Question

A large number of water drops, each of radius $r$, combine to have a drop of radius $R$. If the surface tension is $T$ and mechanical equivalent of heat is $J,$ the rise in heat energy per unit volume will be:

• A. $\frac{2T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)$
• B. $\frac{2T}{rJ}$
• C. $\frac{3T}{rJ}$
• D. $\frac{3T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)$

Answer 1

Answer: (D)

$n\times \frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$
$\therefore n^{1/3}r=R$
$\therefore$ Total change in surface energy
$=\left(n\left(4\pi r^2\right)-4\pi R^2\right)T$
$\Rightarrow 4\pi T\left(n{r}^2-R^2\right)$
$\therefore$ Heat energy
$=\frac{4\pi T\left(n{r}^2-R^2\right)}{J\times \frac{4}{3}\pi R^3}=\frac{3T}{J}\left(\frac{n{r}^2}{R^3}-\frac{1}{R}\right)$
Put $n{r}^3=R^3$
$\therefore \frac{3T}{J}\left(\frac{1}{r}-\frac{1}{R}\right)$