Twenty seven solid iron spheres, each of radius r and surface area S are melted to form a sphere with surface area S′. Find the

(i) radius r′ of the new sphere,

(ii) ratio of S and S′.

By formula,

The volume of a sphere = $\dfrac{4}{3}πr^3$

The volume of 27 solid spheres with radius r = $27 \times \dfrac{4}{3}πr^3 = 36πr^3$ ……(1)

The volume of the new sphere with radius r' = $\dfrac{4}{3}π(r')^3$ …….(2)

(i) Volume of the new sphere = Volume of 27 solid spheres

$\Rightarrow \dfrac{4}{3}π(r')^3 = 36πr^3 (\text{From equation (1) and (2)}) \\[1em] \Rightarrow (r')^3 = 36r^3 \times \dfrac{3}{4} \\[1em] \Rightarrow (r')^3 = 9r^3 \times 3 \\[1em] \Rightarrow (r')^3 = 27r^3 \\[1em] \Rightarrow r' = \sqrt[3]{27r^3} \\[1em] \Rightarrow r' = 3r$

Hence, radius of the new sphere r' = 3r.

(ii) By formula,

Surface area of each iron sphere (S) = 4πr²

Surface area of the new sphere (S') = $4π(r')^2 = 4π(3r)^2 = 36πr^2$

Required ratio = $\dfrac{4πr^2}{36πr^2}$

The ratio of the S and S’ = $\dfrac{1}{9}$

Hence, the ratio of S and S’ is 1 : 9.