64 small solid iron spheres of radius 'r' are melted to form a big sphere of radius R. If S and S' are surface areas of the small and the big sphere respectively, then find the ratio S' : S.
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 16 : 1
When solids are melted and recast, the volume stays the same. The volume of the big sphere gives \(R\) in terms of \(r\), and surface areas of spheres are in the ratio of the squares of their radii (\(S=4\pi r^{2}\)).
The radius of the big sphere (volume unchanged):
\(\displaystyle 64\times\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi R^{3}\)
\(R^{3}=64r^{3}\), so \(R=4r\)
The ratio of the surface areas:
\(\displaystyle\frac{S'}{S}=\frac{4\pi R^{2}}{4\pi r^{2}}=\frac{(4r)^{2}}{r^{2}}=16\), so \(S':S=16:1\)
Tip: Here S is the surface area of one small sphere. The total surface of all 64 small spheres would be \(64\times4\pi r^{2}\), four times that of the big sphere, so melting into one sphere reduces the total surface.