Q 11-06-234JEE MainJEE Main 2019 (12 Apr, Shift 2)Easy
Three particles of masses $50$ g, $100$ g and $150$ g are placed at the vertices of an equilateral triangle of side $1$ m (as shown in the figure). The $(x, y)$ coordinates of the centre of mass will be:
Answer: (A) $\left(\dfrac{7}{12}\ \text{m}, \dfrac{\sqrt3}{4}\ \text{m}\right)$
Positions: $m_1 = 50$ g at $(0, 0)$, $m_2 = 100$ g at $(1, 0)$, $m_3 = 150$ g at $\left(0.5, \dfrac{\sqrt3}{2}\right)$.
$$x = \frac{100(1) + 150(0.5)}{300} = \frac{7}{12}\ \text{m},\qquad y = \frac{150\times\frac{\sqrt3}{2}}{300} = \frac{\sqrt3}{4}\ \text{m}$$
Solution by Sreeraj P, M.Sc Physics