Metallic spheres of radii 6 cm, 8 cm and 10 cm respectively are melted and recasted into a single solid sphere. Taking π = 3.1, find the surface area of the solid sphere formed.

Let radius of spheres be r₁, r₂ and r₃ cm respectively.

Let r be the radius of new sphere formed.

Volume of new sphere formed = Total volume of the three spheres melted

$\Rightarrow \dfrac{4}{3}πr^3 = \dfrac{4}{3}π(r$1)^3 + \dfrac{4}{3}π(r2)^3 + \dfrac{4}{3}π(r3)^3 \\[1em] \Rightarrow \dfrac{4}{3}πr^3 = \dfrac{4}{3}π(r1^3 + r2^3 + r3^3) \\[1em] \Rightarrow r^3 = r1^3 + r2^3 + r_3^3 \\[1em] \Rightarrow r^3 = (6)^3 + (8)^3 + (10)^3 \\[1em] \Rightarrow r^3 = 216 + 512 + 1000 \\[1em] \Rightarrow r^3 = 1728 \\[1em] \Rightarrow r^3 = (12)^3 \\[1em] \Rightarrow r = 12 \text{ cm.}

By formula,

Surface area of sphere = 4πr²

= 4 × 3.1 × 12²

= 1785.6 cm².