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 - Physics MCQ AIEEE 2026 | ExamTest.live

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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:

Solution & Step-by-step Explanation

Moment of inertia about CM axis perpendicular to plane: I_{CM} = \frac{m ( a ^{2} + a ^{2} )}{12} = \frac{m a ^{2}}{6} .Distance from CM (a /2, a /2) to corner (0, 0) is r = (a /2)^{2} + (a /2)^{2} = a^{2} /2 = a / 2 .Using parallel axis theorem: I = I_{CM} + m r^{2} . I = \frac{m a ^{2}}{6} + m (\frac{a}{2})^{2} = \frac{m a ^{2}}{6} + \frac{m a ^{2}}{2} = \frac{m a ^{2} + 3 m a ^{2}}{6} = \frac{4 m a ^{2}}{6} = \frac{2}{3} m a^{2}

Try it yourself before checking the explanation above.

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:

\frac{5}{6} m a^{2}

\frac{1}{12} m a^{2}

\frac{7}{12} m a^{2}

\frac{2}{3} m a^{2}

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