From a sphere of radius 1 m, a sphere of radius 0.5 m is removed from the edge. The shift in the C.M. is

Consider ${\mathrm{m}}_1$ as a sphere with a radius of 0.5 and ${\mathrm{m}}_2$ as the remaining piece of the sphere after ${\mathrm{m}}_1$ has been removed.
Let the origin be the centre of a sphere with a radius of 1 m, and let ${\mathrm{x}}_1$ and ${\mathrm{x}}_2$ be the x components of the corresponding centres of mass ${\mathrm{m}}_1$ and ${\mathrm{m}}_2$.

${\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{b}}^3\mathrm{a}}{{\mathrm{R}}^3-{\mathrm{b}}^3}=\frac{{\displaystyle \frac{{\mathrm{R}}^3}{8}\times \frac{\mathrm{R}}{2}}}{{\mathrm{R}}^3-{\displaystyle \frac{{\mathrm{R}}^3}{8}}}$

${\mathrm{x}}_{\mathrm{cm}}=\frac{{\mathrm{R}}^4}{10}\times \frac{8}{7{\mathrm{R}}^3}=\frac{\mathrm{R}}{14}=\frac{1}{14}\mathrm{m}$

Hence the correct answer is 1/14m.

x₁ = 0.5

x₂ = x

Let density of sphere be d

${\mathrm{m}}_1 = \mathrm{d}\frac{4}{3}\times \mathrm{\pi }\times 0.5^3$

${\mathrm{m}}_2 = \mathrm{d}\frac{4}{3}\times \mathrm{\pi }\times (1^3-0.5^3)$

centre of mass of m₁ and m₂ is at origin

$0=\mathrm{d}\frac{4}{3}\times \mathrm{\pi }\times (0.5^3)\times 0.5+\mathrm{d}\frac{4}{3}\times \mathrm{\pi }\times (1^3-0.5^3)\times \mathrm{x}$

$\Rightarrow \mathrm{x}=-\frac{1}{14}\mathrm{m}$