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. `35(N)/(m^-2)`
B. `70(N)/(m^2)`
C. `140(N)/(m^2)`
D. zero

Answer

C

Explanation

`triangleP=(2T)/(R )=(2xx70xx10^-3)/(1xx10^-3)=140(N)/(m^2)`