Percentage of free space in a body-centred cubic unit cell is ___ (nearest 2 digits)

Given, Free space
We need to find the percentage of free space in a bcc unit cell.

$$\begin{gathered} r=\frac{\sqrt{3}}{4}a \\ \text{ packing efficiency }=\frac{Z\times \text{ volume of sphere }}{\text{ volume of cube }} \\ Number of atoms Z = 2, \\ Sphere volume ==\frac{4}{3}{\pi }^3 \\ \text{cube volume }=a^3 \\ =\frac{2\times \frac{4}{3}\pi r^3}{a^3} \\ =\frac{2\times \frac{4}{3}\times \pi \times {\left[\frac{\sqrt{3}}{4}a\right]}^3}{a^3} \\ =2\times \frac{4}{3}\times \frac{22}{7}\times \frac{\sqrt{3}\times \sqrt{3}\times \sqrt{3}}{4\times 4\times 4}=0.68 \\ So, 0.68\times 100 = 68\% \\ = 100-68=32 \end{gathered}$$

Therefore, the correct answer is 32