Two capillary tubes of radii r 1 = 0.3 cm and r 2 = 0.6 cm are dipped in the same liquid. The ratio of heights through which the liquid will rise in the tubes ( h 2 h 1 ) is

Question

Two capillary tubes of radii r 1 ​ = 0.3 cm and r 2 ​ = 0.6 cm are dipped in the same liquid. The ratio of heights through which the liquid will rise in the tubes ( h 2 ​ h 1 ​ ​ ) is

TestHub · Accepted Answer

The height of capillary rise is given by Jurin's law:

$h = \frac{2 T cos θ}{ρ g r}$

For the same liquid, surface tension $T$ , angle of contact $θ$ , and density $ρ$ are constant. Therefore, the height $h$ is inversely proportional to the radius $r$ of the capillary tube:

$h \propto \frac{1}{r}$

The ratio of the heights is:

$\frac{h _{1}}{h _{2}} = \frac{r _{2}}{r _{1}}$

Substituting the given values $r_{1} = 0.3$ cm and $r_{2} = 0.6$ cm:

$\frac{h _{1}}{h _{2}} = \frac{0.6}{0.3} = \frac{2}{1}$

The ratio is $2 : 1$ .

Answer: $2 : 1$

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