A glass rod of radius ' \(r\) ' ' is inserted symmetrically into a vertical capillary tube of radius ' \(\mathrm{r}_2\) ' \(\left(\mathrm{r}_1 < \mathrm{r}_2\right)\) such that their lower ends are at same level. The arrangement is dipped in water. The height to which water will rise into the tube will be ( \(\rho=\) density of water, \(\mathrm{T}=\) surface tension in water, \(\mathrm{g}\) = acceleration due to gravity)

Vertical component of total force of surface tension
\(
\mathrm{F}=\left(\mathrm{r}_2+\mathrm{r}_1\right) 2 \pi \mathrm{T} \cos \theta
\)
Weight of the liquid in the capillary
\(
\mathrm{W}=\pi\left(\mathrm{r}_2^2-\mathrm{r}_1^2\right) \mathrm{h} \rho \mathrm{g}
\)
This is balanced by the vertical component of the face due to the surface tension
\(
\therefore \pi\left(\mathrm{r}_2^2-\mathrm{r}_1^2\right) \mathrm{h} \rho \mathrm{g}=\left(\mathrm{r}_2+\mathrm{r}_1\right) \times 2 \pi \mathrm{T} \cos \theta
\)
Simplifying and solving for \(\mathrm{h}\) we get,
\(
\mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\left(\mathrm{r}_2-\mathrm{r}_1\right) \rho g}=\frac{2 \mathrm{~T}}{\left(\mathrm{r}_2-\mathrm{r}_1\right) \rho g}
\)
(For pure water, \(\theta=0^{\circ}, \cos \theta=1\) )