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

● ● ● ● ●

Consider elastic collision of a particle of mass m moving with a velocity u with another particle of the same mass at rest. After the collision the projectile and the struck particle move in directions making angles θ₁ and θ₂respectively with the initial direction of motion. The sum of the angles. θ₁ + θ₂, is

A.

45°

B.

90°

C.

135°

D.

180°

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

Explanation

If the masses are equal and target is at rest and after collision both masses moves in different direction. Then angle between direction of velocity will be 90°, if collision is elastic.

● ● ● ● ●

A rope is wound around a hollow cylinder of mass 3 kg and radius 40cm. What is the angular acceleration of the cylinder,if the rope is pulled with a force of 30 N?

A.

25 $m / s^{2}$

B.

0.25 $r a d / s^{2}$

C.

25 $r a d / s^{2}$

D.

5 $m / s^{2}$

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

Explanation

The force applied on the rope provides a torque about the axis of the cylinder, given by F Ã— r = 30 Ã— 0.4 = 12 Nm. The moment of inertia of a hollow cylinder about its axis is MR^2/2. Using torque = I Ã— Î±, we get Î± = 12/(3Ã—0.4^2/2) = 25 rad/s^2.

● ● ● ● ●

Two discs of same moment of inertia rotating about their regular axis passing through centre and perpendicular to the plane of disc with angular velocities $ω_{1} a n d ω_{2}$ . They are brought into contact face to face coinciding the axis of rotation. The expression for loss of energy during this proces is

A.

$\frac{1}{2} I {(ω_{1} + ω_{2})}^{2}$

B.

$\frac{1}{4} I {(ω_{1} - ω_{2})}^{2}$

C.

$I {(ω_{1} - ω_{2})}^{2}$

D.

$\frac{1}{8} I {(ω_{1} - ω_{2})}^{2}$

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

Explanation

(d) Thinking Process

When no external torque acts on system then, angular momentum of system remains constant.

Angular momentum before contact = $I_{1} ω_{1} + I_{2} ω_{2}$

Angular momentum after the discs brought into contact.

$= I_{n e t} ω = (I_{1} + I_{2}) ω$

So, final angular speed of system= $ω$

$= \frac{I_{1} ω_{1} + I_{2} ω_{2}}{I_{1} + I_{2}}$

Now, to calculate loss of energy, we subtract initial and final energies of system. $\Rightarrow L o s s o f e n e r g y = \frac{1}{2} I ω_{1}^{2} + \frac{1}{2} I ω_{2}^{2} - \frac{1}{2} (2 I) ω^{2} = \frac{1}{4} I {(ω_{1} - ω_{2})}^{2}$

● ● ● ● ●

A bullet of mass 10g moving horizontal with a velocity of 400 m/s strikes a wood block of mass 2 kg which is suspended by light inextensible string of length 5 m. As result, the centre of gravity of the block found to rise a vertical distance of 10 cm. The speed of the bullet after it emerges of horizontally from the block wiil be

A.

100 m/s

B.

80 m/s

C.

120 m/s

D.

160 m/s

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

Explanation

(c) According to the law of consevation of momentum.

$p_{i} = p_{f} \Rightarrow (0.01) \times 400 + 0 = 2 v + (0.01) v^{'} . . (i)$

Also velocity v of the block just after the collision is

$v = \sqrt{2 g h} = \sqrt{2 \times 10 \times 0.1} = \sqrt{2} . . . (i i)$

$\Rightarrow F r o m E q s . (i) a n d (i i), w e h a v e v^{'} = 120 m / s$

● ● ● ● ●

Two identical balls A and B having velocities of 0.5 m/s and -0.3 m/s respectively collide elastically in one dimension. The velocities of B and A after the collision respectively will be

A.

-0.5 m/s and 0.3 m/s

B.

0.5 m/s and -0.3 m/s

C.

-0.3 m/s and 0.5 m/s

D.

0.3 m/s and 0.5 m/s

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

Explanation

(b) Key Idea :

In elastic collision,kinetic energy of the system remains unchanged and momentum is also conserved.

It is given that mass of balls are same and collision is perfectly elastic (e=1) so their velocities will be interchanged.

Thus, $v_{A}^{'} = v_{B} = - 0.3 m / s, v_{B}^{'} = v_{A} = 0.5 m / s$

● ● ● ● ●

Two rotating bodies A and B of masses m and 2m with moments of inertia $I_{A}$ and $I_{B} (I_{B} > I_{A})$ have equal kinetic energy of rotation. If $L_{A}$ and $L_{B}$ be their angular momenta respectively, then

A.

$L_{A} = \frac{L_{B}}{2}$

B.

$L_{A} = 2 L_{B}$

C.

$L_{B} > L_{A}$

D.

$L_{A} > L_{B}$

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

Explanation

(c) As we know that, the kinetic energy of a rotating body,

$K E = \frac{1}{2} I ω^{2} = \frac{1}{2} \frac{I^{2} ω^{2}}{I} = \frac{L^{2}}{2 l}$

Also, angular momentum, $L = l ω$

Thus, $K_{A} = K_{B}$

$\Rightarrow \frac{1}{2} \frac{L_{A}^{2}}{I_{A}} = \frac{1}{2} \frac{L_{B}^{2}}{I_{B}} \Rightarrow {(\frac{L_{A}}{L_{B}})}^{2} = \frac{I_{A}}{I_{B}} \Rightarrow \frac{L_{A}}{L_{B}} = \sqrt{\frac{I_{A}}{I_{B}}}$

$L \propto \sqrt{I} ∴ L_{A} < L_{B [∵ I_{B} > I_{A}]}$

● ● ● ● ●

A solid sphere of mass m and radius R is rotating about its diameter. A soild cyclinder of the same mass and same radius is also rotating about its geometrical axis with an angular speed twice that of the sphere. The ratio of their kinetic energies of rotation $(E_{s p h e r e} / E_{c y l i n d e r})$ will be

A.

2:3

B.

1:5

C.

1:4

D.

3:1

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

Explanation

(b) Key Idea KE of a rotating rigid body, KE= $\frac{1}{2} I ω^{2}$

$∴ K E$ of sphere, $K_{s} = \frac{1}{2} I ω_{1}^{2}$

= $\frac{1}{2} \times \frac{2}{5} m R^{2} ω_{2}^{2} = \frac{1}{5} m R^{2} ω_{1}^{2}$

KE of cylinder, $K_{c} = \frac{1}{2} \frac{1}{2} m R^{2} ω_{2}^{2} = \frac{1}{4} m R^{2} ω_{2}^{2}$

$∴ \frac{K_{s}}{K_{c}} = \frac{\frac{m R^{2} ω_{1}^{2}}{5}}{\frac{m R^{2} ω_{1}^{2}}{4}} = \frac{4}{5} \frac{ω_{1}^{2}}{ω_{2}^{2}}$

$= \frac{4}{5} \frac{ω_{1}^{2}}{{(2 ω_{1})}^{2}} = \frac{1}{5} (g i v e n, ω_{2} = 2 ω_{1})$

● ● ● ● ●

From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre ?

A.

$13 M R^{2} / 32$

B.

$11 M R^{2} / 32$

C.

$9 M R^{2} / 32$

D.

$15 M R^{2} / 32$

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

Explanation

The moment of inertia of a solid disc about its perpendicular axis through the center is MR^2/2. The moment of inertia of a circular hole of radius R/2 is (MÏ€R^4)/(64Ï€(R/2)^2) = MR^2/16. Therefore, the remaining part has a moment of inertia of MR^2/2 - MR^2/16 = 13MR^2/32.

● ● ● ● ●

Two particles A and B. move with constant velocities v₁ and v₂. At the initial moment, their position vectors are $r_{1}$ and $r_{2}$ respectively. The condition for particles A and B for their collision is-

A.

$\frac{r_{1} - r_{2}}{|r_{1} - r_{2}|} = \frac{v_{2} - v_{1}}{|v_{2} - v_{1}|}$

B.

$r_{1} \cdot v_{1} = r_{2} \cdot v_{2}$

C.

_{$r_{1} \times v_{1} = r_{2} \times v_{2}$}

D.

r₁-r₂=v₁-v₂

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

● ● ● ● ●

A force F= $a \hat{i}$ $+ 3 \hat{j}$ $+ 6 \hat{k}$ is acting at a point r= $2 \hat{i}$ - $6 \hat{j}$ -12 $\hat{k}$ . The value of α for which angular momentum about origin is conserved is

A.

-1

B.

2

C.

zero

D.

1

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

Explanation

Key Concept : When the resultant external torque acting on a system is zero, the total angular momentum of a system remains constant. This is the principle of the conservation of angular momentum.

Given, force F=α $\hat{i}$ $+ 3 \hat{j}$ $+ 6 \hat{k}$ is acting at a point r= $2 \hat{i}$ - $6 \hat{j}$ -12 $\hat{k}$
As, angular momentum about origin is conserved i.e. τ=constant

=> Torque,τ=0

=> r x F=0

| $\hat{i}$ $\hat{j}$ $\hat{k}$ |
|2-6 -12| =0
|α 3 6|

=> (-36+36) $\hat{i}$ -(12+12α) $\hat{j}$ +(6+6α) $\hat{k}$ =0

=> 0 $\hat{i}$ -12(1+α) $\hat{j}$ +6(1+α) $\hat{k}$ =0

=> 6(1+α)=0=> α=-1

So, value of α for angular momentum about origin is conserved, α=-1

NEET Practice Questions (MCQs) with Answers & Solutions

Frequently Asked Questions