Question

A hemisphere of lead of radius 9 cm is cast into a right circular cone of height 72 cm.Find the radius of the base of the cone.

Answer 1

Let $rcm$ be the radius of the cone.
Given: radius of hemisphere $r_1=9cm$
Cone of height $=72cm$
Volume of hemisphere = volume of cone
$\Rightarrow \frac{2}{3}\times \pi (r_1{)}^3=\frac{1}{3}\times \pi \times r^2\times h$
$\Rightarrow \frac{2}{3}\times \pi \times 9^3=\frac{1}{3}\times \pi \times r^2\times 72$
$\Rightarrow r^2=\frac{9\times 9\times 9\times 2}{72}$
$r=\sqrt{\frac{81}{4}}$
$\therefore r=\frac{9}{2}=4.5cm$ is the required radius of cone.