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(D) The total mass of the wire is $M = \rho L$. Since the circumference of the loop is $L = 2\pi R$,the radius of the loop is $R = \frac{L}{2\pi}$.The

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

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(D) The total mass of the wire is $M = \rho L$. Since the circumference of the loop is $L = 2\pi R$,the radius of the loop is $R = \frac{L}{2\pi}$.The moment of inertia of a circular loop about its diameter is $I_{diam} = \frac{1}{2}MR^2$.Using the parallel axis theorem,the moment of inertia about the tangent $XX'$ is $I = I_{cm} + Md^2$,where $I_{cm} = I_{diam} = \frac{1}{2}MR^2$ and $d = R$.Thus,$I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.Substituting $M = \rho L$ and $R = \frac{L}{2\pi}$:$I = \frac{3}{2}(\rho L)\left(\frac{L}{2\pi}\right)^2 = \frac{3}{2}\rho L \left(\frac{L^2}{4\pi^2}\right) = \frac{3\rho L^3}{8\pi^2}$.

$A$ thin wire of length $L$ and uniform linear mass density $\rho$ is bent into a circular loop with centre at $O$ as shown. The moment of inertia of the loop about the axis $XX'$ is

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