Question

Six identical particles each of mass 0.5 kg are
arranged at the corners of a regular hexagon of
side length 0.5m. If one of the particle is
removed, the shift in the centre of mass is
(A) 1/8
(B) 1/10
(C) 1/12
(D) 1/14​

Answer 1

I hope answer 1/10......

Answer 2

Go through the attached diagram below to get a picture of the diagram to be drawn.

Now, let us choose to omit $(a,0)$;

$x_c_m = \frac{m[\frac{a}{2}-\frac{a}{2}-a-\frac{a}{2}+\frac{a}{2}]}{5m} = \frac{1}{5}(-a) = \frac{-a}{5}$

In the beginning, the value of $x_c_m = 0$

∴ shift in the centre of mass = $\frac{a}{5} = \frac{0.5}{5} m = \frac{1}{10}m$

Ans) (B) $\frac{1}{10}$