A spherical drop of water has `1mm` radius. If the surface tension of water is `70xx10^(-3)`N//m, then the difference of pressure between inside and outside of the spherical drop is:

Options

A. `35Nm^2`
B. `14(N)/(m^2)`
C. `140(N)/(m^2)`
D. none of these

Answer

C

Explanation

The excess pressure or difference of pressure between inside and outside of the spherical drop is given by
`P=(2T)/(R )`
Here the radius of drop is `R`, and surface tension of water is `T`
Given, `R=1mmxx=1xx10^-3m`
`T=70xx10^-3(N)/(m)`
`P=(2xx70xx10^-3)/(1xx10^-3)=140(N)/(m)`