Question:
A sphere of maximum volume is cut-out from a solid hemisphere of radius r, what is the ratio of the volume of the hemisphere to that of the cut-out sphere?
Solution:
Since, a sphere of maximum volume is cut out from a solid hemisphere of radius.
i.e., radius of sphere
Therefore,
The volume of sphere
$=\frac{4}{3} \pi\left(\frac{r}{2}\right)^{3}$
$v_{1}=\frac{1}{6} \pi r^{3}$.......(i)
The volume of hemisphere $v_{2}=\frac{2}{3} \pi r^{3}$ .......(ii)
Divide (i) by (ii).
$\frac{v_{1}}{v_{2}}=\frac{\frac{1}{6} \pi r^{3}}{\frac{2}{3} \pi r^{3}}$
$=\frac{1}{6} \times \frac{3}{2}$
$\frac{v_{1}}{v_{2}}=\frac{1}{4}$
Hence, $v_{2}: v_{1}=4: 1$