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(D) Given,linear mass density $\rho = \frac{M}{L}$,so the total mass $M = \rho L$.Since the wire is bent into a circular loop,its circumference is $2\

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

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(D) Given,linear mass density $\rho = \frac{M}{L}$,so the total mass $M = \rho L$.Since the wire is bent into a circular loop,its circumference is $2\pi R = L$,which gives the radius $R = \frac{L}{2\pi}$.The axis $XX'$ is a tangent to the circular loop in its plane.Using the parallel axis theorem,the moment of inertia $I$ about the tangent $XX'$ is given by $I = I_{cm} + Md^2$,where $I_{cm}$ is the moment of inertia about the center of mass (axis passing through $O$ and perpendicular to the plane of the loop) and $d = R$.For a circular ring,$I_{cm} = \frac{MR^2}{2}$ (about the diameter axis in the plane of the ring).Thus,$I = \frac{MR^2}{2} + MR^2 = \frac{3}{2}MR^2$.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 $L$ has a uniform linear mass density $\rho$. It is bent into a circular loop with center $O$. Calculate the moment of inertia of the circular loop about the axis $XX'$ as shown in the figure.

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