Surface energy of a drop of liquid is 16mJ. When 8 of such identical drops coalesce to form a single drop

Ui=Initial energy=8×4πr2.T=8×16mJIf R be the radius of the bigger drop, Then43πR3=8×43πr3⇒R=2r∴ Uf=Final surface energy =4πR2T=4×4πr2T=4×16mJ∴ energy released=Ui−Uf=(8×16−4×16)mJ=64mJ