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(C) The moment of inertia $I$ of a disc about its central axis is given by $I = \frac{1}{2} M R^2$. Since the discs are made of the same material and

Vedclass · Accepted Answer

$1: 81$

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(C) The moment of inertia $I$ of a disc about its central axis is given by $I = \frac{1}{2} M R^2$. Since the discs are made of the same material and have the same thickness $t$,their mass $M$ is given by $M = 	ext{density} 	imes 	ext{volume} = ho 	imes (\pi R^2 t)$. Substituting $M$ into the formula for $I$,we get $I = \frac{1}{2} (ho \pi R^2 t) R^2 = \frac{1}{2} ho \pi t R^4$. Since $ho$,$\pi$,and $t$ are constant for both discs,$I \propto R^4$. Therefore,the ratio of the moments of inertia is $\frac{I_A}{I_B} = \left( \frac{R_A}{R_B} ight)^4$. Given $R_A = R$ and $R_B = 3R$,we have $\frac{I_A}{I_B} = \left( \frac{R}{3R} ight)^4 = \left( \frac{1}{3} ight)^4 = \frac{1}{81}$. Thus,the ratio is $1: 81$.

Two discs $A$ and $B$ of same material and thickness have radii $R$ and $3R$ respectively. Their moments of inertia about their axis will be in the ratio

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