A large number of water drops, each of radius r , combine to have a drop of radius R . If the surface tension is T and mechanical equivalent of heat is J , the rise in heat energy per unit volume will be:

Step-by-step solution

n × 4 3 π r 3 = 4 3 π R 3 ∴   n 1 / 3 r = R ∴ Total change in surface energy = n 4 π r 2 - 4 π R 2 T ⇒ 4 π T n r 2 - R 2 ∴ Heat energy = 4 π T n r 2 - R 2 J × 4 3 π R 3 = 3   T   J n r 2 R 3 - 1 R Put n r 3 = R 3 ∴   3 T J 1 r - 1 R