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(D) The moment of inertia of a solid disc of mass $M$ and radius $R$ about its central axis is given by $I = \frac{1}{2} M R^2$.Since the discs are ma

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$16$

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(D) The moment of inertia of a solid disc of mass $M$ and radius $R$ about its central axis is given by $I = \frac{1}{2} M R^2$.Since the discs are made of the same material (iron) and have negligible thickness,they have the same surface mass density $\sigma$.Thus,the mass of a disc is $M = \sigma 	imes 	ext{Area} = \sigma 	imes \pi R^2$.For the first disc: $M_1 = \sigma \pi R_1^2$ and $I_1 = \frac{1}{2} M_1 R_1^2 = \frac{1}{2} (\sigma \pi R_1^2) R_1^2 = \frac{1}{2} \sigma \pi R_1^4$.For the second disc: $M_2 = \sigma \pi R_2^2$ and $I_2 = \frac{1}{2} M_2 R_2^2 = \frac{1}{2} (\sigma \pi R_2^2) R_2^2 = \frac{1}{2} \sigma \pi R_2^4$.Given $R_2 = 2 R_1$,we substitute this into the expression for $I_2$:$I_2 = \frac{1}{2} \sigma \pi (2 R_1)^4 = \frac{1}{2} \sigma \pi (16 R_1^4) = 16 	imes (\frac{1}{2} \sigma \pi R_1^4) = 16 I_1$.Therefore,the ratio $\frac{I_1}{I_2} = \frac{I_1}{16 I_1} = \frac{1}{16}$.Comparing this with $1/x$,we get $x = 16$.

Two iron solid discs of negligible thickness have radii $R_1$ and $R_2$ and moments of inertia $I_1$ and $I_2$,respectively. For $R_2 = 2 R_1$,the ratio of $I_1$ and $I_2$ would be $1 / x$,where $x = $ . . . . . . .

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