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 body takes just twice the time as long to slide down a plane inclined at 30^o to the horizontal as if the plane were frictionless. The coefficient of friction between the body and the plane is

A.

$\frac{\sqrt{3}}{4}$

B.

$\sqrt{3}$

C.

$\frac{4}{3}$

D.

$\frac{3}{4}$

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

Explanation

$W h e n t h e p l a n e i s s m o o t h : t = \sqrt{\frac{2 l}{a}} = \sqrt{\frac{2 l}{g \sin 30^{0}}} w h e n t h e p l a n e i s r o u g h : t' = \sqrt{\frac{2 l}{a'}} = \sqrt{\frac{2 l}{g \sin 30^{0} - μ g \cos 30^{0}}} N o w, t' = 2 t \Rightarrow \sqrt{\frac{2 l}{g \sin 30^{0} - μ g \cos 30^{0}}} = 2 \times \sqrt{\frac{2 l}{g \sin 30^{0}}} \frac{2 l}{g \sin 30^{0} - μ g \cos 30^{0}} = 4 \times \frac{2 l}{g \sin 30^{0}} g \sin 30^{0} - μ g \cos 30^{0} = \frac{g \sin 30^{0}}{4} μ \cos 30^{0} = \frac{3}{8} μ = \frac{\sqrt{3}}{4}$

● ● ● ● ●

A body takes time t to reach the bottom of an inclined plane of angle θ with the horizontal. If the plane is made rough, time taken now is 2t. The coefficient of friction of the rough surface is

A.

$\frac{3}{4} \tan θ$

B.

$\frac{2}{3} \tan θ$

C.

$\frac{1}{4} \tan θ$

D.

$\frac{1}{2} \tan θ$

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

Explanation

$W h e n t h e i n c l i n e i s s m o o t h : A c c e l e r a t i o n = g \sin θ t = \sqrt{\frac{2 l}{g \sin θ}} W h e n t h e i n c l i n e i s r o u g h : A c c e l e r a t i o n = g \sin θ - μ g \cos θ t' = \sqrt{\frac{2 l}{g \sin θ - μ g \cos θ}} t' = 2 t \sqrt{\frac{2 l}{g \sin θ - μ g \cos θ}} = 2 \sqrt{\frac{2 l}{g \sin θ}} g \sin θ - μ g \cos θ = \frac{g \sin θ}{4} \frac{3}{4} \sin θ = μ \cos θ μ = \frac{3}{4} \tan θ$

● ● ● ● ●

A block is kept on an inclined plane of inclination θ of length l. The velocity of particle at the bottom of inclined is (the coefficient of friction is μ)

A.

$\sqrt{2 g l (μ \cos θ - \sin θ)}$

B.

$\sqrt{2 g l (\sin θ - μ \cos θ)}$

C.

$\sqrt{2 g l (\sin θ + μ \cos θ)}$

D.

$\sqrt{2 g l (\cos θ + μ \sin θ)}$

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

Explanation

Acceleration (a) $= g (\sin θ - μ \cos θ)$ and s = l

$v = \sqrt{2 a s} = \sqrt{2 g l (\sin θ - μ \cos θ)}$

● ● ● ● ●

A block of mass 0.1 kg is held against a wall by applying a horizontal force of 5 N on the block. If the coefficient of friction between the block and the wall is 0.5, the magnitude of the frictional force acting on the block is

A.

2.5 N

B.

0.98 N

C.

4.9 N

D.

0.49 N

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

● ● ● ● ●

A lead ball strikes a wall and falls down, a tennis ball having the same mass and velocity strikes the wall and bounces back. Check the correct statement

A.

The momentum of the lead ball is greater than that of the tennis ball

B.

The lead ball suffers a greater change in momentum compared with the tennis ball

C.

The tennis ball suffers a greater change in momentum as compared with the lead ball

D.

Both suffer an equal change in momentum

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

● ● ● ● ●

A ball of mass m falls vertically to the ground from a height h₁ and rebound to a height h₂. The change in momentum of the ball on striking the ground is

A.

$m g (h_{1} - h_{2})$

B.

$m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})$

C.

$m \sqrt{2 g (h_{1} + h_{2})}$

D.

$m \sqrt{2 g} (h_{1} + h_{2})$

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

Explanation

When ball falls vertically downward from height $h_{1}$ its velocity ${\vec{v}}_{1} = \sqrt{2 g h_{1}}$

and its velocity after collision ${\vec{v}}_{2} = \sqrt{2 g h_{2}}$

Change in momentum

$Δ \vec{P} = m ({\vec{v}}_{2} - {\vec{v}}_{1}) = m (\sqrt{2 g h_{1}} + \sqrt{2 g h_{2}})$

(because ${\vec{v}}_{1}$ and ${\vec{v}}_{2}$ are opposite in direction)

● ● ● ● ●

One end of the string of length l is connected to a particle of mass m and the other end is connected to a small peg on a smooth horizontal table. If the particle moves in circle with speed v, the net force on the particle (directed towards centre) will be (T represents the tension in the string)

A.

T

B.

$T + \frac{m v^{2}}{l}$

C.

T- $\frac{m v^{2}}{l}$

D.

Zero

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

● ● ● ● ●

A spring of force constant k is cut into lengths of ratio 1:2:3. They are connected in series and the new force constant is $k^{'}$ . If they are connected in parallel and force constant is $k^{''}, t h e n k^{'} : k^{''}$ is

A.

1:6

B.

1:9

C.

1:11

D.

6:11

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

Explanation

$W h e n t h e s p r i n g i s c u t i n t o p i e c e s, t h e y w i l l h a v e t h e n e w f o r c e c o n s \tan t . T h e s p r i n g i s d i v i d e d i n t o 1 : 2 : 3 r a t i o . L e t t h e l_{1} = x, t h e n l_{2} = 2 x a n d l_{3} = 3 x \Rightarrow x + 2 x + 3 x = l \Rightarrow x = \frac{l}{6} S i n c e s p r i n g c o n s \tan t i s i n v e r s e l y p r o p o r t i o n a l t o l e n g t h, n e w c o n s \tan t s a r e : F o r s p r i n g s, k_{1} l_{1} = k_{2} l_{2} = k_{3} l_{3} = k_{n} l_{n} k_{1} = \frac{k l}{\frac{l}{6}} = 6 k k_{2} = \frac{k l}{\frac{l}{3}} = 3 k k_{3} = \frac{k l}{\frac{l}{2}} = 2 k$

When the pieces are connected in series, the resultant force constant

$\frac{1}{v^{'}} = \frac{1}{k_{1}} + \frac{1}{k_{2}} + \frac{1}{k_{3}} \frac{1}{v^{'}} = \frac{1}{6 k} + \frac{1}{2 k} + \frac{1}{3 k} v^{'} = k$

In parallel,the net force constant

$K^{''} = 6 k + 3 k + 2 k = 11 k$

The requried ratio $\frac{K}{K^{''}} = \frac{k}{11 k} = 1 : 11$

● ● ● ● ●

A car is negotiating a curved road of radius R. The road is banked at angle $θ$ . The coefficient of friction between the tyres of the car and the road is $μ_{s}$ . The maximum safe velocity on this road is

A.

$\sqrt{gR (\frac{μ_{s} + tanθ}{1 - μ_{s} tanθ})}$

B.

$\sqrt{\frac{g}{R} (\frac{μ_{s} + tanθ}{1 - μ_{s} tanθ})}$

C.

$\sqrt{\frac{g}{R^{2}} (\frac{μ_{s} + tanθ}{1 - μ_{s} tanθ})}$

D.

$\sqrt{{gR}^{2} (\frac{μ_{s} + tanθ}{1 - μ_{s} tanθ})}$

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

Explanation

For a vehicle on a banked road, the maximum safe velocity is given by âˆš(gR((Î¼s + tanÎ¸)/(1 - Î¼s tanÎ¸))), where g is the acceleration due to gravity, R is the radius of curvature, Î¼s is the coefficient of static friction, and Î¸ is the angle of banking. This expression takes into account the centripetal force and the frictional force acting on the vehicle.

● ● ● ● ●

What is the minimum velocity with which a body of mass m must enter a vertical loop of radius R so that it can complete the loop?

A.

$\sqrt{2 gR}$

B.

$\sqrt{3 g R}$

C.

$\sqrt{5 g R}$

D.

$\sqrt{g R}$

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

NEET Practice Questions (MCQs) with Answers & Solutions

Frequently Asked Questions