Question

Surface tension (T) of a liquid is equal to the work (W) required to increase the surface area ($\Delta A$) of the liquid film by unity at constant temperature. If the volume of a big drop is $27 \times$ the volume of a small drop, the change in surface energy is:

• A. $2\pi D^{2}T$
• B. $4\pi D^{2}T$
• C. $\pi D^{2}T$
• D. $8\pi D^{2}T$

Hint:

When $n$ small drops coalesce into one big drop, surface area decreases and energy is released.

Answer 1

The Correct Option is A

Step 1: Concept
Surface energy $W = T \Delta A$.
Step 2: Analysis
Volume of big drop $V' = 27V$. Since $V \propto R^{3}$, the radius of the big drop $R = 3r$ or $r = D/6$. Change in energy $= T(A_{2} - A_{1}) = T[27(4\pi r^{2}) - 4\pi R^{2}]$. Substituting $r = D/6$ and $R = D/2$: $4\pi T [\frac{3D^{2}}{4} - \frac{D^{2}}{4}]$.
Step 3: Conclusion
The change in surface energy is $2\pi D^{2}T$.
Final Answer: (A)