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An equilateral triangular lamina ABC of mass M has moment of inertia I₀ about an axis through its centroid perpendicular to its plane. D, E, F are the mid-points of the sides and the medial triangle DEF is cut out. The moment of inertia of the remaining part about the same axis is

Asked in JEE Main 11th Jan 1st Shift 2019 · Composite and cavity bodies

Answer: (1) (15)/(16)I₀

Step-by-step solution

DEF is similar with half the side, so its area is 1/4 (mass /M4) and, sharing the same centroid, its MI scales as the square of its mass-times-size: I_DEF=(1/4)(1/2)² I₀=1/(16)I₀.

Iᵣₑₘ=I₀-1/(16)I₀=(15)/(16)I₀.

Why the other options are wrong

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