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 \(\mathrm{a}^{\prime}\) ' 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 corner is
- A \(\frac{\mathrm{Ma}^{2}}{3}\)
- B \(\frac{3}{\mathrm{Ma}^{2}}\)
- C \(\frac{3 \mathrm{Ma}^{2}}{2}\)
- D \(\frac{2 \mathrm{Ma}^{2}}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 \mathrm{Ma}^{2}}{3}\)
Step-by-step Solution
Detailed explanation
\(I_{o}=\frac{M a^{2}}{6}\)
\(A B=\sqrt{2 a^{2}}=\sqrt{2} a \quad \quad A O=\frac{a}{\sqrt{2}}=h\)
\(\therefore I_{A}=I_{o}+M h^{2}\)
\(=\frac{\mathrm{Ma}^{2}}{6}+\frac{\mathrm{Ma}^{2}}{2}=\frac{\mathrm{Ma}^{2}+3 \mathrm{Ma}^{2}}{6}=\frac{4 \mathrm{Ma}^{2}}{6}=\frac{2 \mathrm{Ma}^{2}}{3}\)

\(A B=\sqrt{2 a^{2}}=\sqrt{2} a \quad \quad A O=\frac{a}{\sqrt{2}}=h\)
\(\therefore I_{A}=I_{o}+M h^{2}\)
\(=\frac{\mathrm{Ma}^{2}}{6}+\frac{\mathrm{Ma}^{2}}{2}=\frac{\mathrm{Ma}^{2}+3 \mathrm{Ma}^{2}}{6}=\frac{4 \mathrm{Ma}^{2}}{6}=\frac{2 \mathrm{Ma}^{2}}{3}\)

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