Question:
A hemisphere of lead of radius 6 cm is cast into a right circular cone of height 75 cm. Find the radius of the base of the cone.
Solution:
We have,
Radius of the hemisphere, $R=6 \mathrm{~cm}$ and
Height of the cone, $h=75 \mathrm{~cm}$
Let the radius of the base of the cone be $r$.
Now,
Volume of the cone = Volume of the hemisphere
$\Rightarrow \frac{1}{3} \pi r^{2} h=\frac{2}{3} \pi R^{3}$
$\Rightarrow r^{2}=\frac{2 R^{3}}{h}$
$\Rightarrow r^{2}=\frac{2 \times 6 \times 6 \times 6}{75}$
$\Rightarrow r^{2}=5.76$
$\Rightarrow r=\sqrt{5.76}$
$\therefore r=2.4 \mathrm{~cm}$
So, the radius of the base of the cone is 2.4 cm.