A capillary tube of the radius 0.5 mm is immersed in a beaker of mercury . The level inside the tube is 0.8 cm below the level in beaker and angle of contact is `120^@` . What is the surface tension of mercury , if the mass density of mercury is `rho= 13.6 xx 10^3 kg m^3` and acceleration due to gravity is g = 10 m `s^(-2)` ?

Options

A. `0.225 "N m"^(-1)`
B. `0.544 "N m"^(-1)`
C. `0.285 "N m"^(-1)`
D. `0.375 "N m"^(-1)`

Answer

To solve the problem, we will use the formula for the height of liquid in a capillary tube, which is given by:

\[ h = \frac{2T \cos \theta}{R \rho g} \]

Where:
- \( h \) = height difference (in meters)
- \( T \) = surface tension (in N/m)
- \( \theta \) = angle of contact (in degrees)
- \( R \) = radius of the capillary tube (in meters)
- \( \rho \) = density of the liquid (in kg/m³)
- \( g \) = acceleration due to gravity (in m/s²)

### Step 1: Convert all given values to the appropriate units
- Radius \( R = 0.5 \, \text{mm} = 0.5 \times 10^{-3} \, \text{m} = 0.0005 \, \text{m} \)
- Height \( h = 0.8 \, \text{cm} = 0.8 \times 10^{-2} \, \text{m} = 0.008 \, \text{m} \)
- Density of mercury \( \rho = 13.6 \times 10^{3} \, \text{kg/m}^3 \)
- Acceleration due to gravity \( g = 10 \, \text{m/s}^2 \)
- Angle of contact \( \theta = 120^\circ \)

### Step 2: Calculate \( \cos \theta \)
\[ \cos 120^\circ = -\frac{1}{2} \]

### Step 3: Rearrange the formula to solve for surface tension \( T \)
Rearranging the formula gives us:
\[ T = \frac{h R \rho g}{2 \cos \theta} \]

### Step 4: Substitute the values into the equation
Substituting the known values:
\[ T = \frac{0.008 \, \text{m} \times 0.0005 \, \text{m} \times 13.6 \times 10^{3} \, \text{kg/m}^3 \times 10 \, \text{m/s}^2}{2 \times -\frac{1}{2}} \]

### Step 5: Simplify the equation
Calculating the numerator:
\[ 0.008 \times 0.0005 \times 13.6 \times 10^{3} \times 10 = 0.544 \]
And since \( 2 \times -\frac{1}{2} = -1 \), we have:
\[ T = \frac{0.544}{-1} = -0.544 \, \text{N/m} \]

Since surface tension is a positive quantity, we take the absolute value:
\[ T = 0.544 \, \text{N/m} \]

### Final Answer
The surface tension of mercury is \( 0.544 \, \text{N/m} \).

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