Question

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

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

Answer 1

The correct option is D

$\frac{r\left({\rho }_2-{\rho }_1\right)gh}{2}$

Step 1: Given data

1. Container contains two immiscible liquids of density ${\rho }_1$​ and

1. ${\rho }_1({\rho }_2>{\rho }_2)$
2. A capillary of radius $r$ is inserted in the liquid
3. Denser liquid rises in capillary and attains height equal to $h$ which is also equal to column length of lighter liquid

Step 2: Formula used

1. $P=\rho gh$ where $P$ is the pressure $\rho$ density of the liquid and $h$ is the height

Step 3: Find the surface tension

The pressure difference at the points $A$and $B$, just above and below the meniscus is

$P_A-P_B=\frac{2T}{r}$

Where $T$ is the surface tension $r$ is the radius of the capillary tube

However at the point $A$ , the atmospheric pressure is acting.

$\Rightarrow P_O=P_A$ and $P_E=P_A=P_O$

According to Pascal's Law $P_C=P_D$

$$\begin{gathered} \Rightarrow P_A-\frac{2T}{r}+{\rho }_{2}g\left(h+h\text{'}\right)=P_E+{\rho }_{1}gh+{\rho }_{2}gh\text{'} \\ \Rightarrow \frac{2T}{r}=\left({\rho }_2-{\rho }_1\right)gh \\ \Rightarrow T=\frac{r\left({\rho }_2-{\rho }_1\right)gh}{2} \end{gathered}$$

Hence, option D is correct