question_answer

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 : [JEE Main 12-4-2019 Afternoon]

A) \[\left( \frac{7}{12}m,\frac{\sqrt{3}}{8}m \right)\]   

B) \[\left( \frac{\sqrt{3}}{4}m,\frac{5}{12}m \right)\]

C) \[\left( \frac{7}{12}m,\frac{\sqrt{3}}{4}m \right)\]

D) \[\left( \frac{\sqrt{3}}{8}m,\frac{7}{12}m \right)\]

Correct Answer: C

Solution :

The co-ordinates of the centre of mass \[{{\vec{r}}_{cm}}=\frac{0+150\times \left( \frac{1}{2}\hat{i}+\frac{\sqrt{3}}{2}\hat{j} \right)+100\times \hat{i}}{300}\] \[{{\vec{r}}_{cm}}=\frac{7}{12}\hat{i}+\frac{\sqrt{3}}{4}\hat{j}\] \[\therefore \]Co-ordinate\[\left( \frac{7}{12},\frac{\sqrt{3}}{4} \right)m\]