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A hole of radius $R / 2$ is cut from a thin circular plate of radius $R$ . The mass of the remaining plate is $M$ . The moment of inertia of the plate about an axis through $O$ perpendicular to the $x y$ -plane (i.e. about the $z$ -axis) is : ( The mass of remaining disc is $M$ )

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$\frac{11}{24} M R^{2}$

$\frac{13}{24} M R^{2}$

$\frac{7}{12} M R^{2}$

$\frac{5}{7} M R^{2}$

answer is D.

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Detailed Solution

Area of the remaining disc = $π R^{2} - π {(\frac{R}{2})}^{2} = \frac{3}{4} π R^{2}$

Mass of whole disc $M' = \frac{M \times π R^{2}}{\frac{3}{4} π R^{2}} = \frac{4 M}{3}$

$∴$ Mass of the removed disc, $m = \frac{M}{3}$

Moment of inertia $I = I_{w h o l e d i s c} - I_{r o u n d d i s c}$

$= (\frac{4 M}{3}) \frac{R^{2}}{2} - [\frac{M}{3} \frac{{(\frac{R}{2})}^{2}}{2} + \frac{M}{3} {(\frac{R}{2})}^{2}] = \frac{13}{24} M R^{2}$

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