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(A) The mass per unit area of the original disc is $\sigma = \frac{9M}{\pi R^2}$.The mass of the removed disc is $m = \sigma 	imes \pi (R/3)^2 = \fra

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(A) The mass per unit area of the original disc is $\sigma = \frac{9M}{\pi R^2}$.The mass of the removed disc is $m = \sigma 	imes \pi (R/3)^2 = \frac{9M}{\pi R^2} 	imes \frac{\pi R^2}{9} = M$.The distance of the center of the removed disc from the center of the original disc is $d = R - R/3 = 2R/3$.The moment of inertia of the removed disc about the axis passing through the center of the original disc is calculated using the parallel axis theorem:$I_1 = I_{cm} + md^2 = \frac{1}{2} m (R/3)^2 + m (2R/3)^2 = \frac{1}{2} M (R^2/9) + M (4R^2/9) = \frac{MR^2}{18} + \frac{8MR^2}{18} = \frac{9MR^2}{18} = \frac{1}{2} MR^2$.The moment of inertia of the original complete disc is $I_2 = \frac{1}{2} (9M) R^2 = \frac{9}{2} MR^2$.The moment of inertia of the remaining portion is $I = I_2 - I_1 = \frac{9}{2} MR^2 - \frac{1}{2} MR^2 = 4 MR^2$.

$A$ circular disc of mass $9M$ and radius $R$ has a smaller disc of radius $R/3$ cut from it. Calculate the moment of inertia of the remaining portion about an axis passing through the center of the original disc and perpendicular to its plane. (in $MR^2$)

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