A hollow sphere of internal and external diameters 4 cm and 8 cm respectively, is melted into a cone of base diameter 8 cm. Find the height of the cone.

For sphere,

Internal radius (r) = $\dfrac{4}{2}$ = 2 cm,

External radius (R) = $\dfrac{8}{2}$ = 4 cm.

For cone,

Base radius (r₁) = $\dfrac{8}{2}$ = 4 cm

Height = h.

Since, the hollow sphere is recasted into cone hence their volume will be equal.

$\therefore \dfrac{4}{3}π(R^3 - r^3) = \dfrac{1}{3}πr_1^2h$

Multiplying both sides by 3 and dividing by π we get,

$\Rightarrow 4(R^3 - r^3) = r_1^2h \\[1em] \Rightarrow 4(4^3 - 2^3) = (4)^2h \\[1em] \Rightarrow 4(64 - 8) = 16h \\[1em] \Rightarrow h = \dfrac{4 \times 56}{16} \\[1em] \Rightarrow h = \dfrac{224}{16} \\[1em] \Rightarrow h = 14 \text{ cm}.$

Hence, the height of the cone = 14 cm.