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(D) The length of the wire is $l$. Let $R$ be the radius of the circular loop. Then, $2 \pi R = l$, which gives $R = \frac{l}{2 \pi}$.The mass of the

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

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(D) The length of the wire is $l$. Let $R$ be the radius of the circular loop. Then, $2 \pi R = l$, which gives $R = \frac{l}{2 \pi}$.The mass of the wire is $m = \rho l$.The moment of inertia of a circular loop about its diameter is $I_{diam} = \frac{1}{2} m R^2$.According to the parallel axis theorem, the moment of inertia about a tangent $AB$ is $I = I_{cm} + m R^2$, where $I_{cm} = I_{diam} = \frac{1}{2} m R^2$.Therefore, $I = \frac{1}{2} m R^2 + m R^2 = \frac{3}{2} m R^2$.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 $l$ having linear mass density $\rho$ is bent into a circular loop with $C$ as its centre, as shown in the figure. The moment of inertia of the loop about the line $AB$ is

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