Question

Mercury has an angle of contact equal to ${140}^{\circ }$ with soda-lime glass. A narrow tube of radius $1.0mm$ 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 $0.465N{m}^{-2}.$ Density of mercury $=13.6\times 10kg{m}^{-3}$

Answer 1

Angle of contact between mercury and soda lime glass, $\theta ={140}^{\circ }$

Radius of the narrow tube, $a=1mm=1\times {10}^{-3}m$

Surface tension of mercury at the given temperature$S=0.465N{m}^{-1}$

Density of mercury, $\rho =13.6\times {10}^{3}kg{m}^{-3}$

Acceleration due to gravity, $g=9.8m{s}^{-2}$

Surface tension is related with the angle of contact and the dip in the height $\left(h\right)$ as:

$h=\frac{2S\cos \theta }{\rho ga}$

Putting the values, we get, $h=\frac{2\times 0.465\times \cos {140}^{\circ }}{13.6\times {10}^3\times 9.8\times {10}^{-3}}$

$=-0.00534m=-5.34mm$

Here, the negative sing shows the decrease in the level of mercury. Hence, the mercury level dips by $5.34mm.$