Question

Several spherical drops of a liquid each of radius $r$ coalesce to form a single drop of radius $R$. If $T$ is the surface tension, then the energy liberated will be :

• A. $4\pi R^{3}T(\frac{1}{r}-\frac{1}{R})$
• B. $2\pi R^{3}T(\frac{1}{r}-\frac{1}{R})$
• C. $\frac{4}{3}\pi r^{3}T(\frac{1}{r}-\frac{1}{R})$
• D. $2\pi R^{3}T(\frac{1}{R}-\frac{1}{r})$

Answer 1

The correct option is A $4\pi R^{3}T(\frac{1}{r}-\frac{1}{R})$

Volume of spherical drops $\Rightarrow n\left(\frac{4}{3}\pi r^3\right)=\frac{4}{3}\pi R^3$

$So,n=\frac{R^3}{r^3}$

$EnergyLiberated,u=T\Delta A$

$=T[{4\pi R}^2-n4\pi r^2]$

$=4\pi T[R^2-{nr}^2]$

$=4\pi T[R^2-\frac{R^3}{r^3}{r}^2]$

$=4\pi R^{3}T[\frac{1}{R}-\frac{1}{r}]$

$=3\left(\frac{4}{3}\pi R^3\right)T(\frac{1}{R}-\frac{1}{r})$

$u=4\pi T{R}^3(\frac{1}{R}-\frac{1}{r})(R>r)$

$u=4\pi T{R}^3(\frac{1}{r}-\frac{1}{R})(r>R)$