The surface tension and vapor pressure of water at `20^(@)`C is `7.28 xx 10^(-2) N m^(-1)` and `2.33 xx 10^(3)` Pa respectively. The radius of the smallest spherical water droplet which can form without evaporating at `25^(@)`C is

Options

A. `1.25 xx 10^(-5)` m
B. `6.25 xx 10^(-5)` m
C. `4.3 xx 10^(8)` m
D. `3.4 xx 10^(3)` m

Answer

B

Explanation

Here, `S=7.28xx10^(-2)N m^(-1),P=2.33xx10^(23)` Pa
The drop will evaporation if the water pressure is greater than the vapour pressure P
As `P=(2S)/(R)`, where R is the radius of drop
`therefore R=(2S)/(P)=(2xx7.28xx10^(-2))/(2.33xx10^(3))=6.25xx10^(-5)m`