A solid sphere of radius 14 cm is melted and recast into 8 equal solid spheres. What is the sum of the surface areas of the 8 small spheres? (Take \(\pi=\frac{22}{7}\))
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 4928 cm2
The metal is the same, so the total volume does not change: \(8\times\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3\), which gives \(r=\frac{R}{2}\).
\(8r^3=14^3\), so \(2r=14\) and \(r=7\) cm
Surface area of one small sphere \(=4\times\displaystyle\frac{22}{7}\times7\times7=616\) cm2
Sum for 8 spheres \(=8\times616=4928\) cm2