JEE PHYSICS MCQ: SYSTEM OF PARTICLES

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Application +4 | -1

The moment of inertia of a uniform rod about a perpendicular axis passing through one end is

{I_1}

. The same rod is bent into a ring and its moment of inertia about a diameter is

{I_2}

. Then

\frac{{{I_1}}}{{{I_2}}}

View answer and explanation

Correct option: D

\frac{{8{\pi ^2}}}{3}

The correct answer is D (

\frac{{8{\pi ^2}}}{3}

) because the ratio of the moment of inertia of a rod about one end to the moment of inertia of a ring about its diameter is

\frac{{{I_1}}}{{{I_2}}} = \frac{{8{\pi ^2}}}{3}

More Questions from SYSTEM OF PARTICLES

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\mathrm{AB}

\mathrm{L}

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x

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

\mathrm{A}

\mathrm{b}

1.0 \mathrm{~g}

8 \mathrm{~cm}

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

t = 1\,sec

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