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(C) The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its centre and perpendicular to its length is

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about an axis passing through its centre and perpendicular to its length is $I = \frac{ML^2}{12}$.When the rod is bent into a ring of radius $R$,the circumference of the ring is equal to the length of the rod,so $L = 2\pi R$,which implies $R = \frac{L}{2\pi}$.The moment of inertia of a ring about its diameter is $I^{\prime} = \frac{MR^2}{2}$.Substituting $R = \frac{L}{2\pi}$ into the expression for $I^{\prime}$,we get $I^{\prime} = \frac{M}{2} \left(\frac{L}{2\pi}\right)^2 = \frac{ML^2}{8\pi^2}$.Now,the ratio $\frac{I}{I^{\prime}}$ is $\frac{\frac{ML^2}{12}}{\frac{ML^2}{8\pi^2}} = \frac{8\pi^2}{12} = \frac{2\pi^2}{3}$.

Moment of Inertia of a thin uniform rod rotating about the perpendicular axis passing through its centre is $I$. If the same rod is bent into a ring and its moment of inertia about its diameter is $I^{\prime}$,then the ratio $\frac{I}{I^{\prime}}$ is

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