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(A) The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about its perpendicular bisector is given by:$I_1 = \frac{ML^2}{12} \quad .

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(A) The moment of inertia of a thin uniform rod of mass $M$ and length $L$ about its perpendicular bisector is given by:$I_1 = \frac{ML^2}{12} \quad ...(1)$When the rod is bent into a semicircular arc,its length $L$ becomes the circumference of a semicircle of radius $r$,so $L = \pi r$,which gives $r = \frac{L}{\pi}$.The moment of inertia of a semicircular arc of mass $M$ and radius $r$ about an axis passing through its centre and perpendicular to its plane is:$I_2 = Mr^2$Substituting $r = \frac{L}{\pi}$:$I_2 = M \left(\frac{L}{\pi}\right)^2 = \frac{ML^2}{\pi^2} \quad ...(2)$Now,calculating the ratio $I_1 : I_2$:$\frac{I_1}{I_2} = \frac{ML^2/12}{ML^2/\pi^2} = \frac{\pi^2}{12}$Since $\pi^2 \approx 9.87$,we have $\frac{\pi^2}{12} \approx \frac{9.87}{12} Therefore,the ratio $I_1 : I_2 < 1$.

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