A body centered cubic lattice is made up of hollow shere of $B$. Sphere of solid $A$ are present in hollow sphere of $B$. Radius of $A$ is half of the radius of $B$. What is the ratio of total volume of sphere $B$ unoccupied by $A$ in unit cell and volume of unit cell?

The correct option is B $\frac{7\pi \sqrt{3}}{64}$
$\text{Effective number of atoms of B present in a unit cell}=2$
$\text{Total volume of B uncoupied by A in a unit cell}=2\times \frac{4}{3}(R^3-r^3)\times \pi =\frac{7}{3}\pi R^3(r=\frac{R}{2})$
$\text{Volume of unit cell}=a^3$
Again, in body centered cubic lattice, $\sqrt{3a}=4R$
$\therefore a^3={\left(\frac{4R}{\sqrt{3}}\right)}^3=\frac{64}{3\sqrt{3}}{R}^3$
Ratio of total volume of sphere $B$ unoccupied by $A$ in unit cell and volume of unit cell
$=\frac{\frac{7}{3}\pi R^3}{\frac{64}{3\sqrt{3}}{R}^3}=\frac{7\pi \sqrt{3}}{64}$