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(C) Let the mass of the smaller disc be $m$. Since the area is proportional to the square of the radius,the mass of the larger disc (radius $2R$) is $

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(C) Let the mass of the smaller disc be $m$. Since the area is proportional to the square of the radius,the mass of the larger disc (radius $2R$) is $4m$.For $I_0$:The moment of inertia of the larger disc about $O$ is $I_{large, O} = \frac{(4m)(2R)^2}{2} = 8mR^2$.The moment of inertia of the smaller disc about its own center is $\frac{mR^2}{2}$. Using the parallel axis theorem,its moment of inertia about $O$ is $I_{small, O} = \frac{mR^2}{2} + mR^2 = \frac{3}{2}mR^2$.Thus,$I_0 = 8mR^2 - \frac{3}{2}mR^2 = \frac{13}{2}mR^2$.For $I_P$:The moment of inertia of the larger disc about $P$ (using parallel axis theorem) is $I_{large, P} = \frac{(4m)(2R)^2}{2} + (4m)(2R)^2 = 8mR^2 + 16mR^2 = 24mR^2$.The moment of inertia of the smaller disc about $P$ (using parallel axis theorem) is $I_{small, P} = \frac{mR^2}{2} + m((2R)^2 + R^2) = \frac{mR^2}{2} + 5mR^2 = \frac{11}{2}mR^2$.Thus,$I_P = 24mR^2 - \frac{11}{2}mR^2 = \frac{37}{2}mR^2$.The ratio $\frac{I_P}{I_0} = \frac{37/2}{13/2} = \frac{37}{13} \approx 2.846$.Rounding to the nearest integer,we get $3$.

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