A hemisphere of lead of radius 12 cm is melted and cast into a right circular cone of height 54 cm. Find the radius of the base of the cone.

Radius of hemisphere, r = 12 cm

Volume of hemisphere = $\dfrac{2}{3}π\text{r}^3$

Radius of the cone = R cm

Height of the cone, h = 54 cm

Volume of cone = $\dfrac{1}{3}π\text{R}^2 \text{h}$

Since, hemisphere is melted and recasted into a cone, the volume remains the same.

$\therefore \dfrac{1}{3}π\text{R}^2 \text{h} = \dfrac{2}{3}π\text{r}^3 \\[1em] \Rightarrow \dfrac{1}{3}\text{R}^2 \text{h} = \dfrac{2}{3}\text{r}^3 \\[1em] \Rightarrow \text{R}^2 = \dfrac{2 \times 3}{3 \times \text{h}} \times \text{r}^3 \\[1em] \Rightarrow \text{R}^2 = \dfrac{2 \times 3}{3 \times 54} \times 12^3 \\[1em] \Rightarrow \text{R}^2 = \dfrac{6}{162} \times 1728 \\[1em] \Rightarrow \text{R}^2 = \dfrac{10368}{162} \\[1em] \Rightarrow \text{R}^2 = 64 \\[1em] \Rightarrow \text{R} = \sqrt{64} \\[1em] \Rightarrow \text{R} = 8 \text{ cm.}$

Hence, the radius of the base of the cone is 8 cm.