A vertical glass capillary with inside diameter $0.50 \mathrm{mm}$ is submerged into water so that the length of its part emerging outside the water surface is equal to $25 \mathrm{mm}$.Find the radius of curvature of the meniscus. Surface tension of water is $73\times {10}^{-3} \mathrm{N}/\mathrm{m}$.

In the capillary tube, the water should rise to a height

$$\begin{gathered} h=\frac{2T}{r\rho g} \text{ Here }T=73\times {10}^{-3} \mathrm{N}/\mathrm{m} \\ \mathrm{r}=\frac{0.50 \mathrm{mm}}{2}=0.25\times {10}^{-3} \mathrm{m} \\ \therefore \mathrm{h}=\frac{2\times 73\times {10}^{-3}}{0.25\times {10}^{-3}\times {10}^3\times 9.8}=59\times {10}^{-3} \mathrm{m}=59 \mathrm{mm} \end{gathered}$$

Now $h>h^{\text{'}}$; i.e., length is outside water surface. Therefore, radius of meniscus $>$ radius of capillary $r$.

If $\mathrm{R}$ is the radius of meniscus, then we have

$\frac{2T}{R}=h^{\text{'}}\rho g\text{ or }R=\frac{2T}{h^{\text{'}}\rho g}$

Here $h^{\text{'}}=25mm=25\times {10}^{-3} \mathrm{m}$

$\therefore R=\frac{2\times 73\times {10}^{-3}}{25\times {10}^{-3}\times 9.8}=0.6\times {10}^{-3} \mathrm{m}$