IIT JEE Advanced 2009 Paper 1

The coordinates of the centre of mass are defined as $\vec{R}_{CM} = \frac{\sum_i m_i \vec{r}_i}{\sum_i m_i}$. Taking only the $y$-components, $Y_{CM} = \frac{\sum_i m_i y_i}{\sum_i m_i}$. The mass and the $y$-coordinate of each part of the drawing are: outer circle: mass $6m$, $y = 0$. left inner circle: mass $m$, $y = a$. right inner circle: mass $m$, $y = a$. vertical line segment: mass $m$, $y = 0$. horizontal line segment: mass $m$, $y = -a$. Total mass of the ink $= 6m + m + m + m + m = 10m$. Thus, $Y_{CM}$ = $= \frac{ma + ma - ma}{10m} = \frac{ma}{10m} = \frac{a}{10}$. Hence the $y$-coordinate of the centre of mass of the ink in the drawing is $\frac{a}{10}$, which is the option marked as correct.