JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

A thin hollow cylinder open at both ends.

(i) Sliding without rolling

(ii) Rolls without slipping, with the same speed.

The ratio of kinetic energy in the two cases is

View answer and explanation

Correct option: C

1:2

Option C (

1:2

) is correct because the ratio of kinetic energy when sliding to when rolling is 1:2, considering the hollow cylinder's moment of inertia.

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It is easier for a swimmer jumping into water from a height to describe a loop in the air by A uniform thin rod

\mathrm{AB}

of length

\mathrm{L}

has linear mass density

\mu(x)=a+\frac{b x}{L}

, where

x

is measured from

\mathrm{A}

. If the

\mathrm{CM}

of the rod lies at a distance of

\left(\frac{7}{12}\right) \mathrm{L}

from

\mathrm{A}

, then a and

\mathrm{b}

1.0 \mathrm{~g}

is rotating in a circle of diameter

8 \mathrm{~cm}

. Its moment of inertia will be The angular velocity of a particle is given by

\omega = {t^4} + 2{t^3}

then its angular acceleration of the particle at

t = 1\,sec

\left( {rad\,{s^{ - 2}}} \right)