Question

A container has two immiscible liquids of densities ${\rho }_1$ and ${\rho }_2(>{\rho }_1)$. A capillary tube of radius $r$ is inserted in the liquid so that its bottom reaches upto the denser liquid. The denser liquid rises in the capillary and attains a height $h$ from the interface of the liquids, which is equal to the column length of the lighter liquid. Assuming angle of contact to be zero, the surface tension of heavier liquid is

• A. $2\pi r{\rho }_2\text{gh}$
• B. $\frac{{\rho }_{2}rgh}{2}$
• C. $\frac{r}{2}({\rho }_2-{\rho }_1)\text{gh}$
• D. $2\pi r({\rho }_2-{\rho }_1)\text{gh}$

Answer 1

The correct option is C $\frac{r}{2}({\rho }_2-{\rho }_1)\text{gh}$


The pressure difference at the points $A$ and $B$, just above and below the meniscus is
$P_A-P_B=\frac{2T}{r}$
However at the point $A$, the atmospheric pressure is acting.
$\Rightarrow P_0=P_A$
and $P_E=P_A=P_0$
According to Pascal's law, $P_C=P_D$
$\Rightarrow P_A-\frac{2T}{r}+{\rho }_{2}g(h+h^{\prime })=P_E+{\rho }_{1}gh+{\rho }_{2}g{h}^{\prime }$
$\Rightarrow \frac{2T}{r}=({\rho }_2-{\rho }_1)gh$
$\Rightarrow T=\frac{r}{2}({\rho }_2-{\rho }_1)gh$