Question

A big water drop is formed by the combination of n small water drop of eqaul radii. The ratio of the surface energy of n drop to surface energy of big drop is

Answer 1

surface energy, r u saying about new capacity or potential that the drop will have

if asking about new potential then

V= (n^1/3)v

Answer 2

volume of big drop = n × volume of small drop

or, 4/3π R³ = n × 4/3 πr³

or, R³ = nr³

or, $R=\sqrt[3]{n}r$

surface area of small drop ,a = 4πr²

let T is surface tension of water

so, surface energy of n small water drops = nT.a= nT × 4πr² .....(1)

surface area of big water drop , A = 4πR²

so, surface energy of big water drop = T.A = T × 4πR² = T × 4π($\sqrt[3]{n}r)^2$......(2)

now, ratio of the surface energy of n drops to surface energy of big drop = n.T × 4πr²/T.4π($n^{2/3}$)r²

= $\sqrt[3]{n}$

hence, answer is $\bf{\sqrt[3]{n}}$