JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Rotational Mechanics

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Moment of inertia of a square plate of side / about the axis passing through one of the corner and perpendicular to the plane of square plate is given by:

(A) $MI^{2}$
(B) $\frac{MI ^{2}}{12}$
(C) $\frac{MI ^{2}}{6}$
(D) $\frac{2}{3} M I^{2}$ ✓ Correct

Answer: (D)

Solution

About the axis through the centre perpendicular to its plane, a square plate of side

ℓ

has

I_{C} = \frac{M ℓ ^{2}}{6}

. The centre-to-corner distance is

d = \frac{ℓ}{2}

, so by the parallel-axis theorem the moment of inertia about the parallel axis through a corner is

I = I_{C} + M d^{2} = \frac{M ℓ ^{2}}{6} + \frac{M ℓ ^{2}}{2} = \frac{2 M ℓ ^{2}}{3} .

JEE Main · 2021 · 27 Aug 2021 · Shift 1 · Single correct option

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