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

The centre of mass will be in the centre of a square when all identical particles are at its vertices. When a particle is eliminated, as seen below,

${\mathrm{X}}_{\mathrm{CM}}=\frac{\mathrm{m}\left(0\right)+\mathrm{m}\left(\mathrm{a}\right)+\mathrm{m}\left(0\right)}{3\mathrm{m}}=\frac{\mathrm{a}}{3}$

${\mathrm{Y}}_{\mathrm{CM}}=\frac{\mathrm{m}\left(0\right)+\mathrm{m}\left(0\right)+\mathrm{m}\left(\mathrm{a}\right)}{3\mathrm{m}}=\frac{\mathrm{a}}{3}$

the new position of the centre of mass from origin

$\mathrm{c}\text{'}=\sqrt{\frac{{\mathrm{a}}^2}{4}+\frac{{\mathrm{a}}^2}{4}}=\frac{\sqrt{2\mathrm{a}}}{3}$

$\mathrm{shift}=\mathrm{c}-\mathrm{c}\text{'}=\frac{\mathrm{a}}{\sqrt{2}}-\frac{\mathrm{a}\sqrt{2}}{3}=\frac{2\sqrt{2}}{3\sqrt{2}}=\frac{2}{3}\mathrm{m}$

Hence the correct answer is 2/3m.