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. Also’ the formula for volume of different shapes are already defined. So’ you only need to equate them.