Consider a uniform square plate of side $a$ and mass $m$. The moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is
Answer: (D) $\dfrac{2}{3}ma^2$
Concept. Find $I$ about the normal axis through the centre with the perpendicular-axis theorem, then shift to the corner with the parallel-axis theorem.
Step 1. Centre, normal axis. For a square plate, the axis through the centre parallel to a side gives $\dfrac{ma^2}{12}$ (like a rod of length $a$). Two such axes are perpendicular, so
$$I_{O}=\frac{ma^2}{12}+\frac{ma^2}{12}=\frac{ma^2}{6}$$
Step 2. Distance from the centre to a corner: $d=\dfrac{a}{\sqrt2}$ (half the diagonal).
Step 3. Parallel axis.
$$I_{\text{corner}}=\frac{ma^2}{6}+m\frac{a^2}{2}=\frac{2}{3}ma^2$$
Answer. $\dfrac23ma^2$ (Option D).
Solution by Sreeraj P, M.Sc Physics
