A circular hole of diameter R is cut from a d | vedclass

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(D) The moment of inertia $(M.I.)$ of the complete disc of mass $M$ and radius $R$ about its center $O$ is:$I_{Total} = \frac{1}{2} M R^2$  --- $(i)$T

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$\left( \frac{13}{32} \right) M R^2$

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(D) The moment of inertia $(M.I.)$ of the complete disc of mass $M$ and radius $R$ about its center $O$ is:$I_{Total} = \frac{1}{2} M R^2$  --- $(i)$The mass of the removed circular hole (radius $r = R/2$) is:$m = \frac{M}{\pi R^2} 	imes \pi (R/2)^2 = \frac{M}{4}$The $M.I.$ of the removed hole about its own central axis is:$I_{cm} = \frac{1}{2} m r^2 = \frac{1}{2} \left( \frac{M}{4} ight) \left( \frac{R}{2} ight)^2 = \frac{M R^2}{32}$Using the parallel axis theorem,the $M.I.$ of the removed hole about the center $O$ of the original disc is:$I_{removed} = I_{cm} + m d^2 = \frac{M R^2}{32} + \left( \frac{M}{4} ight) \left( \frac{R}{2} ight)^2 = \frac{M R^2}{32} + \frac{M R^2}{16} = \frac{3 M R^2}{32}$The $M.I.$ of the remaining portion is:$I_{remaining} = I_{Total} - I_{removed} = \frac{1}{2} M R^2 - \frac{3 M R^2}{32} = \frac{16 M R^2 - 3 M R^2}{32} = \frac{13}{32} M R^2$

$A$ circular hole of diameter $R$ is cut from a disc of mass $M$ and radius $R$; the circumference of the cut passes through the centre of the disc. The moment of inertia of the remaining portion of the disc about an axis perpendicular to the disc and passing through its centre is

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