MCQ
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
  • A
    $\left( {\frac{{\sqrt 3 }}{7}\,m,\,\frac{7}{{12}}\,m} \right)$
  • B
    $\left( {\frac{7}{{12}}m,\,\frac{{\sqrt 3 }}{8}\,m} \right)$
  • C
    $\left( {\frac{{\sqrt 3 }}{4}\,m,\,\frac{5}{{12}}\,m} \right)$
  • $\left( {\frac{7}{{12}}m,\,\frac{{\sqrt 3 }}{4}\,m} \right)$

Answer

Correct option: D.
$\left( {\frac{7}{{12}}m,\,\frac{{\sqrt 3 }}{4}\,m} \right)$
d
The co-ordinates of the center of mass 

${\overrightarrow r _{cm}} = \frac{{0 + 150 \times \left( {\frac{1}{2}i + \frac{{\sqrt 3 }}{2}\hat j} \right) + 100 \times \hat i}}{{300}}$

${\overrightarrow r _{cm}} = \frac{7}{{12}}\hat i + \frac{{\sqrt 3 }}{4}\hat j$

$\therefore \,\,Co - ordinate\left( {\frac{7}{{12}},\frac{{\sqrt 3 }}{4}} \right)m$

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