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

Answer 1

Answer: (C)

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

Since intial volume=final volume

Let's say number of drops $=N$

Then energy released $=T,\Delta A$

$A_i=\left(4\pi r^2\right).n$

$n=$number of drops

Total volume $=n.\left(\frac{4\pi }{3}{r}^3\right)=\left(n.4\pi r^2\right).\frac{r}{3}$

$\begin{array}{c}or,A_i=\left(n.4\pi r^2\right)=\frac{3V}{r}\\ A_f=n.4\pi R^2=\frac{3V}{R}\\ \therefore Energyreleased=T\left(\frac{3V}{R}-\frac{3V}{r}\right)\\ =3VT\left(\frac{1}{r}-\frac{1}{R}\right)\end{array}$