Question

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

Answer 1

Volume of smaller drop $=\frac{4}{3}\pi r^3$

two smaller drops merge to forms larger drop of radius R.

their volumes add up, i.e.

$V^{\prime }=2V$

$\Rightarrow \frac{4}{3}\pi R^3=2\times \frac{4}{3}\pi r^3\Rightarrow R^{\prime }=2^{1/3}r.$

thus, new surface energy,

$u=2T.A$ [A = Surface area of drop]

$=2T.4\pi R^{\prime }2$

$=8T.\pi 2^{2/3}{r}^2$

$\Rightarrow u=2^{11/3}\pi T{r}^2$ (Ans)