Question

A number of water droplets each of radius $r$ coalesce to form a droplet of radius $R .$ The rise in temperature is (Surface tension of water =S )

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

Answer 1

Answer: (B)

Let $n$ drops coalesce to form big drop. Then
$n \times \frac{4}{3} \pi r^{3}=\frac{4}{3} \pi R^{3} $
or $ n=\frac{R^{3}}{r^{3}}$
As heat produeed $=\frac{\text { work done }}{J}$
$\therefore m s \Delta T=\frac{\text { surface tension } \times \text { decrease in area }}{J}$
or $m s \Delta T=\frac{S}{J} \times\left[n \times 4 \pi r^{2}-4 \pi R^{2}\right]$
or $\frac{4}{3} \pi R^{3} \times 1 \times 1 \times \Delta T=\frac{S}{J} \times\left[n \times 4 \pi r^{2}-4 \pi R^{2}\right]$
or $ \Delta T=\frac{3 S \times 4 \pi}{J 4 \pi R^{3}}\left[n r^{2}-R^{2}\right]=\frac{3 S}{J R^{3}}\left[\frac{R^{3}}{r^{3}} r^{2}-R^{2}\right]$
$=\frac{3 S R^{3}}{J R^{3}}\left[\frac{1}{r}-\frac{1}{R}\right]=\frac{3 S}{J}\left[\frac{1}{r}-\frac{1}{R}\right]$