Mercury has an angle of contact of `120^(@)` with glass. A narrow tube of radius `1.0mm` made of the glass is dipped in a through containing mercury. By what amount does the mercury dip down in the tube relative to the liquid surface outside. Surface tension of mercury at the temperature of the experiment os `0.5 N//m` and density of mercury is `13.6xx10^(3) kg//m^(3)`.
(Take `g = 9.8 m//s^(2))`.

Explanation

To solve the problem of how much mercury dips down in a tube relative to the liquid surface outside, we can use the formula related to capillarity. The height of the liquid column (h) in the tube can be calculated using the following formula:

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

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

### Step 1: Identify the given values
- Surface tension, \( T = 0.5 \, \text{N/m} \)
- Angle of contact, \( \theta = 120^\circ \)
- Density of mercury, \( \rho = 13.6 \times 10^3 \, \text{kg/m}^3 \)
- Radius of the tube, \( R = 1.0 \, \text{mm} = 1.0 \times 10^{-3} \, \text{m} \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 2: Convert the angle of contact to cosine
The angle \( \theta \) is given as \( 120^\circ \). We need to find \( \cos(120^\circ) \):
\[ \cos(120^\circ) = -\frac{1}{2} \]

### Step 3: Substitute the values into the formula
Now, we can substitute the values into the height formula:
\[ h = \frac{2 \times 0.5 \times \cos(120^\circ)}{(13.6 \times 10^3) \times (9.8) \times (1.0 \times 10^{-3})} \]

Substituting \( \cos(120^\circ) \):
\[ h = \frac{2 \times 0.5 \times \left(-\frac{1}{2}\right)}{(13.6 \times 10^3) \times (9.8) \times (1.0 \times 10^{-3})} \]

### Step 4: Simplify the equation
Calculating the numerator:
\[ 2 \times 0.5 \times \left(-\frac{1}{2}\right) = -0.5 \]

Calculating the denominator:
\[ (13.6 \times 10^3) \times (9.8) \times (1.0 \times 10^{-3}) = 133.28 \, \text{N/m}^2 \]

Now substituting back:
\[ h = \frac{-0.5}{133.28} \]

### Step 5: Calculate the height
\[ h = -3.75 \times 10^{-3} \, \text{m} = -3.75 \, \text{mm} \]

### Conclusion
The negative sign indicates that the mercury dips down in the tube relative to the liquid surface outside. Therefore, the amount by which mercury dips down in the tube is:
\[ \text{Dip} = 3.75 \, \text{mm} \]