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A massless equilateral triangle EFG of side a has three particles of mass m at its vertices. The moment of inertia of the system about the line EX perpendicular to EG and lying in the plane of EFG is N/(20)ma². The value of N is ______.

Asked in JEE Main 3rd Sept 2nd Shift 2020 · Radius of gyration and numericals

Answer: 25

Step-by-step solution

Take EX as the perpendicular to EG through the vertex E. The mass at E is on the line (distance 0).

The mass at G is at perpendicular distance a: contributes ma². The apex F is at horizontal distance /a2: contributes m(/a2)²=(ma²)/4.

I=ma²+(ma²)/4=5/4 ma²=(25)/(20)ma², so N=25.

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