Two discs of same mass and same thickness hav | vedclass

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(D) The moment of inertia of a disc is $I = \frac{1}{2} M r^2$.Since mass $M = V \rho = (\pi r^2 t) \rho$,we can write $r^2 = \frac{M}{\pi t \rho}$.Su

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$\rho_2 : \rho_1$

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(D) The moment of inertia of a disc is $I = \frac{1}{2} M r^2$.Since mass $M = V \rho = (\pi r^2 t) \rho$,we can write $r^2 = \frac{M}{\pi t \rho}$.Substituting $r^2$ into the moment of inertia formula: $I = \frac{1}{2} M \left( \frac{M}{\pi t \rho} \right) = \frac{M^2}{2 \pi t \rho}$.Given that both discs have the same mass $M$ and same thickness $t$,the moment of inertia is inversely proportional to the density: $I \propto \frac{1}{\rho}$.Therefore,the ratio of their moments of inertia is $\frac{I_1}{I_2} = \frac{\rho_2}{\rho_1}$.

Two discs of same mass and same thickness have densities $\rho_1$ and $\rho_2$. The ratio of their moment of inertia about their central axes is .........

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