Two discs have the same thickness. Their radi | vedclass

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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 disc has thickness $t$,the mass $M$ is gi

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$R_2 > R_1$ and $d_2 > d_1$

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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 disc has thickness $t$,the mass $M$ is given by $M = 	ext{Volume} 	imes 	ext{Density} = (\pi R^2 t) 	imes d$.Substituting $M$ into the formula for $I$,we get $I = \frac{1}{2} (\pi R^2 t d) R^2 = \frac{1}{2} \pi t d R^4$.Given that the thickness $t$ is the same for both discs,the moment of inertia $I$ is proportional to $d R^4$.For the second disc to have a greater moment of inertia than the first $(I_2 > I_1)$,we must have $d_2 R_2^4 > d_1 R_1^4$.This condition is satisfied if $R_2 > R_1$ and $d_2 > d_1$.

Two discs have the same thickness. Their radii are $R_1$ and $R_2$ and their densities are $d_1$ and $d_2$. The moment of inertia of the second disc is greater than the first if:

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