Q 11-06-198JEE MainJEE Main 2020 (4 Sep, Shift 2)Medium
For a uniform rectangular sheet of sides $60\ \text{cm}$ and $80\ \text{cm}$, the ratio of moments of inertia about the axes perpendicular to the sheet and passing through $O$ (the centre of mass) and $O'$ (a corner point) is:
Answer: (B) $1/4$
About the centre: $I_O = \dfrac{M(a^{2}+b^{2})}{12}$. The corner is at distance $\dfrac{\sqrt{a^{2}+b^{2}}}{2}$ from the centre, so
$$I_{O'} = \frac{M(a^{2}+b^{2})}{12} + \frac{M(a^{2}+b^{2})}{4} = \frac{M(a^{2}+b^{2})}{3}$$
$\dfrac{I_O}{I_{O'}} = \dfrac14$ (independent of the actual side lengths).
Solution by Sreeraj P, M.Sc Physics