Question:
An 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 \(\dfrac{N}{20}ma^2\) where N is an integer. The value of N is _____.
An 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 \(\dfrac{N}{20}ma^2\) where N is an integer. The value of N is _____.
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Find each particle's perpendicular distance from the axis EX (perpendicular to EG at E) and add \(mr^2\). The distances are 0, \(a/2\) and \(a\).
Updated On: Nov 5, 2026
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Correct Answer: 25
Solution and Explanation
Take E as origin with EG along the x-axis, so the axis EX is the line x = 0 in the plane.
E is on the axis (distance 0). G is at \(x = a\), so its distance is \(a\). F lies above the midpoint of EG, at \(x = a/2\), so its distance is \(a/2\).
\(I = m(0)^2 + m\left(\dfrac a2\right)^2 + m(a)^2 = \dfrac{ma^2}{4}+ma^2 = \dfrac54 ma^2\).
\(\dfrac54 = \dfrac{25}{20}\), so \(N = 25\).
Answer: 25
E is on the axis (distance 0). G is at \(x = a\), so its distance is \(a\). F lies above the midpoint of EG, at \(x = a/2\), so its distance is \(a/2\).
\(I = m(0)^2 + m\left(\dfrac a2\right)^2 + m(a)^2 = \dfrac{ma^2}{4}+ma^2 = \dfrac54 ma^2\).
\(\dfrac54 = \dfrac{25}{20}\), so \(N = 25\).
Answer: 25
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