1. Surface energy of a liquid drop

For a liquid drop, surface energy is proportional to surface area:

$E = T \cdot A$

where $T$ is surface tension and $A$ is surface area.

For a spherical drop of radius $r$:

$A = 4\pi r^2$

So,

$E = T(4\pi r^2)$

2. Initial surface energy

There are two small drops, each of radius $R$.

Surface energy of one drop:

$E_1 = T(4\pi R^2)$

For two drops:

$E_{\text{before}} = 2 \times 4\pi T R^2 = 8\pi T R^2$

3. Radius of the large drop after coalescence

Volume is conserved.

Initial total volume:

$2\left(\frac{4}{3}\pi R^3\right)$

Let the radius of the large drop be $R'$. Then,

$\frac{4}{3}\pi R'^3 = 2\left(\frac{4}{3}\pi R^3\right)$

$R'^3 = 2R^3$

$R' = 2^{1/3}R$

4. Final surface energy

Surface area of the large drop:

$A' = 4\pi R'^2 = 4\pi (2^{1/3}R)^2 = 4\pi 2^{2/3}R^2$

Thus final surface energy is:

$E_{\text{after}} = T \cdot 4\pi 2^{2/3}R^2$

5. Ratio of surface energies before and after

$\frac{E_{\text{before}}}{E_{\text{after}}} = \frac{8\pi T R^2}{4\pi T 2^{2/3}R^2}$

$= \frac{2}{2^{2/3}} = 2^{1/3}$

Therefore,

$E_{\text{before}} : E_{\text{after}} = 2^{1/3} : 1$

6. Option check

• A: $2^{1/3}:1$ ✅
• B: $1:2^{1/3}$ ❌
• C: $2:1$ ❌
• D: $1:2$ ❌

Hence the correct answer is A.