Surface tension of two liquids (having same densities), T_ 1 and T_...

For capillary rise, the height of liquid in a tube is given by: $h = \frac{2T\cos\theta}{\rho g r}$, where $T$ is surface tension and $r$ is the tube radius.

For two liquids with equal density and contact angles:
$h_1 = \frac{2T_1\cos\theta}{\rho g r_1}$ and $h_2 = \frac{2T_2\cos\theta}{\rho g r_2}$

Taking the ratio: $\frac{h_1}{h_2} = \frac{T_1}{T_2} \times \frac{r_2}{r_1}$

Since $r_1 > r_2$, we have $\frac{r_2}{r_1} < 1$. The capillary rise is inversely proportional to tube radius, so the smaller radius tube will have greater height. Since the liquids are measured to have heights $h_1$ and $h_2$ with $r_1 > r_2$, and they have the same density, the empirical relation shows that both liquids rise to different heights inversely proportional to their radii. For liquids with the same density measured in different tubes, if the heights and radii satisfy the capillary rise equation, the surface tensions must be equal: $T_1 = T_2$, and $h_1 < h_2$ since $r_1 > r_2$.