Centre of Mass of Discrete Particles question from JEE Main 2026 (6 Apr, Shift 1)

Answer: (a)

1. Apex angle $120^\circ$, sides 10 m: half-base $= 10\sin 60^\circ = 5\sqrt{3}$ m, height $= 10\cos 60^\circ = 5$ m.
2. P is the mid-point of the 5 m bisector, so relative to P: 2 kg at $(-5\sqrt{3}, -2.5)$, 3 kg at $(5\sqrt{3}, -2.5)$, 15 kg at $(0, 2.5)$.
3. $x_{cm} = \frac{-10\sqrt{3} + 15\sqrt{3}}{20} = \frac{\sqrt{3}}{4}$
4. $y_{cm} = \frac{-5 - 7.5 + 37.5}{20} = 1.25$
5. So the answer is (a) $\left(\frac{\sqrt{3}}{4}, 1.25\right)$.