The moment of inertia of thin square plate PQRS of uniform thickness, about an axis passing through centre ' O ' and perpendicular to the plane of the plate is (I₁, I₂, I₃, I₄ are respectively the moments of inertia about axis 1, 2, 3, 4 which are in the plane of the plate as shown in figure) (a) I₁ + I₂ + I₃ (b) I₁ + I₃ + I₄ (c) I₁ + I₂ + I₃ + I₄ (d) I₁ + I₃

Answer: (d). Perpendicular axis theorem: Iz equals the sum of the moments of inertia about any two perpendicular in-plane axes through O. Axes 1 and 2 (diagonals) are perpendicular, and so are 3 and 4: Iz = I₁ + I₂ = I₃ + I₄. For a square, all in-plane axes through O have equal I by symmetry, so I₁ = I₂ = I₃ = I₄ and Iz = 2I₁ = I₁ + I₃. So the answer is (d) I₁ + I₃.

The moment of inertia of thin square plate PQRS of uniform thickness… | MHT CET 2024 (9 May, Shift 1)

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Moment of Inertia & Radius of Gyration

The moment of inertia of thin square plate PQRS of uniform thickness, about an axis passing through centre ' O ' and perpendicular to the plane of the plate is $(I_1, I_2, I_3, I_4$ are respectively the moments of inertia about axis $1, 2, 3, 4$ which are in the plane of the plate as shown in figure)

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Answer: (d)

Perpendicular axis theorem: $I_z$ equals the sum of the moments of inertia about any two perpendicular in-plane axes through $O$ .
Axes 1 and 2 (diagonals) are perpendicular, and so are 3 and 4: $I_z = I_1 + I_2 = I_3 + I_4$ .
For a square, all in-plane axes through $O$ have equal $I$ by symmetry, so $I_1 = I_2 = I_3 = I_4$ and $I_z = 2I_1 = I_1 + I_3$ .
So the answer is (d) $I_1 + I_3$ .

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