(2012 Mains)
Solution:
Use the centre of mass formula $x_{cm} = \frac{\sum m_i x_i}{\sum m_i}$. Substituting the given values: $x_{cm} = \frac{300(0) + 500(40) + 400(70)}{300 + 500 + 400} = \frac{48000}{1200} = 40\text{ cm}$. Option (a) is correct.
Three masses are placed on the $x$-axis : $300\text{ g}$ at origin, $500\text{ g}$ at $x = 40\text{ cm}$ and $400\text{ g}$ at $x = 70\text{ cm}$. The distance of the center of mass from the origin is :
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