Four cubes of side 'a' each of mass 40 g, 20 g, 10 g and 20 g are arranged in XY plane as shown in the figure. The coordinates of COM of the combination with respect to point O is:

\frac{19 a}{18}, \frac{17 a}{18}

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\frac{17 a}{18}, \frac{11 a}{18}

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\frac{17 a}{18}, \frac{13 a}{18}

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\frac{13 a}{18}, \frac{17 a}{18}

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Solution

The correct option is A $\frac{19 a}{18}, \frac{17 a}{18}$
Since there are 4 cubes,
$X_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3} + m_{4} x_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
$Y_{c m} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3} + m_{4} y_{4}}{m_{1} + m_{2} + m_{3} + m_{4}}$
Length of cube $= a$ ;
$x_{1} = \frac{a}{2}$ , $x_{2} = \frac{3 a}{2}$ , $x_{3} = \frac{5 a}{2}$ , $x_{4} = \frac{a}{2}$
$y_{1} = \frac{a}{2}$ , $y_{2} = \frac{a}{2}$ , $y_{3} = \frac{a}{2}$ , $y_{4} = \frac{3 a}{2}$
$m_{1} = 20$ , $m_{2} = 10$ , $m_{3} = 20$ , $m_{4} = 40$ :
$X_{c m} = \frac{20 \times \frac{a}{2} + 10 \times \frac{3 a}{2} + 20 \times \frac{5 a}{2} + 40 \times \frac{a}{2}}{20 + 10 + 20 + 40} = \frac{19}{18} a$ ;
$Y_{c m} = \frac{20 \times \frac{a}{2} + 10 \times \frac{a}{2} + 20 \times \frac{a}{2} + 40 \times \frac{3 a}{2}}{20 + 10 + 20 + 40} = \frac{17}{18} a$

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