Blog Log in Get Started Free

NEET Practice Questions (MCQs) with Answers & Solutions

Practice free NEET NEET multiple-choice questions online with instant answers and detailed explanations. No login required.

All Physics Chemistry Botany Zoology

Language English हिंदी

● ● ● ● ●

A cylinder of 500 g and radius 10 cm has moment of inertia (about its natural axis)

$2.5 \times 10^{- 3} k g - m^{2}$

$2 \times 10^{- 3} k g - m^{2}$

$5 \times 10^{- 3} k g - m^{2}$

$3.5 \times 10^{- 3} k g - m^{2}$

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

Explanation

$I = \frac{1}{2} M R^{2} = \frac{1}{2} \times 0.5 \times {(0.1)}^{2} = 2.5 \times 10^{- 3} k g - m^{3}$

● ● ● ● ●

Two rings of the same radius and mass are placed such that their centers are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the center and perpendicular to the plane of one of the rings is (mass of the ring and radius = r)

$\frac{1}{2} {mr}^{2}$

${mr}^{2}$

$\frac{3}{2} {mr}^{2}$

$2 {mr}^{2}$

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

Explanation

${mr}^{2} + \frac{{mr}^{2}}{2} = \frac{3}{2} {mr}^{2}$

● ● ● ● ●

The angular velocity of a body changes from $\frac{1}{25}$ rev./sec to 1 rev./sec. without applying any torque. The ratio of its radius of gyration in two situations is :

5: 1

1: 5

1: 25

25: 1

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

Explanation

$I_{1} ω_{1} = I_{2} ω_{2}, o r M K_{1}^{2} (2 π n_{1}) = M K_{2}^{2} (2 π n_{2}) \Rightarrow \frac{K_{1}^{2}}{K_{2}^{2}} = \frac{n_{2}}{n_{1}} \Rightarrow \frac{K_{1}}{K_{2}} = \sqrt{\frac{n_{2}}{n_{1}}} = \frac{\sqrt{1}}{\sqrt{\frac{1}{25}}} = \frac{5}{1}$

● ● ● ● ●

A solid cylinder and a solid sphere both having the same mass and radius, are released from a rough inclined plane. Both roll without slipping. Then,

the force of friction that acts on the two is the same

the force of friction is greater in case of a sphere than for a cylinder

the force of friction is greater in case of a cylinder than for a sphere

the force of friction will depend on the nature of the surface of the body that is moving and that of the inclined surface and is independent of the shape and size of the moving body

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

Explanation

$In case of pure rolling . Force of friction f = \frac{mg \sin θ}{1 + (\frac{{mR}^{2}}{I})}$

Since a moment of inertia of disc is more, the force of friction on cylinder will be more.

● ● ● ● ●

A rigid body rotates about a fixed axis with variable angular velocity equal to $α - β t$ , at the time t, where $α, β$ are constants. The angle through which it rotates before its stops

$\frac{α^{2}}{2 β}$

$\frac{α^{2} - β^{2}}{2 α}$

$\frac{α^{2} - β^{2}}{2 β}$

$\frac{(α - β) α}{2}$

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

Explanation

$ω = α - βt . Comparing with ω = ω_{0} - αt initial angular velocity = α Angular retardation = β ∴ Angle rotated before it stops is \frac{α^{2}}{2 β} from 0 α^{2} - 2 βθ$

● ● ● ● ●

An automobile moves on a road with a speed of 54 kmh $^{- 1}$ . The radius of its wheel is 0.45 m and moment of inertia of the wheel about its axis of rotation is 3 kg m $^{2}$ . If the vehicle is brought to rest in 15s, the magnitude of average torque transmitted by its brakes to the wheel is

8.58 $m^{2} s^{- 2}$

10.86 $m^{2} s^{- 2}$

2.86 $m^{2} s^{- 2}$

6.66 $m^{2} s^{- 2}$

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

Explanation

$ω_{i} = \frac{15}{0.45} = \frac{100}{3} ω_{i} = 0 ω_{i} = ω_{i} + αt 0 = \frac{100}{3} + (- α) (15) α = \frac{100}{45} τ = (I) (α) = 3 x \frac{100}{45} = 6.66 N . m .$

● ● ● ● ●

A rigid body rotates with an angular momentum L. If its rotational kinetic energy is made 4 times, its angular momentum will become

16L

$\sqrt{2} L$

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

Explanation

$L = \sqrt{2 K I} (J u s t l i k e p = \sqrt{2 k m})$

K is made 4 times. Hence L will become as $L \propto \sqrt{K}$

● ● ● ● ●

Angular momentum of a body is defined as the product of

Mass and angular velocity

Centripetal force and radius

Linear velocity and angular velocity

Moment of inertia and angular velocity

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

Explanation

$L = I ω$

● ● ● ● ●

A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity $ω$ . Four objects each of mass m, are kept gently to the opposite ends of two perpendicular diameters of the ring. The angular velocity of the ring will be

$\frac{M ω}{M + 4 m}$

$\frac{(M + 4 m) ω}{M}$

$\frac{(M - 4 m) ω}{M + 4 m}$

$\frac{M ω}{4 m}$

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

Explanation

Initial angular momentum of ring, $L = I ω = M r^{2} ω$

Final angular momentum of ring and four particles system $L = (M r^{2} + 4 m r^{2}) ω'$

As there is no torque on the system therefore angular momentum remains constant.

$M r^{2} ω = (M r^{2} + 4 m r^{2}) ω' \Rightarrow ω' = \frac{M ω}{M + 4 m}$

● ● ● ● ●

Let g be acceleration due to gravity on the surface of earth and T be the rotational kinetic energy of the earth. Suppose the earth's radius decreases by 2%. Keeping all other quantities same (even $ω$ )

g decreases by 2% and T decreases by 4%

g decreases by 4% and T decreases by 2%

g increases by 4% and T decreases by 4%

g decreases by 4% and T increases by 4%

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

Explanation

$\begin{aligned} g & = \frac{GM}{R^{2}} \\ (KE)_{rotalon} & = T = \frac{1}{2} {Iω}^{2} \\ = \frac{1}{2} \times \frac{2}{5} {MR}^{2} ω^{2} \\ \frac{Δg}{g} & = \frac{- 2 ΔR}{R} and \frac{ΔT}{T} = \frac{2 ΔR}{R} \end{aligned}$

As R decreases by 2%, g increases by 4% and KE decrease by 4%.

Ready to ace NEET?

Free access · No credit card required

Browse All Questions Create Free Account

Frequently Asked Questions

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.

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.