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 $\left(n{r}^3=R^3\right)$