Question

A hemisphere of lead of radius $7cm$ is cast into a right circular cone of height $49cm$. Find the radius of the base.

Hint: Try to find out the volume of both hemispheres and cones and equate them.

Given,
Radius of Hemisphere $R = 7cm$
Height of cone $h = 49cm$
Volume of hemisphere $V = \dfrac{2}{3}\pi {R^3}$
$= \dfrac{2}{3}\pi \times {\left( 7 \right)^3}{\text{ c}}{{\text{m}}^3} \\ = \dfrac{2}{3} \times 343 \times \pi {\text{ c}}{{\text{m}}^3} \\ = \dfrac{{686}}{3}\pi {\text{ c}}{{\text{m}}^3} \\$
Volume of cone $V = \dfrac{1}{3}\pi {r^2}h$
$= \dfrac{1}{3}\pi {r^2} \times 49{\text{ cm}}$
The hemisphere is cast into a right circular cone. So the volume of the hemisphere will be equal to the volume of the cone.
$\therefore$Volume of hemisphere = Volume of cone
$\dfrac{{686}}{3}\pi c{m^3} = \dfrac{{49}}{3}\pi {r^2}cm \\ {r^2} = \dfrac{{686}}{3}{\text{ }}c{m^2} \\ {r^2} = 14{\text{ }}c{m^2} \\ r = \sqrt {14{\text{ }}c{m^2}} \\ r = \sqrt {14} {\text{ }}cm \\ r = 3.74{\text{ }}cm \\$
Hence, radius of base of cone $r = 3.74{\text{ }}cm$

Note: Whenever there is one shape converted into another, always keep in mind that their volumes will always be the same. Alsoâ€™ the formula for volume of different shapes are already defined. Soâ€™ you only need to equate them.