The excess of pressure, due to surface tension, on a spherical liquid drop of radius 'R' is proportional to

Options

A. `R^(-1)`
B. R
C. `R^(-2)`
D. `R^(2)`

Answer

A

Explanation

The excess of pressure due to surface tension on a spherical liquid drop is give by
`P = (2T)/(R) " "…….(i)`
where T = surface tension
and R radius of liquid drop.
So form Eq.(i) we , get
`P prop (1)/(R) or P prop R^(-1)`