Question

Two drops of equal radius coalesce to form a bigger drop. What is ratio of surface energy of bigger drop to smaller one ?

• A. $2^{1/2} : 1$
• B. $1 :1$
• C. $2^{2/3} : 1$
• D. None of these

Answer 1

Answer: (C)

$2^{2/3} : 1$

Answer 2

Volume remains constant after coalescing. Thus,
$\frac{4}{3} \pi R^{3}=2 \times \frac{4}{3} \pi r^{3}$
where $R$ is radius of bigger drop and $r$ is radius of each smaller drop
$\therefore R=2^{1 / 3} r$
Now, surface energy per unit surface area is the surface tension. So, Surface energy, $W=T \Delta \,A$
or $W=4 \pi R^{2} T$
Therefore, surface energy of bigger drop
$W_{1} =4 \pi\left(2^{1 / 3} r\right)^{2} T$
$=\left(2^{2 / 3}\right) 4 \pi r^{2} T$
Surface energy of smaller drop
$W_{2}=4 \pi r^{2} T$
Hence, required ratio
$\frac{W_{1}}{W_{2}}=2^{2 / 3}: 1$