[પાણીનું પૃષ્ઠતાણ ${T}=7.3 \times 10^{-2} \, {Nm}^{-1}$, સંપર્કકોણ $=0, {g}=10\, {ms}^{-2}$ અને પાણીની ઘનતા $\left.=1.0 \times 10^{3} \,{kg} \,{m}^{-3}\right]$
\(\therefore \mathrm{P}_{\mathrm{atm}}-\frac{2 \mathrm{~T}}{\mathrm{r}_{1}}+\rho \mathrm{g}(\mathrm{x}+\Delta \mathrm{h})=\mathrm{P}_{2 \mathrm{tm}}-\frac{2 \mathrm{~T}}{\mathrm{r}_{2}}+\rho g \mathrm{x}\)
\(\therefore \rho g \Delta \mathrm{h}=2 \mathrm{~T}\left[\frac{1}{\mathrm{r}_{1}}-\frac{1}{\mathrm{I}_{2}}\right]\)
\(=2 \times 7.3 \times 10^{-2}\left[\frac{1}{2.5 \times 10^{-3}}-\frac{1}{4 \times 10^{-3}}\right]\)
\(\therefore \Delta \mathrm{h}=\frac{2 \times 7.3 \times 10^{-2} \times 10^{3}}{10^{3} \times 10}\left[\frac{1}{2.5}-\frac{1}{4}\right]\)
\(=2.19 \times 10^{-3} \mathrm{~m}=2.19 \mathrm{~mm}\)