The angle of contact of mercury with soda lime glass is `140^(@)`. A capillary tube of radius 1.0 mm is dipped in a trough containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside?
Surface tension of mercury `=0.465Nm^(-1)`
Density of mercury `=13.6xx10^(3)kgm^(-3)`

Explanation

`{:("Angle of"),("contact"):}}theta=140^(@)`
`costheta-cos140^(@)=cos(180^(@)-40^(@))`
`=-cos40^(@)=-0.7660`
`{:("Radius of"),("capillary tube"):}}r=1.0mm=1xx10^(-3)m`
`{:("Surface tension"),("of mercury"):}}T=0.465Nm^(-1)`
`{:("Density of"),("mercury"):}}rho=13.6xx10^(3)kgm^(-3)`
`{:("Dip in the"),("mercury level"):}}h=(2rhocostheta)/(rrhog)`
`thereforeh=(2xx0.465xx(-0.766))/(1xx10^(-3)xx13.6xx10^(3)xx9.8)`
`=-5.34xx10^(-3)m=5.34mm`