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(A) For a uniform thin rod of mass $m$ and length $\ell$,the moment of inertia about an axis passing through one end and perpendicular to its length i

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(A) For a uniform thin rod of mass $m$ and length $\ell$,the moment of inertia about an axis passing through one end and perpendicular to its length is $I_{1} = \frac{m \ell^{2}}{3}$.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 $\ell = 2 \pi r$,which implies $r = \frac{\ell}{2 \pi}$.The moment of inertia of a ring of mass $m$ and radius $r$ about its diameter is $I_{2} = \frac{1}{2} m r^{2}$.Substituting $r = \frac{\ell}{2 \pi}$ into the expression for $I_{2}$,we get $I_{2} = \frac{1}{2} m \left(\frac{\ell}{2 \pi}\right)^{2} = \frac{m \ell^{2}}{8 \pi^{2}}$.Now,calculating the ratio $\frac{I_{1}}{I_{2}}$:$\frac{I_{1}}{I_{2}} = \frac{\frac{m \ell^{2}}{3}}{\frac{m \ell^{2}}{8 \pi^{2}}} = \frac{8 \pi^{2}}{3}$.Comparing this with the given expression $\frac{x \pi^{2}}{3}$,we find $x = 8$.

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