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\times7.28\times10^{-2}}{2.33\times10^{3}}$
$=6.25\times10^{-5}\,m$