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

Answer

Given:
Radius of capillary tube $r=0.5 \mathrm{~mm}=5 \times 10^{-4} \mathrm{~m}$
Depth (where pressure is to be found) $h=5.0 \mathrm{~cm}=5 \times 10^{-2} \mathrm{~m}$
Surface tension of water $T=0.075 \mathrm{~N} / \mathrm{m}$
Excess pressure at 5 cm before the surface:
$\mathrm{P}=\rho_{\mathrm{hg}}=1000 \times\left(5 \times 10^{-2}\right) \times 9.8=490 \mathrm{~N} / \mathrm{m}^2$
Excess pressure at the surface is given by:
$P_0=\frac{2 T}{r}=\frac{2 \times(0.75)}{\left(5 \times 10^{-4}\right)}$
$=300 \mathrm{~N} / \mathrm{m}^2$
Difference in pressure: $P_0-P=490=300=190 \mathrm{~N} / \mathrm{m}^2$
Hence, the required difference in pressure is $190 \mathrm{~N} / \mathrm{m}^2$.

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