Moment of inertia of a thin uniform rod rotating about a perpendicular axis passing through its center is $I$. If the same rod is bent into a ring and its moment of inertia about its diameter is $F$, then the ratio $\frac{I}{F}$ will be:

Question

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

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

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

$\frac{5}{3} \pi^2$

Solution

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