To solve the problem step by step, we will use the principles of capillarity and the relationship between surface tension, density, and height of the liquid column.
### Step 1: Understand the problem
We have a capillary tube with a radius of 0.5 mm immersed in mercury. The height difference between the mercury level in the beaker and the level inside the tube is given as 0.8 cm. The contact angle is 120 degrees. We need to find the surface tension of mercury.
### Step 2: Convert units
Convert the radius and height into meters for consistency in SI units:
- Radius \( r = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} = 5 \times 10^{-4} \, \text{m} \)
- Height \( h = 0.8 \, \text{cm} = 0.8 \times 10^{-2} \, \text{m} = 8 \times 10^{-3} \, \text{m} \)
### Step 3: Identify the relevant formula
The pressure difference due to the height of the liquid column can be expressed as:
\[
\Delta P = \rho g h
\]
where:
- \( \Delta P \) is the pressure difference,
- \( \rho \) is the density of mercury,
- \( g \) is the acceleration due to gravity,
- \( h \) is the height difference.
The pressure difference can also be related to surface tension \( T \) by the formula:
\[
\Delta P = \frac{2T}{r}
\]
where \( r \) is the radius of the capillary tube.
### Step 4: Equate the two pressure differences
Since both expressions represent the same pressure difference, we can equate them:
\[
\rho g h = \frac{2T}{r}
\]
### Step 5: Solve for surface tension \( T \)
Rearranging the equation to solve for \( T \):
\[
T = \frac{\rho g h r}{2}
\]
### Step 6: Substitute the known values
Substituting the known values into the equation:
- \( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \)
- \( g = 10 \, \text{m/s}^2 \)
- \( h = 8 \times 10^{-3} \, \text{m} \)
- \( r = 5 \times 10^{-4} \, \text{m} \)
\[
T = \frac{(13.6 \times 10^3) \cdot (10) \cdot (8 \times 10^{-3}) \cdot (5 \times 10^{-4})}{2}
\]
### Step 7: Calculate the surface tension
Calculating the values:
\[
T = \frac{(13.6 \times 10^3) \cdot (10) \cdot (8 \times 10^{-3}) \cdot (5 \times 10^{-4})}{2}
\]
\[
= \frac{(13.6 \times 10^3) \cdot (10) \cdot (8) \cdot (5)}{2 \cdot 10^3}
\]
\[
= \frac{(13.6 \cdot 10 \cdot 8 \cdot 5)}{2}
\]
\[
= \frac{5440}{2} = 2720 \, \text{N/m}
\]
### Final Answer
The surface tension of mercury is \( T = 2720 \, \text{N/m} \).
