The surface tension and vapour pressure of water at `20^(@)C` is `7.28xx10^(-2)Nm^(-1)` and `2.33xx10^(3)Pa`, respectively. What is the radius of the smallest spherical water droplet which can form without evaporating at `20^(@)C`?

Explanation

Here, `S = 7.28 xx 10^(-2) N//m , p = 2.33 xx 10^3 Pa` The drop will evaporate if the water pressure is greater then the vapour pressure p (i.e.excess pressure in a drop) Using `p = (2 S)/(R ),` where R is the radius of drop, we have `R = (2S)/(P) = (2xx7.28 xx 10^(-2))/(2.33 xx 10^3) = 6.25 xx 10^(-5)m`