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. Density of mercury = 13.6x10^–3kgm^–3.)

Solution

Here,

Angle of contact, $θ = 140 °$

Radius of the narrow tube, r = 1.0mm=10^-3m
Surface tension, T = 0.465 N/m
Density of mercury, $ρ$ = 13600 kg/m³
We know the height by which liquid rises in the tube is,

$h = \frac{2 T cosθ}{rpg} = \frac{2 \times 0.465 \cos 140}{10^{- 3} \times 13600 \times 9.8} = - 0.00535 m = - 5.35 m$

Negative sign means that mercury will dip down.

$∴ Mercury will dip down by 5.35 mm .$