Three particles of masses 50 g, 100 g and 150 g are placed at the… | JEE Main 2019 (12 Apr, Shift 2)

Q18JEE Main 2019 · 12 APR · SHIFT 2Medium

Centre of Mass of Discrete Particles

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:

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Answer: (c)

Positions: 50 g at $(0, 0)$ , 100 g at $(1, 0)$ , 150 g at $\left(0.5, \frac{\sqrt{3}}{2}\right)$ m.
$x_{cm} = \frac{100(1) + 150(0.5)}{300} = \frac{175}{300} = \frac{7}{12}$ m
$y_{cm} = \frac{150 \cdot \frac{\sqrt{3}}{2}}{300} = \frac{\sqrt{3}}{4}$ m
So the answer is (c) $\left(\frac{7}{12}m, \frac{\sqrt{3}}{4}m\right)$ .

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Centre of Mass of Discrete Particles question from JEE Main 2019 (12 Apr, Shift 2)

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