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Q.

Four identical particles are each of mass 1kg are arranged at the corners of a square of side length $2 \sqrt{2} m$ . If one of the particles is removed, then the shift in the centre of mass is

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a

$\frac{2}{3} m$

b

$\frac{4}{3} m$

c

2m

d

$\frac{8}{3} m$

answer is B.

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Detailed Solution

$\begin{array}{l} d = \frac{2 \sqrt{2}}{\sqrt{2}} = 2 m, m_{1} = 4 m, m_{2} = m \\ Shift = \frac{m_{2} d}{m_{1} - m_{2}} = \frac{2}{3} m \end{array}$

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