A glass U-tube is such that the diameter of one limb is $3.0 \mathrm{mm}$ and that of the other is $6.0 \mathrm{mm}$. The tube is inverted vertically with the open ends below the surface of water in a beaker. What is the difference between the heights to which water rises in the two limbs? Surface tension of water is $0.07{\mathrm{Nm}}^{-1}$. Assume that the angle of contact between water and glass is $0°$.

Suppose pressure at the points, A, B, C and D be ${\mathrm{P}}_{\mathrm{A}}$, ${\mathrm{p}}_{\mathrm{B}},{\mathrm{p}}_{\mathrm{C}}$ and ${\mathrm{p}}_{\mathrm{D}}$ respectively.

The pressure on the concave side of the liquid surface is greater than that on the other side by $\frac{2 T}{R}$.

An angle of contact $\mathrm{\theta }$ is given to be $0°$, hence

$$\begin{gathered} \mathrm{Rcos}0°=\mathrm{r} \\ \mathrm{orR}=\mathrm{r} \\ \therefore {\mathrm{p}}_{\mathrm{A}}={\mathrm{p}}_{\mathrm{B}}+\frac{2\mathrm{T}}{{\mathrm{r}}_1} \\ \mathrm{and} {\mathrm{p}}_{\mathrm{C}}={\mathrm{p}}_{\mathrm{D}}+\frac{2\mathrm{T}}{{\mathrm{r}}_2} \end{gathered}$$

where, ${\mathrm{r}}_1$and ${\mathrm{r}}_2$ are the radii of the two limbs

$$\begin{gathered} \mathrm{But} {\mathrm{p}}_{\mathrm{A}}={\mathrm{p}}_{\mathrm{C}} \\ \therefore {\mathrm{p}}_{\mathrm{B}}+\frac{2\mathrm{T}}{{\mathrm{r}}_1}={\mathrm{p}}_{\mathrm{D}}+\frac{2\mathrm{T}}{{\mathrm{r}}_2} \\ \text{ or }{\mathrm{p}}_{\mathrm{D}}-{\mathrm{p}}_{\mathrm{B}}=\mathrm{\rho gh}=2\mathrm{T}\left(\frac{1}{{\mathrm{r}}_1}-\frac{1}{{\mathrm{r}}_2}\right) \\ \text{where, h is the difference in water levels in the two limbs} \\ \mathrm{Now}, \mathrm{h}=\frac{2\mathrm{T}}{\mathrm{\rho g}}\left(\frac{1}{{\mathrm{r}}_1}-\frac{1}{{\mathrm{r}}_2}\right) \\ \mathrm{Given} \mathrm{that} \mathrm{T}=0.07{\mathrm{Nm}}^{-1},\mathrm{\rho }=1000{\mathrm{kgm}}^{-3} \\ {\mathrm{n}}_1=\frac{3}{2} \mathrm{mm}=\frac{3}{20} \mathrm{cm}=\frac{3}{20\times 100} \mathrm{m} \\ =1.5\times {10}^{-3} \mathrm{m},{\mathrm{r}}_2=3\times {10}^{-3} \mathrm{m} \\ \therefore \mathrm{h}=\frac{2\times 0.07}{1000\times 9.8}\left(\frac{1}{1.5\times {10}^{-3}}-\frac{1}{3\times {10}^{-3}}\right)\mathrm{m} \\ =4.76\times {10}^{-3} \mathrm{m}=4.76 \mathrm{mm} \end{gathered}$$