The correct option is D 1.8π×10−12 J
Let r be the radius of smaller drop and R of the bigger one.
Equating the initial and final volumes, we have
43πR3=1000×43πr3
⇒R=10r=10×10−7 m
As we know,
Water drops have only one free surface.
∴ Net Change in surface area;ΔA=4πR2−1000×4πr2
=4π(10−6)2−1000×4π(10−7)2
=−36π×10−12 m2
Here, negative sign implies that surface area is decreasing. Hence energy released in the process,
W=T×|ΔA|=0.05×36π×10−12 m2=1.8π×10−12 J