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

radius of the capillary tube is 0.5m = $5\times {10}^{-4}m$

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

surface tension of water T is $0.075N/M$

excess pressure at 5 cm $P=\rho hg=1000\times (5\times {10}^{-2})\times 9.8=490N/m^2$

excess the pressure

$p_0=\frac{2T}{r}=\frac{2\times \left(0.75\right)}{5\times {10}^{-4}}$

$p_0=300N/m^2$

difference in the pressure

$p_0-p=490-300=190N/m^2$