MHT CET · Physics · Rotational Motion
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
- A \(\frac{\mathrm{Ma}^2}{6}\)
- B \(\frac{2 \mathrm{Ma}^2}{3}\)
- C \(\frac{\mathrm{Ma}^2}{3}\)
- D \(\frac{2 \mathrm{Ma}^2}{5}\)
Answer & Solution
Correct Answer
(B) \(\frac{2 \mathrm{Ma}^2}{3}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} \mathrm{I}_0 & =\frac{\mathrm{Ma}^2}{6} \\ \mathrm{AB} & =\sqrt{2 \mathrm{a}^2}=\sqrt{2} \mathrm{a} \\ \therefore \quad \mathrm{AO} & =\frac{\mathrm{a}}{\sqrt{2}}=\mathrm{h} \\ \therefore \quad \mathrm{I}_{\mathrm{A}} & =\mathrm{I}_0+\mathrm{Mh}^2 \\ & =\frac{\mathrm{Ma}^2}{6}+\frac{\mathrm{Ma}^2}{2} \\ & =\frac{8 \mathrm{Ma}^2}{12}=\frac{2}{3} \mathrm{Ma}^2\end{aligned}\)


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