Four cubes of side ' $a$ ' each of mass $40 g m, 20 g m, 10 g m, 20 g m$ are arranged in $X Y$ plane as shown in figure. The coordinates of the centre of mass of the combination with respect to origin are:

19 a / 18, 17 a / 18

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17 a / 18, 11 a / 18

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17 a / 18, 13 a / 18

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13 a / 18, 17 a / 18

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Solution

The correct option is B $19 a / 18, 17 a / 18$
Here, $m_{4} = 40 g m$ , $m_{1} = 20 g m$ , $m_{2} = 10 g m$ and $m_{3} = 20 g m$
Center of mass of each cube lies at its geometrical centre.
Also, $x_{1} = 0.5 a$ , $x_{2} = 1.5 a$ , $x_{3} = 2.5 a$ and $x_{4} = 0.5 a$
Also, we get $y_{1} = y_{2} = y_{3} = 0.5 a$ and $y_{4} = 1.5 a$
X-coordinate of centre of mass $X_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$⟹$ $X_{c m} = \frac{(20) (0.5 a) + (10) (1.5 a) + (20) (2.5 a) + (40) (0.5 a)}{20 + 10 + 20 + 40} = \frac{19}{18} a$
Y-coordinate of centre of mass $Y_{c m} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$⟹$ $Y_{c m} = \frac{(20) (0.5 a) + (10) (0.5 a) + (20) (0.5 a) + (40) (1.5 a)}{20 + 10 + 20 + 40} = \frac{17}{18} a$

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