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(B) The length of the wire is $\ell = \pi r$,which implies $r = \frac{\ell}{\pi}$.The moment of inertia of a semicircular wire about the axis passing

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$\frac{M\ell^2}{2\pi^2}$

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(B) The length of the wire is $\ell = \pi r$,which implies $r = \frac{\ell}{\pi}$.The moment of inertia of a semicircular wire about the axis passing through its diameter $(XX')$ is given by $I = Mr^2$.Substituting the value of $r$ into the formula:$I_{XX'} = M \left( \frac{\ell}{\pi} \right)^2 = \frac{M\ell^2}{\pi^2}$.Wait,let's re-evaluate the moment of inertia of a semicircular wire about its diameter. For a thin ring of mass $M$ and radius $r$,the moment of inertia about a diameter is $\frac{1}{2}Mr^2$. Since a semicircle is half of a ring,the moment of inertia of the semicircular wire about the diameter $XX'$ is also $\frac{1}{2}Mr^2$.Therefore,$I_{XX'} = \frac{1}{2} M \left( \frac{\ell}{\pi} \right)^2 = \frac{M\ell^2}{2\pi^2}$.

$A$ wire of length $\ell$ and mass $M$ is bent into a semicircle of radius $r$ as shown in the figure. Calculate the moment of inertia about the axis $XX'$.

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