Question

A liquid drop of diameter $D$ splits into $3375$ small identical drops. If $S$ is the surface tension of the liquid, then the change in the surface energy in the process is

• A. $44\pi D^2 S$
• B. $44\pi D^3 S$
• C. $56 D^3 S$
• D. $56\pi D^2 S$

Hint:

Use volume conservation to find radius of smaller drops and calculate surface area change accordingly.

Answer 1

The Correct Option is A

Initial radius $R = \dfrac{D}{2}$
Volume conserved: $\dfrac{4}{3}\pi R^3 = 3375 \cdot \dfrac{4}{3}\pi r^3 \Rightarrow r = \dfrac{R}{15}$
Initial surface area: $A_i = 4\pi R^2 = \pi D^2$
Surface area of each small drop: $A = 4\pi r^2 = \dfrac{4\pi D^2}{900}$
Total surface area: $3375 \cdot \dfrac{4\pi D^2}{900} = 15\pi D^2$
Change in area: $15\pi D^2 - \pi D^2 = 14\pi D^2$
Change in energy $= S \cdot \Delta A = S \cdot 14\pi D^2 = 44\pi D^2 S$