Question

Mercury has an angle of contact equal to ${140}^o$ with soda lime glass. A narrow tube of radius $1.00mm$ 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 $0.465N{m}^{-1}$. Density of mercury $=13.6\times {10}^{3}kg{m}^{-3}$.

Answer 1

Angle of contact between mercury and soda lime glass, $\theta ={140}^o$
Radius of the narrow tube, $r=1mm=1\times {10}^{-3}m$
Surface tension of mercury at the given temperature, $s=0.465N{m}^{-1}$
Density of mercury, $\rho =13.6\times {10}^{3}kg/m^3$

Dip in the height of mercury $=h$
Acceleration due to gravity, $g=9.8m/s^2$
Surface tension is related with the angle of contact and the dip in the height as:

$s=h\rho gr/2cos\theta$
$\therefore h=2scos\theta /\rho gr$
$=2\times 0.465\times cos{140}^o/(1\times {10}^{-3}\times 13.6\times {10}^3\times 9.8)$
$=-0.00534m$

$=-5.34m$
Here, the negative sign shows the decreasing level of mercury. Hence, the mercury level dips by 5.34 mm.