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(C) The moment of inertia of a circular ring of mass $m$ and radius $R$ about its diameter is given by $I = \frac{1}{2} mR^2$.Given that the linear ma

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

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(C) The moment of inertia of a circular ring of mass $m$ and radius $R$ about its diameter is given by $I = \frac{1}{2} mR^2$.Given that the linear mass density is $\rho = \frac{m}{L}$, where $L$ is the circumference of the loop $(L = 2 \pi R)$.Therefore, $\rho = \frac{m}{2 \pi R}$, which implies $R = \frac{m}{2 \pi \rho}$.Substituting the value of $R$ into the formula for the moment of inertia:$I = \frac{1}{2} m \left( \frac{m}{2 \pi \rho} \right)^2$$I = \frac{1}{2} m \left( \frac{m^2}{4 \pi^2 \rho^2} \right)$$I = \frac{m^3}{8 \pi^2 \rho^2}$.

$A$ thin uniform wire of mass $m$ and linear mass density $\rho$ is bent in the form of a circular loop. The moment of inertia of the loop about its diameter is

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