Two water droplets combine to form a large drop. In this process, the energy is :

Hint: When two water droplets combine to form a large and a single drop, the net surface area decreases because the sum of areas of two drops is certainly bigger than a single drop. Hence, its surface tension also decreases.

Complete step by step answer:
Let us consider two drops of radius r.
Let the radius of the big drop be R.
Since the volume remains constant in this entire process we can say that,
$
  2\times \dfrac{4}{3}\pi {{r}^{3}}=\dfrac{4}{3}\pi {{R}^{3}} \\
  \Rightarrow 2{{r}^{3}}={{R}^{3}} \\
$
Hence, $R={{2}^{1/3}}r$
Now, we will compare the area in both the cases
Before combination the area is considered as $4\pi {{r}^{2}}n$
After combination it is $4\pi {{R}^{2}}$
We know that surface energy is the product of surface tension and area.
The change in surface energy can be calculated as
$
 \Delta E=S\times \Delta A \\ =S(4\pi {{R}^{2}}-2\times 4\pi {{r}^{2}}) \\
 \Rightarrow \Delta E =4\pi S({{R}^{2}}-2{{r}^{2}}) \\
 \Rightarrow \Delta E=4\pi S({{2}^{2/3}}{{r}^{2}}-2{{r}^{2}}) \\
 \Rightarrow \Delta E =4\pi S{{r}^{2}}({{2}^{2/3}}-2) \\
$
$({{2}^{2/3}}-2)$ is a negative value. All the other quantities are positive.
Hence, the value of surface energy has faced a decrease.
Hence we can say that the system has released its extra energy and has moved towards a more stable state.
The liberation of energy is the end result that we obtain by combining two or more drops to form a single droplet of liquid.


Note: The work done is stored in the unit surface area and this is done in the form of potential energy. This potential energy is referred to as the surface energy of a system. This is dependent on temperature. As temperature increases, the net cohesive force decreases and surface energy decreases.