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(D) The moment of inertia of a circular disc about its central axis is given by $I = \frac{1}{2} M R^2$.Since mass $M = 	ext{density} (ho) 	imes \

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$64$

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(D) The moment of inertia of a circular disc about its central axis is given by $I = \frac{1}{2} M R^2$.Since mass $M = 	ext{density} (ho) 	imes 	ext{volume} (V) = ho (\pi R^2 t)$,we have $I = \frac{1}{2} (ho \pi R^2 t) R^2 = \frac{1}{2} ho \pi R^4 t$.For disc $X$: $I_X = \frac{1}{2} ho \pi R^4 t$.For disc $Y$: $I_Y = \frac{1}{2} ho \pi (4R)^4 (\frac{t}{4}) = \frac{1}{2} ho \pi (256 R^4) (\frac{t}{4}) = \frac{1}{2} ho \pi (64 R^4 t) = 64 	imes (\frac{1}{2} ho \pi R^4 t)$.Therefore,$\frac{I_Y}{I_X} = \frac{64 I_X}{I_X} = 64$.

$A$ circular disc $X$ of radius $R$ is made from an iron plate of thickness $t$,and another disc $Y$ of radius $4R$ is made from an iron plate of thickness $\frac{t}{4}$. The ratio of the moment of inertia $\frac{I_Y}{I_X}$ is:

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