Calculate the diameter of a capillary tube in which mercury is derpessed by 1.21 cm. Given surface tension for mercury is `540 xx 10^(-3) "Nm"^(-1).` the angle of contact with glass is `140^(@)` and density of mercury is `13.6 xx 10^(3) "kg" m^(-3).`

Answer

To calculate the diameter of a capillary tube in which mercury is depressed by 1.21 cm, we can use the formula for capillary action. The formula for the height of the liquid column in a capillary tube is given by:

\[ h = \frac{2 \sigma \cos \theta}{\rho g r} \]

Where:
- \( h \) = height of the liquid column (depression in this case)
- \( \sigma \) = surface tension of the liquid
- \( \theta \) = angle of contact
- \( \rho \) = density of the liquid
- \( g \) = acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \))
- \( r \) = radius of the capillary tube

### Step 1: Convert the given values into appropriate units
- The depression \( h = 1.21 \, \text{cm} = 1.21 \times 10^{-2} \, \text{m} \)
- Surface tension \( \sigma = 540 \times 10^{-3} \, \text{N/m} = 0.540 \, \text{N/m} \)
- Angle of contact \( \theta = 140^\circ \)
- Density of mercury \( \rho = 13.6 \times 10^{3} \, \text{kg/m}^3 \)
- Acceleration due to gravity \( g = 9.81 \, \text{m/s}^2 \)

### Step 2: Calculate \( \cos \theta \)
Using the angle of contact:
\[ \cos(140^\circ) = -\cos(40^\circ) \approx -0.766 \]

### Step 3: Rearrange the formula to solve for \( r \)
From the formula:
\[ h = \frac{2 \sigma \cos \theta}{\rho g r} \]
We can rearrange it to find \( r \):
\[ r = \frac{2 \sigma \cos \theta}{\rho g h} \]

### Step 4: Substitute the values into the equation
Substituting the known values into the equation:
\[ r = \frac{2 \times 0.540 \, \text{N/m} \times (-0.766)}{(13.6 \times 10^{3} \, \text{kg/m}^3) \times (9.81 \, \text{m/s}^2) \times (1.21 \times 10^{-2} \, \text{m})} \]

### Step 5: Calculate \( r \)
Calculating the numerator:
\[ 2 \times 0.540 \times (-0.766) \approx -0.826 \]
Calculating the denominator:
\[ (13.6 \times 10^{3}) \times (9.81) \times (1.21 \times 10^{-2}) \approx 1.628 \times 10^{3} \]
Now substituting these values:
\[ r = \frac{-0.826}{1.628 \times 10^{3}} \approx -5.07 \times 10^{-4} \, \text{m} \]
Since we are interested in the magnitude:
\[ r \approx 5.07 \times 10^{-4} \, \text{m} \]

### Step 6: Calculate the diameter \( d \)
The diameter \( d \) is twice the radius:
\[ d = 2r = 2 \times 5.07 \times 10^{-4} \approx 1.014 \times 10^{-3} \, \text{m} \approx 1.014 \, \text{mm} \]

### Final Answer
The diameter of the capillary tube is approximately \( 1.014 \, \text{mm} \).

---