The moment of inertia of a uniform square plate about an axis perpendicular to its plane and passing through…

The moment of inertia of a uniform square plate about an axis perpendicular to its plane and passing through the centre is $\frac{\mathrm{Ma}^2}{6}$, where ' $\mathrm{M}$ ' is the mass and ' $a$ ' is the side of square plate. Moment of inertia of this plate about an axis perpendicular to its plane and passing through one of its corners is
  1. $\frac{\mathrm{Ma}^2}{6}$
  2. $\frac{2 \mathrm{Ma}^2}{3}$
  3. $\frac{\mathrm{Ma}^2}{3}$
  4. $\frac{2 \mathrm{Ma}^2}{5}$

Solution

$\begin{aligned} \mathrm{I}_0 & =\frac{\mathrm{Ma}^2}{6} \\ \mathrm{AB} & =\sqrt{2 \mathrm{a}^2}=\sqrt{2} \mathrm{a} \\ \therefore \quad \mathrm{AO} & =\frac{\mathrm{a}}{\sqrt{2}}=\mathrm{h} \\ \therefore \quad \mathrm{I}_{\mathrm{A}} & =\mathrm{I}_0+\mathrm{Mh}^2 \\ & =\frac{\mathrm{Ma}^2}{6}+\frac{\mathrm{Ma}^2}{2} \\ & =\frac{8 \mathrm{Ma}^2}{12}=\frac{2}{3} \mathrm{Ma}^2\end{aligned}$

Asked in: MHT CET 2023 (10 May Shift 2)

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