A sphere of maximum volume is cut-out from a solid

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, $v_{2}: v_{1}=4: 1$