Question

A certain number of spherical drops of a liquid of radius 'r' coalesce to form a single drop of radius 'R' and volume 'V'. If 'T' is the surface tension of the liquid then:

• A. $Energy=4VT(\frac{1}{r}-\frac{1}{R})$ is released
• B. $Energy=3VT(\frac{1}{r}+\frac{1}{R})$ is released
• C. $Energy=3VT(\frac{1}{r}-\frac{1}{R})$ is released
• D. Energy is neither released nor absorbed

Answer 1

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

$\Delta U=\left(T\right)\left(\Delta A\right)$
$A\left(initial\right)=\left(4\pi r^2\right)n$
$A\left(final\right)=4\pi R^2$
$\Delta A=\left(4\pi r^2\right)n-4\pi R^2$
$\left(\frac{4}{3}\pi r^3\right)n=\frac{4}{3}\pi R^3$
$n=\frac{R^3}{r^3}$
$\Delta A=4\pi [\frac{R^3}{r^3}\cdot r^2-R^2]=4\pi [\frac{R^3}{r}-\frac{R^3}{R}]=\left(\frac{4\pi R^3}{3}\right)3[\frac{1}{r}-\frac{1}{R}]$
$\Delta A=3V[\frac{1}{r}-\frac{1}{R}]$
$\Delta U=3VT[\frac{1}{r}-\frac{1}{R}]$