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(B) $1$. For a thin uniform rod of mass $M$ and length $L$,the moment of inertia about the perpendicular axis passing through its centre is $I = \frac

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

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(B) $1$. For a thin uniform rod of mass $M$ and length $L$,the moment of inertia about the perpendicular axis passing through its centre is $I = \frac{ML^2}{12}$.$2$. When the rod is bent into a ring of radius $R$,the circumference of the ring equals the length of the rod: $2\pi R = L$,so $R = \frac{L}{2\pi}$.$3$. The moment of inertia of a ring of mass $M$ and radius $R$ about its diameter is $I_1 = \frac{1}{2}MR^2$.$4$. Substituting $R = \frac{L}{2\pi}$ into the expression for $I_1$: $I_1 = \frac{1}{2}M(\frac{L}{2\pi})^2 = \frac{1}{2}M(\frac{L^2}{4\pi^2}) = \frac{ML^2}{8\pi^2}$.$5$. Given $I_1 = xI$,we have $\frac{ML^2}{8\pi^2} = x(\frac{ML^2}{12})$.$6$. Solving for $x$: $x = \frac{12}{8\pi^2} = \frac{3}{2\pi^2}$.

Moment of inertia of a thin uniform rod of length $L$ and mass $M$ rotating about the perpendicular axis passing through its centre is $I$. If the same rod is bent in the form of a ring,its moment of inertia about the diameter is $I_1$. If $I_1 = xI$,then the value of $x$ is:

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