What is the center of mass of the object below?
The shape is uniformly dense.
All answers are stated with respect to the origin, which is at the lower left corner of the object.

(\frac{2}{3} a, \frac{4}{3} a)

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(a, a)

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(\frac{2}{3} a, \frac{2}{3} a)

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(\frac{4}{3} a, \frac{4}{3} a)

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(\frac{5}{6} a, \frac{5}{6} a)

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Solution

The correct option is B $(\frac{5}{6} a, \frac{5}{6} a)$
Let the object is made by three same squares of side $a$ , the centre of mass of each square will be at the intersection of diagonals of square ,
now coordinates of CM of upper square , $(a / 2, 3 a / 2)$ ,
coordinates of CM of lower square at origin , $(a / 2, a / 2)$
coordinates of CM of lower square (rightmost) , $(3 a / 2, a / 2)$ ,
therefore ,
$x_{c m} = \frac{m_{1} x_{1} + m_{2} x_{2} + m_{3} x_{3}}{m_{1} + m_{2} + m_{3}}$ , $y_{c m} = \frac{m_{1} y_{1} + m_{2} y_{2} + m_{3} y_{3}}{m_{1} + m_{2} + m_{3}}$ ,
mass of all the squares will be same as they have same density and same dimensions (volume) , let mass of each square is $m$ ,
$x_{1} = a / 2$ , $x_{2} = a / 2$ , $x_{3} = 3 a / 2$ , $y_{1} = 3 a / 2$ , $y_{1} = a / 2$ , $y_{1} = a / 2$ ,
$x_{c m} = \frac{m a / 2 + m a / 2 + 3 m a / 2}{m + m + m} = 5 a / 6$ ,
$y_{c m} = \frac{3 m a / 2 + m a / 2 + m a / 2}{m + m + m} = 5 a / 6$ ,

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