The given figure shows a disc of mass M and r | vedclass

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(B) The moment of inertia of a disc of mass $M$ and radius $R$ about its diameter is given by $I_{diameter} = \frac{MR^2}{4}$.In the given figure,the

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$M\left( \frac{R^2}{4} \right)$

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(B) The moment of inertia of a disc of mass $M$ and radius $R$ about its diameter is given by $I_{diameter} = \frac{MR^2}{4}$.In the given figure,the disc lies in the $x-y$ plane,and its center is at $(a, 0)$.The $x$-axis passes through the center of the disc and lies in the plane of the disc.Therefore,the $x$-axis acts as a diameter for the disc.Since the moment of inertia about any diameter of a disc is $\frac{MR^2}{4}$,the moment of inertia of the disc about the $x$-axis is simply $\frac{MR^2}{4}$.

The given figure shows a disc of mass $M$ and radius $R$ lying in the $x-y$ plane with its center on the $x$-axis at a distance $a$ from the origin. Then,the moment of inertia of the disc about the $x$-axis is:

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