Question

Two small drops of liquid of same radius coalesce to form a big drop. The ratio of the total surface energies after and before the change is

• A. $2^{\frac{1}{3}}:1$
• B. $2^{-\frac{1}{3}}:1$
• C. $2^{-\frac{2}{3}}:1$
• D. $2^{\frac{2}{3}}:1$

Hint:

When drops merge, volume is conserved but surface area decreases.

Answer 1

The Correct Option is B

Step 1: Volume conservation.
Let radius of each small drop be $r$.
\[ 2 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3 \] \[ R = 2^{1/3} r \]
Step 2: Surface energy relation.
Surface energy $\propto$ surface area $\propto r^2$.

Step 3: Total surface energy before coalescence.
\[ E_{\text{before}} \propto 2 \times 4\pi r^2 = 8\pi r^2 \]
Step 4: Surface energy after coalescence.
\[ E_{\text{after}} \propto 4\pi R^2 = 4\pi (2^{1/3}r)^2 = 4\pi 2^{2/3}r^2 \]
Step 5: Ratio of surface energies.
\[ \frac{E_{\text{after}}}{E_{\text{before}}} = \frac{4\pi 2^{2/3}r^2}{8\pi r^2} = 2^{-1/3} \]
Step 6: Conclusion.
The required ratio is $2^{-1/3}:1$.