A solid sphere is placed inside a cube such that it touches all six faces. What percentage of the cube's volume is not occupied by the sphere?

Correct option is A

Given:
A solid sphere is placed inside a cube such that it touches all six faces of the cube.
Formula Used:
Volume of cube = a³
Volume of sphere = $$\frac 43$$​πr³
Percentage not occupied = $$\left(\frac{\text{Volume of cube} - \text{Volume of sphere}}{\text{Volume of cube}}\right) \times 100$$​​
Solution:
Let the side of the cube be a.
Since the sphere touches all six faces, diameter of sphere = a, so radius r = $$\frac a2$$​.
Volume of cube = a³​​
Volume of sphere  $$= \frac{4}{3}\pi\left(\frac{a}{2}\right)^3= \frac{4}{3}\pi\left(\frac{a^3}{8}\right)= \frac{\pi a^3}{6}$$​​
​$$\text{Volume not occupied}= a^3 - \frac{\pi a^3}{6}= a^3\left(1 - \frac{\pi}{6}\right) \\\text{Percentage not occupied}= \left(1 - \frac{\pi}{6}\right) \times 100 \\\text{Taking } \pi = 3.14, \\= \left(1 - \frac{3.14}{6}\right) \times 100= 47.67\% \;(\text{approx.})$$​​
Therefore, approximately 47.67% of the cube’s volume is not occupied by the sphere.