The figure shows two solid discs with radius $R$ and $r$ respectively. If mass per unit area is the same for both, what is the ratio of MI of bigger disc around axis $AB$ (which is $\perp$ to the plane of the disc and passing through its centre) to MI of smaller disc around one of its diameters lying on its plane? Given $M$ is the mass of the larger disc.
(MI stands for moment of inertia)
Answer: (D) $2R^4 : r^4$
Same mass per unit area $\sigma$: $M = \sigma\pi R^2$, $m = \sigma\pi r^2 = M\dfrac{r^2}{R^2}$.
Bigger disc about the perpendicular axis through the centre: $I_1 = \dfrac{MR^2}{2}$.
Smaller disc about a diameter: $I_2 = \dfrac{mr^2}{4} = \dfrac{Mr^4}{4R^2}$.
$$\frac{I_1}{I_2} = \frac{MR^2/2}{Mr^4/(4R^2)} = \frac{2R^4}{r^4}$$
Solution by Sreeraj P, M.Sc Physics