A spherical drop of water as `1mm` radius. If the surface tension of the the water is `50xx10^-3(N)/(m)`, then the difference of pressure between inside and outside the spherical drop is:

Options

A. `25(N)/(m^2)`
B. `10000(N)/(m^2)`
C. `100(N)/(m^2)`
D. `50(N)/(m^2)`

Answer

C

Explanation

`P_(excess)=(2T)/(R )=(2(50xx10^-3))/((10^-3))=100(N)/(m^2)`