Question

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

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

Answer 1

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

Excess pressure due to miniscus is,
$P_1=P_0-\frac{2T}{r}$
Pressure at the interface is
$P_c=P_1+{\rho }_{2}gh$
$P_c=P_0-\frac{2T}{r}+{\rho }_{2}gh$
Pressure at the interface can also be written as,
$P_c={\rho }_{1}gh+P_0$
${\rho }_{1}gh+P_0=P_0-\frac{2T}{r}+{\rho }_{2}gh$
${\rho }_{2}gh-{\rho }_{1}gh=\frac{2T}{r}$
$({\rho }_2-{\rho }_1)gh=\frac{2T}{r}$
$T=\frac{rgh}{2}({\rho }_2-{\rho }_1)$