Question

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

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

Answer 1

The correct option is C $2^{2/3}:1$

Let us suppose the surface tension is $T$, the radius of smaller drop is $r$ and that of bigger drop is $R$.
As total volume of fluid will be same
$\therefore$ Volume of two smaller drops $=$ Volume of bigger drop
$\Rightarrow 2\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
$\Rightarrow R=2^{1/3}r$ ......(1)
Now, ratio of surface energies
$\frac{E_{big}}{E_{small}}=\frac{T\times 4\pi R^2}{T\times 4\pi r^2}$
$[$ surface energy $=$surface tension $\times$ surface area $]$
$=\frac{R^2}{r^2}=2^{2/3}:1$ $[$ from (1) $]$