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 Sand S’.

Solution:

Volume of the solid sphere = (4/3)πr3

Volume of twenty seven solid sphere = 27×(4/3)πr3 = 36 π r3

(i) New solid iron sphere radius = r’

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

(4/3)π(r’)3 = 36 π r3

(r’)3 = 27r3

r’= 3r

Radius of new sphere will be 3r (thrice the radius of original sphere)

(ii) Surface area of iron sphere of radius r, S =4πr2

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

Now

S/S’ = (4πr2)/( 4π (r’)2)

S/S’ = r2/(3r’)2 = 1/9

The ratio of S and S’ is 1: 9.