another body of same mass and same velocity moving towards north collides with former and coalsces and moves towards north-east. Its velocity is

A.

10 m/s

B.

5 m/s

C.

2.5 m/s

D.

$5 \sqrt{2} m / s$

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

Explanation

Initial momentum = $\vec{P} = m v \hat{i} + m v \hat{j}$

$| \vec{P} | = \sqrt{2} m v$

Final momentum = 2m × V

By the law of conservation of momentum

$2 m \times V = \sqrt{2} m v$ ⇒ $V = \frac{v}{\sqrt{2}}$

In the problem v = 10 m/s (given)

∴ $V = \frac{10}{\sqrt{2}} = 5 \sqrt{2} m / s$

● ● ● ● ●

Which of the following is not a perfectly inelastic collision

A.

Striking of two glass balls

B.

A bullet striking a bag of sand

C.

An electron captured by a proton

D.

A man jumping onto a moving cart

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

Explanation

Because in perfectly inelastic collision the colliding bodies stick together and move with common velocity

● ● ● ● ●

A neutron having mass of $1 .67 \times 10^{- 27} k g$ and moving at $10^{8} m / s$ collides with a deutron at rest and sticks to it. If the mass of the deutron is $3 .34 \times 10^{- 27} k g$ then the speed of the combination is

A.

$2 .56 \times 10^{3} m / s$

B.

$2 .98 \times 10^{5} m / s$

C.

$3 .33 \times 10^{7} m / s$

D.

$5 .01 \times 10^{9} m / s$

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

Explanation

According to law of conservation of momentum.

Momentum of neutron = Momentum of combination

⇒ $1 .67 \times 10^{- 27} \times 10^{8} = (1 .67 \times 10^{- 27} + 3 .34 \times 10^{- 27}) v$

∴ $v = 3 .33 \times 10^{7} m / s$

● ● ● ● ●

A body of mass m₁ is moving with a velocity V. It collides with another stationary body of mass m₂. They get embedded. At the point of collision, the velocity of the system

A.

Increases

B.

Decreases but does not become zero

C.

Remains same

D.

Become zero

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

Explanation

By momentum conservation before and after collision.

$m_{1} V + m_{2} \times 0 = (m_{1} + m_{2}) v$ ⇒ $v = \frac{m_{1}}{m_{1} + m_{2}} V$

i.e. Velocity of system is less than V.

● ● ● ● ●

A bullet of mass m moving with velocity v strikes a block of mass M at rest and gets embedded into it. The kinetic energy of the composite block will be

A.

$\frac{1}{2} m v^{2} \times \frac{m}{(m + M)}$

B.

$\frac{1}{2} m v^{2} \times \frac{M}{(m + M)}$

C.

$\frac{1}{2} m v^{2} \times \frac{(M + m)}{M}$

D.

$\frac{1}{2} M v^{2} \times \frac{m}{(m + M)}$

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

Explanation

By conservation of momentum, $m v + M \times 0 = (m + M) V$

Velocity of composite block $V = (\frac{m}{m + M}) v$

K.E. of composite block $= \frac{1}{2} (M + m) V^{2}$

$= \frac{1}{2} (M + m) {(\frac{m}{M + m})}^{2} v^{2} = \frac{1}{2} m v^{2} (\frac{m}{m + M})$

● ● ● ● ●

A shell is fired from a cannon with velocity v m/sec at an angle θ with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon. The speed (in m/sec) of the other piece immediately after the explosion is

A.

3v cos θ

B.

2v cos θ

C.

$\frac{3}{2} v \cos θ$

D.

$\frac{\sqrt{3}}{2} v \cos θ$

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

● ● ● ● ●

Two particles of masses m₁ and m₂ in projectile motion have velocities ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ respectively at time t = 0. They collide at time t₀. Their velocities become ${\vec{v}}_{1}'$ and ${\vec{v}}_{2}'$ at time 2t₀ while still moving in air. The value of $|(m_{1} \vec{v_{1}}' + m_{2} \vec{v_{2}}') - (m_{1} \vec{v_{1}} + m_{2} \vec{v_{2}})|$ is

A.

Zero

B.

$(m_{1} + m_{2}) g t_{0}$

C.

$2 (m_{1} + m_{2}) g t_{0}$

D.

$\frac{1}{2} (m_{1} + m_{2}) g t_{0}$

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

Explanation

The momentum of the two-particle system, at t = 0 is

${\vec{P}}_{i} = m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2}$

Collision between the two does not affect the total momentum of the system.

A constant external force $(m_{1} + m_{2}) g$ acts on the system.

The impulse given by this force, in time t = 0 to $t = 2 t_{0}$ is $(m_{1} + m_{2}) g \times 2 t_{0}$

∴ |Change in momentum in this interval|

$= |m_{1} \vec{v}'_{1} + m_{2} \vec{v}'_{2} - (m_{1} {\vec{v}}_{1} + m_{2} {\vec{v}}_{2})| = 2 (m_{1} + m_{2}) g t_{0}$

NEET Practice Questions (MCQs) with Answers & Solutions

Frequently Asked Questions