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(C) The moment of inertia of a disc of mass $M$ and radius $R$ about its diameter is given by $I = \frac{1}{4}MR^2$.Since the masses $M_1$ and $M_2$ a

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$5$

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(C) The moment of inertia of a disc of mass $M$ and radius $R$ about its diameter is given by $I = \frac{1}{4}MR^2$.Since the masses $M_1$ and $M_2$ are equal,the ratio of the moments of inertia is $\frac{I_1}{I_2} = \frac{R_1^2}{R_2^2}$.The mass of a disc is $M = ho \cdot V = ho \cdot \pi R^2 t$,where $ho$ is density and $t$ is thickness.Given $M_1 = M_2$,we have $ho_1 R_1^2 t_1 = ho_2 R_2^2 t_2$.Thus,$\frac{R_1^2}{R_2^2} = \frac{ho_2 t_2}{ho_1 t_1}$.Given $\frac{ho_1}{ho_2} = \frac{3}{5}$,so $\frac{ho_2}{ho_1} = \frac{5}{3}$.Given $t_1 = 1\,cm$ and $t_2 = 0.5\,cm$,so $\frac{t_2}{t_1} = \frac{0.5}{1} = \frac{1}{2}$.Substituting these values,$\frac{R_1^2}{R_2^2} = \frac{5}{3} 	imes \frac{1}{2} = \frac{5}{6}$.Therefore,the ratio of the moments of inertia is $\frac{5}{6}$.Comparing this with $\frac{x}{6}$,we get $x = 5$.

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