Question

Mercury has an angle of contact equal to 140° with soda lime glass. A narrow tube of radius 1.00 mm made of this glass 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 at the temperature of the experiment is 0.465 N m^-1. Density of mercury = 13.6 x 10³ km m^-3.

Answer 1

Here angle of contact,

θ = 140°

cosθ = cos 140° = -0.7660

Also, r = 1.00 mm = 1 x 10^-3 m

Surface tension, T = .465 Nm^-1

Density, ρ = 13.6 x 10³ kg m^-3

Using formula,

h = {2 s cosθ}/{rρg}

we get,

h = {2 x 0.465 x (-0.7660)}/{1 x 10^-3 x 13.6 x 10³ x 9.8}

= -5.34 x 10^-3 m = -5.34 mm

[Negative sign shows that the liquid mercury is depressed below the actual level.]