The height of liquid column raised in a capillary tube of certain radius when dipped in liquid 'A' vertically is \(5 \mathrm{~cm}\). If the tube is dipped in a similar manner in another liquid ' \(\mathrm{B}\) ' of surface tension and density double the values of liquid ' \(A\) ', the height of liquid column raised in liquid 'B' would be (Assume angle of contact same)

\(\begin{array}{ll} & \text { Height of liquid column, } \mathrm{h}=\frac{2 \mathrm{~T} \cos \theta}{\mathrm{r} \rho \mathrm{g}} \\ & \text { Given: } \mathrm{h}_1=5 \mathrm{~cm} \\ & \text { also } \mathrm{T}_2=2 \mathrm{~T}_1 ; \rho_2=2 \rho_1 \\ \therefore \quad & \mathrm{h}_1=\frac{2 \mathrm{~T}_1 \cos \theta}{\mathrm{r} \rho_1 \mathrm{~g}} \\ & \mathrm{~h}_2=\frac{2 \mathrm{~T}_2 \cos \theta}{\mathrm{r} \rho_2 \mathrm{~g}}=\frac{2\left(2 \mathrm{~T}_1\right) \cos \theta}{\mathrm{r}\left(2 \rho_1\right) \mathrm{g}}=\frac{4 \mathrm{~T}_1 \cos \theta}{\mathrm{r}\left(2 \rho_1\right) \mathrm{g}} \\ \therefore \quad & \frac{\mathrm{h}_1}{\mathrm{~h}_2}=\frac{5}{\mathrm{~h}_2}=\frac{2 \mathrm{~T}_1 \cos \theta}{\mathrm{r} \rho_1 \mathrm{~g}} \times \frac{\mathrm{r} 2 \rho_1 \mathrm{~g}}{4 \mathrm{~T}_1 \cos \theta} \\ & \frac{5}{\mathrm{~h}_2}=1 \\ \therefore \quad & \mathrm{h}_2=5 \mathrm{~cm}=0.05 \mathrm{~m}\end{array}\)