Two solid spheres of radii 2 cm and 4 cm are melted and recast into a cone of height 8 cm. Find the radius of the cone so formed.

Given,

Radius of first solid sphere (r₁) = 2 cm

Radius of second solid sphere (r₂) = 4 cm

Height of cone (h) = 8 cm

Let radius of cone be r cm.

Since, two spheres are melted and recasted into cone.

∴ Volume of 1st sphere + Volume of 2nd sphere = Volume of cone

$\Rightarrow \dfrac{4}{3}π(r$1)^3 + \dfrac{4}{3}π(r2)^3 = \dfrac{1}{3}πr^2h \\[1em] \Rightarrow 4(r1)^3 + 4(r2)^3 = r^2h \\[1em] \Rightarrow r^2 = \dfrac{4r1^3 + 4r2^3}{h} \\[1em] \Rightarrow r^2 = \dfrac{4 \times 2^3 + 4 \times 4^3}{8} \\[1em] \Rightarrow r^2 = \dfrac{4 \times 8 + 4 \times 64}{8} \\[1em] \Rightarrow r^2 = \dfrac{32 + 256}{8} \\[1em] \Rightarrow r^2 = \dfrac{288}{8} \\[1em] \Rightarrow r^2 = 36 \\[1em] \Rightarrow r = \sqrt{36} \\[1em] \Rightarrow r = 6 \text{ cm}.

Hence, the radius of cone formed = 6 cm.