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JEE Main 2020 (3 Sep, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2020 (3 Sep, Shift 2)

Q144JEE Main 2020 · 3 SEP · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

A massless equilateral triangle EFG of side ‘a’ (As shown in figure) has three particles of mass m situated at its vertices. The moment of inertia of the system about the line EX perpendicular to EG in the plane of EFG is $\frac{N}{20}$ ma2 where N is an integer. The value of N is _____.

Numerical answer type. Work it out, then open the answer.

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Answer: 25

$I = \sum m r^2$ , with $r$ the perpendicular distance from line EX.
E lies on the axis: $r = 0$ . F is above the midpoint of EG: $r = \frac{a}{2}$ . G: $r = a$ .
$I = m\frac{a^2}{4} + ma^2 = \frac{5}{4}ma^2 = \frac{25}{20}ma^2$
So $N = 25$ .

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