Two capillary tubes of same diameter are kept vertically in two liquids whose densities are in the ratio 4 : 3 . If their surface tensions are in the ratio 6:5, the ratio of heights ( h 2 h 1 )

Question

Two capillary tubes of same diameter are kept vertically in two liquids whose densities are in the ratio 4 : 3 . If their surface tensions are in the ratio 6:5, the ratio of heights ( h 2 ​ h 1 ​ ​ ) of liquids in the two capillary tubes is (Their angle of contacts are same)

TestHub · Accepted Answer

The height of a liquid in a capillary tube is given by the formula:

$h = \frac{2 T cos θ}{r ρ g}$

Since the capillary tubes have the same diameter, $r_{1} = r_{2}$ . The angle of contact is also the same, so $θ_{1} = θ_{2}$ . Therefore, the height $h$ is directly proportional to the surface tension $T$ and inversely proportional to the density $ρ$ :

$h \propto \frac{T}{ρ}$

The ratio of the heights of the liquids in the two capillary tubes is:

$\frac{h _{1}}{h _{2}} = \frac{T _{1}}{T _{2}} \times \frac{ρ _{2}}{ρ _{1}}$

Given that $\frac{T _{1}}{T _{2}} = \frac{6}{5}$ and $\frac{ρ _{1}}{ρ _{2}} = \frac{4}{3}$ , we get:

$\frac{h _{1}}{h _{2}} = \frac{6}{5} \times \frac{3}{4}$

$\frac{h _{1}}{h _{2}} = \frac{18}{20} = \frac{9}{10}$

Answer: $\frac{9}{10}$

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