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

The correct option is D $\frac{7\pi \sqrt{3}}{64}$
In a bcc unit cell, the spheres are present at the corners as well as at the centre, the spheres at corners does not touch each other. But the spheres at body centre touches all the spheres at the corners of the unit cell.
Let us consider, the radius of sphere B be 'R' and the radius of sphere A be 'r'.
$\therefore$
$r=\frac{R}{2}$

The total number of spheres B present in the unit cell will be 2. Similarly, the total number of spheres A present in the unit cell will also be 2.
In order to find out the volume of sphere B not occupied by the sphere A, we need to substact the total volume of the spheres A from the total volume of spheres B.
$\text{Volume of spheres of B unoccupied by A}=\text{Total volume of sphere B}-\text{Total volume of sphere A}$

Total volume of B unoccupied by A
in a unit cell
$=(2\times \frac{4}{3}\pi R^3)-(2\times \frac{4}{3}\pi r^3)$
Since, $r=\frac{R}{2}$
Total volume of B unoccupied by A
in a unit cell is
$=2\times \frac{4}{3}(R^3-\frac{R^3}{8})\times \pi =\frac{7\pi R^3}{3}$

For bcc,
$\sqrt{3}a=4R$
$\Rightarrow a=\frac{4R}{\sqrt{3}}$
$\text{Volume of unit cell}=a^3=\frac{64}{3\sqrt{3}}{R}^3$
$\therefore \text{Required ratio}=\frac{7\pi R^3/3}{\frac{64}{3\sqrt{3}}{R}^3}=\frac{7\pi \sqrt{3}}{64}$