Solid metallic spheres of diameters 12 cm, 16 cm and 20 cm respectively, are melted to form a single solid sphere. Find the radius of the resulting sphere.

By formula,

Radius = $\dfrac{\text{Diameter}}{2}$

So,

r₁ = $\dfrac{12}{2}$ = 6 cm,

r₂ = $\dfrac{16}{2}$ = 8 cm,

r₃ = $\dfrac{20}{2}$ = 10 cm.

Let radius of resulting sphere be r cm.

Since, solid metallic spheres of diameters 12 cm, 16 cm and 20 cm respectively, are melted to form a single solid sphere.

∴ Volume of resulting sphere = Sum of volume of three smaller spheres

$\therefore \dfrac{4}{3}πr^3 = \dfrac{4}{3}π(6)^3 + \dfrac{4}{3}π(8)^3 + \dfrac{4}{3}π(10)^3 \\[1em] \Rightarrow \dfrac{4}{3}πr^3 = \dfrac{4}{3}π\Big(6^3 + 8^3 + 10^3\Big) \\[1em] \Rightarrow r^3 = 216 + 512 + 1000 \\[1em] \Rightarrow r^3 = 1728 \\[1em] \Rightarrow r = \sqrt[3]{1728} \\[1em] \Rightarrow r = 12 \text{ cm}.$

Hence, radius of resultant sphere = 12 cm.