A thin metal wire of length L and mass M is b | vedclass

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(C) $1$. The length of the wire is $L$. Since it is bent into a semicircular ring,the circumference of a full circle would be $2 \pi R = 2L$,where $R$

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

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(C) $1$. The length of the wire is $L$. Since it is bent into a semicircular ring,the circumference of a full circle would be $2 \pi R = 2L$,where $R$ is the radius of the semicircle. Thus,$\pi R = L$,which gives $R = \frac{L}{\pi}$.$2$. The moment of inertia of a complete circular ring of mass $M'$ and radius $R$ about its diameter is $I_{diam} = \frac{1}{2} M' R^2$.$3$. For a semicircular ring of mass $M$ and radius $R$,the moment of inertia about the axis passing through its ends (the diameter) is the same as that of a full ring of the same radius and mass $M$ about its diameter. Therefore,$I = \frac{1}{2} M R^2$.$4$. Substituting $R = \frac{L}{\pi}$ into the formula,we get $I = \frac{1}{2} M \left(\frac{L}{\pi}\right)^2 = \frac{M L^2}{2 \pi^2}$.

$A$ thin metal wire of length $L$ and mass $M$ is bent to form a semicircular ring as shown. The moment of inertia about the axis $XX^1$ passing through its ends is:

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