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(D) The moment of inertia of a circular disc is given by $I = \frac{1}{2}MR^2$.Since mass $M = 	ext{density} (ho) 	imes 	ext{volume} (V) = ho \

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$16$

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(D) The moment of inertia of a circular disc is given by $I = \frac{1}{2}MR^2$.Since mass $M = 	ext{density} (ho) 	imes 	ext{volume} (V) = ho 	imes (\pi R^2 t)$,where $t$ is the thickness.Substituting $M$ into the formula: $I = \frac{1}{2} (\pi R^2 t ho) R^2 = \frac{1}{2} \pi t ho R^4$.Given that the discs are made of the same material (same $ho$) and have the same thickness $(t)$,we have $I \propto R^4$.Since diameter $D = 2R$,it follows that $I \propto D^4$.Therefore,$\frac{I_A}{I_B} = \left( \frac{D_A}{D_B} ight)^4$.Given $D_A = 2 D_B$,we get $\frac{I_A}{I_B} = (2)^4 = 16$.Thus,the moment of inertia of $A$ is $16$ times that of $B$.

Two circular discs of iron have the same thickness. The diameter of disc $A$ is twice that of disc $B$. The moment of inertia of $A$ will be ........ times that of $B$.

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