Question

A certain number of spherical drops of a liquid of radius $r$ each, coalesce to form a single drop of radius $R$ and volume $V$. If $T$ is the surface tension of the liquid, then

• A. Energy $=3VT(\frac{1}{r}+\frac{1}{R})$ is released
• B. Energy $=3VT(\frac{1}{r}+\frac{1}{R})$ is absorbed
• C. Energy $=3VT(\frac{1}{r}-\frac{1}{R})$ is released
• D. Energy $=3VT(\frac{1}{r}-\frac{1}{R})$ is absorbed

Answer 1

The correct option is C Energy $=3VT(\frac{1}{r}-\frac{1}{R})$ is released

Given,
Radius of smaller liquid drop $=r$
Radius of larger drop $=R$
Volume of larger drop $=V$
Surface tension of the liquid $=T$
Let us suppose $n$ drops coaslece to form a bigger drop.
As volume of liquid will be same,
Volume of $n$ smaller drops $=$ volume of single bigger drop
$\Rightarrow n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3=V$
$\Rightarrow n\times 4\pi r^3=4\pi R^3=3V$ ........(1)
We know, surface energy $=$ surface tension $\times$ change in surface area
$\Rightarrow E=T\Delta A$
$=T[4\pi R^2-n\times 4\pi r^2]$
$=T[\frac{4\pi R^3}{R}-\frac{n\times 4\pi r^3}{r}]$
$=T[\frac{3V}{R}-\frac{3V}{r}]$ ......$[$ from (1) $]$
$=3VT[\frac{1}{R}-\frac{1}{r}]$
As $R>r$ $\Rightarrow \frac{1}{R}<\frac{1}{r}$
$\therefore E$ is negative.
Hence, $3VT[\frac{1}{r}-\frac{1}{R}]$ energy will be released.