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MHT CET 2020 (19 Oct, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2020 (19 Oct, Shift 2)

Q141MHT CET 2020 · 19 OCT · SHIFT 2Easy

Moment of Inertia & Radius of Gyration

Three point masses, each of mass $m$ are placed at the corners of an equilateral triangle of side $\ell$ . The moment of inertia of the system about an axis passing through one of the vertices and parallel to the side joining other two vertices, will be

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Answer: (c)

The mass at the vertex is on the axis and contributes zero.
The other two masses are at height $\frac{\sqrt{3}}{2}\ell$ from the axis.
$I = 2m \cdot \frac{3\ell^2}{4} = \frac{3}{2}m\ell^2$
So the answer is (c) $\frac{3}{2}m\ell^2$ .

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