The radius of a spherical drop of water is 1 mm. If surface tension of water be `70 xx 10^(-3) Nm^(-1)`, the pressure difference between inside and outside the drop will be

Options

A. `70 Nm^(-2)`
B. `140 Nm^(-2)`
C. `280 Nm^(-2)`
D. zero

Answer

B

Explanation

The excess pressure p is given by `, p = (2S)/(R)`
According to the question, `p = (2 xx 70 xx 10^(-3))/(10^(-3)) = 140 Nm^(-2)`