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.

Given,

External diameter of the hollow sphere = 8 cm

So, external radius (R) = $\dfrac{8}{2}$ = 4 cm

Internal diameter of the hollow sphere = 4 cm

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

Diameter of cone = 8 cm

Radius of cone (r₁) = 4 cm

Let height of the cone = h cm.

Since, hollow sphere is melted and recasted into cone.

∴ Volume of sphere = Volume of cone

$\Rightarrow \dfrac{4}{3}π(R^3 - r^3) = \dfrac{1}{3}π(r$1)^2h \\[1em] \Rightarrow 4(R^3 - r^3) = r1^2h \\[1em] \Rightarrow 4 \times [(4)^3 - (2)^3] = (4)^2 \times h \\[1em] \Rightarrow 4 \times [64 - 8] = 16h \\[1em] \Rightarrow h = \dfrac{4 \times 56}{16} \\[1em] \Rightarrow h = 14 \text{ cm}.

Hence, the height of the cone = 14 cm.