Let the radius of the bigger drop be R. Since the total volume remains conserved during the process:
8×34πr3=34πR3
R3=8r3⇒R=2r
Initial surface energy of the 8 drops is:
Ui=8×(4πr2S)=32πr2S
Final surface energy of the bigger drop is:
Uf=4πR2S=4π(2r)2S=16πr2S
The surface energy released in the process is:
ΔU=Ui−Uf=32πr2S−16πr2S=16πr2S
Answer: 16πr2S