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

Answer 1

Answer: (A)

Energy liberated during this process is equal to change in surface energy of liquid.
$E=T\left(A_{1} - A_{2}\right)$ , where $A_{1}$ is initial surface area and $A_{2}$ is final surface area.
As the volume of all small droplets will be equal to one big drop.
$n\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi R^{3}\Rightarrow n=\frac{R}{r^{3}}$
$E=T\left(n \left(4 \pi r^{2}\right) - 4 \pi R^{2}\right)E=T\left(\frac{R^{3}}{r^{3}} 4 \pi r^{2} - 4 \pi R^{2}\right)E=4\pi R^{3}T\left(\frac{1}{r} - \frac{1}{R}\right)$ and $\left(n r^{3} = R^{3}\right)$