A solid sphere of mass 5 text{ kg} and a disc | vedclass

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(C) The moment of inertia of a solid sphere of mass $M_s$ and radius $R$ about its diameter is $I_{cm} = \frac{2}{5} M_s R^2$.Using the parallel axis

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$7, 5$

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(C) The moment of inertia of a solid sphere of mass $M_s$ and radius $R$ about its diameter is $I_{cm} = \frac{2}{5} M_s R^2$.Using the parallel axis theorem, the moment of inertia about its tangent is $I_{sphere} = I_{cm} + M_s R^2 = \frac{2}{5} M_s R^2 + M_s R^2 = \frac{7}{5} M_s R^2$.Given $M_s = 5 	ext{ kg}$, $I_{sphere} = \frac{7}{5} 	imes 5 	imes R^2 = 7 R^2$.The moment of inertia of a disc of mass $M_d$ and radius $R$ about its diameter is $I_{cm} = \frac{1}{4} M_d R^2$.Using the parallel axis theorem, the moment of inertia about a tangent in its plane is $I_{disc} = I_{cm} + M_d R^2 = \frac{1}{4} M_d R^2 + M_d R^2 = \frac{5}{4} M_d R^2$.Given $M_d = 4 	ext{ kg}$, $I_{disc} = \frac{5}{4} 	imes 4 	imes R^2 = 5 R^2$.The ratio is $\frac{I_{sphere}}{I_{disc}} = \frac{7 R^2}{5 R^2} = \frac{7}{5}$.Thus, $x = 7$ and $y = 5$.

$A$ solid sphere of mass $5 \text{ kg}$ and a disc of mass $4 \text{ kg}$ have the same radius. The ratio of the moment of inertia of the sphere about its tangent to the moment of inertia of the disc about a tangent in its plane will be $x : y$. The value of $x$ and $y$ respectively is

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