Surface tension of mercury is `0.465 N m^(-1)`. The excess pressure inside a mercury drop of diameter 6mm is

Options

A. 310 pa
B. 410 pa
C. 510 pa
D. 610 pa

Answer

A

Explanation

Here, D=6 mm
`r=(D)/(2)=(6mm)/(2)=3 mm =3xx10^(-3)`m
Excess pressure inside the mercury drop is
`P=(2S)/(r)=(2xx0.465 N m^(-1))/(3xx10^(-3)m)=310 Pa`