Question

The surface tension and vapour pressure of water at $20^{\circ }C$ is $7.28\times 10^{-2}N{m}^{-1}$ and $2.33\times 10^{3}Pa$ respectively. The radius of the smallest spherical water droplet which can form without evaporating at $25^{\circ }C$ is

• A. $1.25\times 10^{-5}m$
• B. $6.25\times 10^{-5}m$
• C. $4.3\times 10^{8}m$
• D. $3.4\times 10^{3}m$

Answer 1

Answer: (B)

Here, $S=7.28\times 10^{-2}N{m}^{-1}$, $P=2.33\times 10^{3}Pa$
The drop will evaporate if the water pressure is greater than the vapour pressure $P$
As $P=\frac{2S}{R}$, where $R$ is the radius of drop
$\therefore R=\frac{2S}{P}=\frac{2\times 7.28\times 10^{-2}}{2.33\times 10^3}$
$=6.25\times 10^{-5}m$