Question

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

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

Answer 1

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

Radius of smaller drop $=r$
Radius of bigger drop $=R$
$A_i=4\pi r^2$
$A_j=4\pi R^2$
We know, volume will remain constant.
$V_i=V_j$
$\Rightarrow 2\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3$
$\Rightarrow R=2^{1/3}r$
Also,
Surface energy; $U=T\times A$
$\therefore \frac{\text{Surface energy of bigger drop}}{\text{Surface energy of smaller drop}}=\frac{T\times A_j}{T\times A_i}$
$=\frac{T\times 4\pi R^2}{T\times 4\pi r^2}=\frac{2^{2/3}}{1}$