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′.

We know that volume of the solid sphere = (4/3)πr³

Therefore, volume of 27 solid sphere = 27×(4/3)πr³ = 36 π r³ _________(1)

(i) Let the radius of the new solid iron sphere = r’

Volume of this new sphere = (4/3)π(r’)³_________(2)

Comparing these 2 equations,

(4/3)π(r’)³ = 36 π r³

(r’)³ = 27r³

r’= 3r

The radius of the new sphere will be 3r.

(ii) We know that surface area sphere = 4πr²

SA of the iron sphere of radius r =4πr²

SA of iron sphere of radius r’= 4π (r’)²

Taking ratio,

S/S’ = (4πr²)/( 4π (r’)²)

= r²/(3r’)² 

= 1/9

The ratio of S and S’ is 1: 9.