Question

An open capillary tube contains a drop of water. When the tube is in its vertical position, the drop forms a column with a length of $2cm$. The internal diameter of the capillary tube is $1mm$. Determine the radii of curvature of the upper and lower meniscus in each case. Consider the wetting to be complete. Surface tension of water $=0.0075N/m$

Answer 1

When capillary is in vertical position, thge upper meniscus is concave and pressure due to surface tension

$P_s=\frac{2T}{R_1}$

is directed vertically upward, $R_1$ being radius of curvature of upper

meniscus. When wetting is complete

$P_1=\frac{2T}{r}$

$r$ being radius of the capillary tube $=\frac{D}{2}=\frac{1}{2}mm$

$R_1=r=0.5mm$ in each case.

We have $P_1=\frac{2T}{r}=\frac{2\times 75\times {10}^{-3}}{0.5\times {10}^{-3}}=300N/m^2$

The hydrostatic pressure $P_2=h\rho g$ is always directed downwards.

At $h=2cm$

$P_2=h\rho g=2\times {10}^{-2}\times 10\times {10}^3=200N/m^2$

Clearly $P_1>P_2$. That is resulting pressure is directed upward. For

equilibrium, the pressure due to lower meniscus should be downward. This

makes the lower meniscus concave downwards. Then,

$P_1-P_2=\frac{2T}{R_2}$

$300-200=\frac{2\times 75\times {10}^{-3}}{R_2}$

$R_2=\frac{2\times 75\times {10}^{-3}}{100}=1.50\times {10}^{-3}m=1.50mm$