Two discs of same thickness but of different | vedclass

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

Vedclass · Accepted Answer

$3:1$

Vedclass · Answer

(B) The moment of inertia of a disc about its central axis 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.Given that $M$ and $t$ are the same for both discs,we have $M = ho \pi R^2 t$,which implies $R^2 = \frac{M}{\pi ho t}$.Substituting this into the expression for $I$:$I = \frac{1}{2} M \left( \frac{M}{\pi ho t} ight) = \frac{M^2}{2 \pi t ho}$.Since $M$ and $t$ are constant,$I \propto \frac{1}{ho}$.Therefore,the ratio of the moments of inertia is $\frac{I_1}{I_2} = \frac{ho_2}{ho_1}$.Given $\frac{ho_1}{ho_2} = \frac{1}{3}$,we have $\frac{ho_2}{ho_1} = \frac{3}{1}$.Thus,the ratio of the moments of inertia is $3:1$.

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