A solid metallic hemisphere of radius 8 cm is melted and recasted into right circular cone of base radius 6 cm. Determine the height of the cone.

Radius of solid hemisphere (r) = 8 cm.

Volume of the hemisphere (V) = $\dfrac{2}{3}πr^3$

Radius of cone (R) = 6 cm.

Let height of cone = h cm.

Volume of cone = $\dfrac{1}{3}πR^2h$

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

$\therefore \dfrac{1}{3}πR^2h = \dfrac{2}{3}πr^3 \\[1em] \Rightarrow h = \dfrac{2π \times 8^3 \times 3}{3 \times π \times 6^2} \\[1em] \Rightarrow h = \dfrac{2 \times 512}{36} \\[1em] \Rightarrow h = \dfrac{1024}{36} \\[1em] \Rightarrow h = \dfrac{256}{9} \\[1em] \Rightarrow h = 28\dfrac{4}{9} \text{ cm}.$

Hence, the height of the cone is $28\dfrac{4}{9}$ cm.