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Rotational Motion MCQs for NEET — Physics Questions with Answers

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

16L

$\sqrt{2} L$

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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 $L \propto \sqrt{K}$

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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

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Explanation

$L = I ω$

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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

$\frac{M ω}{M + 4 m}$

$\frac{(M + 4 m) ω}{M}$

$\frac{(M - 4 m) ω}{M + 4 m}$

$\frac{M ω}{4 m}$

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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.

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

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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%

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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%.

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The angular momentum of a system of particles is conserved

when no external force acts upon the system

when no external torque acts upon the system

when no external impulse acts upon the system

when axis of rotation remains same

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Explanation

$τ = \frac{d L}{d t} . I f τ = 0, L = c o n s \tan t$

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One hollow and one solid cylinder of the same radius roll down on a smooth inclined plane. Then the foot of the inclined plane is reached by

solid cylinder earlier

hollow cylinder earlier

simultaneously both

the heavier earlier irrespective of being solid or hollow

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Explanation

Since friction is absent, rolling cannot take place and both bodies slide

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A circular platform is mounted a frictionless vertical axle. Its radius R = 2m and moment of inertia about the axle are 200 ${kgm}^{2}$ . It is initially at rest. A 50 kg man on the edge of the platform and begins to walk along the edge at the speed of 1 $m s^{- 1}$ relative to the ground. Time taken by the man to complete one revolution is

$πs$

$\frac{3 π}{2} s$

$2 πs$

$\frac{π}{2} s$

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Explanation

The angular speed of the platform is given by the linear speed of the man divided by the radius of the platform (Ï‰ = v/R = 1/2 rad/s). The time period for one revolution is T = 2Ï€/Ï‰ = 2Ï€/(1/2) = 4Ï€ seconds.

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The motor of an engine is rotating about its axis with an angular velocity of 100 rpm. It comes to rest is 15 s, after being switched off. Assuming constant angular deceleration. What are the numbers of revolutions made by it before coming to rest?

12.5

32.6

15.6

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Explanation

$0 = ω_{0} - α t ∴ α = \frac{ω_{0}}{t} = \frac{(100 \times 2 π) / 60}{15} = 0.7 r a d / s^{2}$

Now angle rotated before coming to rest $θ$ = $\frac{ω_{0}^{2}}{2 α}$

$o r θ = \frac{{(100 \times 2 π / 60)}^{2}}{2 \times 0.7} = 78.33 r a d o r n u m b e r o f r o t a t i o n n = \frac{θ}{2 π} = 12.5$

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The moment of inertia of a loop of radius R and mass M, about any tangent line in its plane will be

$\frac{3 M R^{2}}{2}$

$\frac{M R^{2}}{2}$

$M R^{2}$

$\frac{M R^{2}}{4}$

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Explanation

Use theorem of paralle axes.

$I = \frac{M R^{2}}{2} + M R^{2} = \frac{3}{2} M R^{2}$

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A hollow sphere of diameter 0.2 m and mass 2 kg is rolling on an inclined plane with velocity v = 0.5 m/ s. The kinetic energy of the sphere is

0.1 J

0.3 J

0.5 J

0.42 J

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Explanation

$K_{N} = \frac{1}{2} {mv}^{2} (\frac{K^{2}}{R^{2}} + 1) = \frac{1}{2} \times 2 \times (0 .5)^{2} \times (\frac{2}{3} + 1) = 0 .42 J$

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