Four particles of mass $m_{1} = 2 m, m_{2} = 4 m$ and $m_{3} = m$ and $m_{4}$ are placed at four corners of the square. What should be the value of $m_{4}$ so that the center of mass of all the four particles are exactly at the center of the square?

2 m

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8 m

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6 m

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n o n e o f t h e s e

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Solution

The correct option is D $n o n e o f t h e s e$
Let center be the reference point and side of the square is 2a
$x_{c m}$ coordinate of the center of mass of the system is 0.
$m_{1}$ , $m_{2}$ , $m_{3}$ and $m_{4}$ is at -a,a,a and -a.
$x_{c m} = \frac{m_{1} \times - a + m_{2} \times a + m_{3} \times a + m_{4} \times - a}{m_{1} + m_{2} + m_{3} + m_{4}} = 0$
$\Rightarrow 2 m \times - a + 4 m \times a + m \times a + m_{4} \times - a = 0$
$\Rightarrow m_{4} = 3 m$
hence D option is correct

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