A certain number of spherical drops of a liquid of radius $r$ coalesce to form a single drop of radius $R$ and

Question

A certain number of spherical drops of a liquid of radius $r$ coalesce to form a single drop of radius $R$ and volume $V$. If $T$ is the surface tension of the liquid, then:

Accepted Answer

1. **Principle:** When smaller drops coalesce to form a larger drop, the total volume remains conserved, but the total surface area decreases. The decrease in surface energy appears as released energy .
2. **Formulas:** 
   - Volume of a sphere: $V = \frac{4}{3}\pi R^3 = n \cdot \frac{4}{3}\pi r^3$ (where $n$ is the number of small drops).
   - Surface Area: $A = 4\pi R^2$.
   - Surface Energy: $E = T 	imes A$.
3. **Derivation:**
   - Initial Surface Area ($A_i$): $n(4\pi r^2)$. Using $V = n\frac{4}{3}\pi r^3$, we can substitute $n = \frac{3V}{4\pi r^3}$. Thus, $A_i = \left(\frac{3V}{4\pi r^3}ight)4\pi r^2 = \frac{3V}{r}$.
   - Final Surface Area ($A_f$): $4\pi R^2$. Using $V = \frac{4}{3}\pi R^3$, we get $A_f = \frac{3V}{R}$.
   - Change in Energy ($\Delta E$): $T(A_i - A_f) = T\left(\frac{3V}{r} - \frac{3V}{R}ight) = 3VT\left(\frac{1}{r} - \frac{1}{R}ight)$.
4. **Conclusion:** Since $r < R$, the term $(\frac{1}{r} - \frac{1}{R})$ is positive, indicating energy is released due to the decrease in surface area.

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