Question

A capillary tube of radius 0.50 mm is dipped vertically in a pot of water. Find the difference between the pressure of the water in the tube 5.0 cm below the surface and the atmospheric pressure. Surface tension of water = 0.075 $N{m}^{-1}$.

Answer 1

r = 0.5 mm = $5\times {10}^{-3}m$

h = 5.0 cm = $5\times {10}^{-2}m$

T = 0.075 N/m

The excess pressure at 5 cm before the surface

P =$h\rho g=1000\times (5\times {10}^{-2})\times 9.8$

= $490N/m^2$

Again excess pressure at the surface is

P = $\frac{2T}{r}=\frac{2\times \left(0.75\right)}{(5\times {10}^{-4})}$

$=300N/m^2$

Difference in pressure

$=P=\left(\frac{F}{r}\right)$

= 490 -300 = 190 $N/m^2$