(C) Total mass $M = m \times L$.The radius $R$ is found from $L = 2\pi R$, so $R = \frac{L}{2\pi}$.The moment of inertia of a ring about its diameter

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(C) Total mass $M = m \times L$.The radius $R$ is found from $L = 2\pi R$, so $R = \frac{L}{2\pi}$.The moment of inertia of a ring about its diameter is $I_{diam} = \frac{1}{2}MR^2$.Using the parallel axis theorem, the moment of inertia about the tangent $yy'$ is $I = I_{cm} + MR^2$, where $I_{cm} = I_{diam} = \frac{1}{2}MR^2$.Thus, $I = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$.Substituting $M = mL$ and $R = \frac{L}{2\pi}$:$I = \frac{3}{2} (mL) \left(\frac{L}{2\pi}\right)^2 = \frac{3}{2} mL \left(\frac{L^2}{4\pi^2}\right) = \frac{3mL^3}{8\pi^2}$.Therefore, the correct option is $C$.

Class 11 Physics System of Particles and Rotational Motion Moment of Inertia of Compound Bodies and Theorem of Moment of Inertia

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$A$ thin wire of length $L$ and linear mass density $m$ is bent into a circular ring (in $x-y$ plane) with centre $C$ as shown in the figure. The moment of inertia of the ring about an axis $yy'$ will be:

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A
$\frac{3mL^3}{8\pi}$
B
$\frac{3mL^2}{8\pi^2}$
C
$\frac{3mL^3}{8\pi^2}$
D
$\frac{3mL^2}{8\pi}$

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System of Particles and Rotational Motion

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Moment of Inertia of Compound Bodies and Theorem of Moment of Inertia

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$A$ thin wire of length $L$ and linear mass density $m$ is bent into a circular ring (in $x-y$ plane) with centre $C$ as shown in the figure. The moment of inertia of the ring about an axis $yy'$ will be:

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