Two small drops each of radius $r$ coalesce to form a larger drop. Find energy of the new drop. (surface tension is $T$ )

Hint: Use the formula of the volume of the raindrop given below and the relation between the volume of the larger rain drop to that of the smaller rain drop. From that calculate the radius of the raindrop substitute in the energy formula to find the energy equation.

Formula used:
(1) The formula of the volume of the drop of the rain is given by
$V = \dfrac{4}{3}\pi {R^3}$
Where $V$ is the volume of the spherical raindrop, $R$ is the radius of the raindrop.
(2) The new surface energy is calculated by
$U = 2T.A$
Where $U$ is the energy, $T$ is the surface tension of the raindrop and $A$ is the area of the spherical raindrop.

Complete step by step solution:
It is given that the two small rain drops merge together to form the larger rain drops.
Hence substituting the volume of the smaller and the larger rain drop.
$\Rightarrow 2 \times \dfrac{4}{3}\pi {r^3} = \dfrac{4}{3}\pi {R^3}$
By canceling the similar terms in both the left hand side and right hand side of the equation.
$\Rightarrow R = {2^{1/3}}r$
The new surface energy formed is calculated by using the formula.
$\Rightarrow U = 2T.A$
$\Rightarrow U = 2T.4\pi {R^2}$
By substituting the value of the radius of the larger rain drop,
$\Rightarrow U = 8T.\pi {2^{2/3}}{r^2}$
By doing the basic arithmetic operation in the above step, we get
$\Rightarrow U = {2^{11/3}}\pi T{r^2}$

The volume of the larger rain drop is ${2^{11/3}}\pi T{r^2}$.

Note: The shape of the rain drop is bulkier at the bottom and lesser at the top of the raindrop. This is mainly due to the surface tension of the raindrop to reduce its surface only to the particular area. The rain drops will be greater size at the top and while reaching the soil, it gets separated into small droplets by the wind velocity.