Two capillary tubes of the same diameter are kept vertically in two different liquids whose density are in the ratio 4 : 3 . Rise of liquid in both the capillaries is h 1 and h 2 respectively. The

Question

Two capillary tubes of the same diameter are kept vertically in two different liquids whose density are in the ratio 4 : 3 . Rise of liquid in both the capillaries is h 1 ​ and h 2 ​ respectively. Their surface tensions are in the ratio 6 : 5 , the ratio of the heights ( h 2 ​ h 1 ​ ​ ) is (Assume that their angles of contact are same)

TestHub · Accepted Answer

The height of liquid in a capillary tube is given by Jurin's law:

$h = \frac{2 T cos θ}{r ρ g}$

Since the diameter of the tubes and the angle of contact are the same for both liquids, $r$ and $θ$ are constant. 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 is:

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

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

$\frac{h _{1}}{h _{2}} = \frac{6}{5} \times \frac{3}{4} = \frac{18}{20} = 0.9$

Answer: $0.9$

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