Question

A glass tube of 1 mm bore is dipped vertically into a container of mercury, with its lower end 5 cm below the mercury surface. What must be the gauge pressure of air in the tube to a hemispherical bubble at its lower end? Given density of mercury ` = 13.6 xx 10^(3) kg//m^(3)`, surface tension of mercury = `440 xx 10^(-3) Nm^(-1)` and `g = 10m//s^(2)`

Answer 1

Here, `R=1.2 mm = 0.5 mm`
`= 5 xx 10^(-4)m`,
`h= 5 cm = 0.05 m , rho = 13.6 xx 10^(3) kg//m^(3)`
`S= 440 xx 10^(-3) Nm^(-1)`
Gauge pressure in hemispherical bubble
`= h rho g + (2S)/(R )`
`= 0.05 xx (13.6 xx 10^(3)) xx 10 + ( 2 xx 440 xx10^(3))/(5 xx 10^(-4))`
`= 6800+1760 = 8560 Pa`.