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(C) Let $\sigma$ be the mass per unit area (surface mass density). Since $\sigma$ is the same for both discs:Mass of larger disc,$M = \sigma \pi R^{2}

Vedclass · Accepted Answer

$2R^{4}:r^{4}$

Vedclass · Answer

(C) Let $\sigma$ be the mass per unit area (surface mass density). Since $\sigma$ is the same for both discs:Mass of larger disc,$M = \sigma \pi R^{2}$Mass of smaller disc,$m = \sigma \pi r^{2}$The moment of inertia of the larger disc about axis $AB$ (perpendicular to the plane through the centre) is $I_{1} = \frac{1}{2} MR^{2} = \frac{1}{2} (\sigma \pi R^{2}) R^{2} = \frac{1}{2} \sigma \pi R^{4}$.The moment of inertia of the smaller disc about its diameter is $I_{2} = \frac{1}{4} mr^{2} = \frac{1}{4} (\sigma \pi r^{2}) r^{2} = \frac{1}{4} \sigma \pi r^{4}$.The ratio of the moments of inertia is $\frac{I_{1}}{I_{2}} = \frac{\frac{1}{2} \sigma \pi R^{4}}{\frac{1}{4} \sigma \pi r^{4}} = \frac{1/2}{1/4} \cdot \frac{R^{4}}{r^{4}} = 2 \frac{R^{4}}{r^{4}}$.Thus,the ratio is $2R^{4}:r^{4}$.

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