Let the radius of bigger droplet is R, then by volume conservation:
(34πR3)=1000(34π(1)3)
⇒R=10mm
Final potential energy:
Uf=1000(4πR2T)=1000×4π(10×10−3)2T
Initial potential energy:
Ui=4π×(r)2T=4π×(10−3)2T
Therefore, change in potential energy,
ΔE=T×1000×4π(10−3)2−T×4π(10×10−3)2
ΔE=4×π×7×10−2[1000−100]×10−6
ΔE=7.92×10−4J