A long capillary tube of radius `0.2 mm` is placed vertically inside a beaker of water.
If the tube is now pushed into water so that only `5.0 cm` of its length is above the surface, then determine the angle of contact between the liquid and glass surface.

Options

A. `cos^(-1)(4/5)`
B. `cos^(-1)(5/7)`
C. `cos^(-1)(3/5)`
D. `cos^(-1)(5/4)`

Explanation

To solve the problem, we need to determine the angle of contact (θ) between the water and the glass surface of the capillary tube. We will use the balance of forces acting on the water column inside the capillary tube.

### Step-by-Step Solution:

1. **Identify the Forces Acting on the Water Column**:
- The weight of the water column acts downward.
- The upward force due to surface tension acts along the contact line of the water with the glass.

2. **Weight of the Water Column**:
- The weight (W) of the water column can be expressed as:
\[ W = \text{mass} \times g = \rho \times V \times g \]
- The volume (V) of the water column in the capillary tube is given by:
\[ V = \pi r^2 h \]
- Here, \( r = 0.2 \, \text{mm} = 0.2 \times 10^{-3} \, \text{m} \) and \( h = 5.0 \, \text{cm} = 0.05 \, \text{m} \).
- Therefore, the weight can be written as:
\[ W = \rho \times \pi r^2 h \times g \]

3. **Surface Tension Force**:
- The surface tension force (F_T) acts along the contact line and can be expressed as:
\[ F_T = \gamma \times L \]
- Where \( L \) is the contact length, which is the circumference of the tube:
\[ L = 2 \pi r \]
- Thus, the surface tension force can be rewritten as:
\[ F_T = \gamma \times 2 \pi r \]

4. **Force Balance**:
- At equilibrium, the upward force due to surface tension equals the downward weight of the water column:
\[ F_T \cos \theta = W \]
- Substituting the expressions for \( F_T \) and \( W \):
\[ \gamma \times 2 \pi r \cos \theta = \rho \times \pi r^2 h \times g \]

5. **Simplifying the Equation**:
- Canceling \( \pi r \) from both sides:
\[ 2 \gamma \cos \theta = \rho r h g \]
- Rearranging gives:
\[ \cos \theta = \frac{\rho r h g}{2 \gamma} \]

6. **Substituting Values**:
- Assume the density of water \( \rho = 1000 \, \text{kg/m}^3 \), gravitational acceleration \( g = 10 \, \text{m/s}^2 \), and surface tension of water \( \gamma \approx 0.072 \, \text{N/m} \).
- Now substituting the values:
\[ r = 0.2 \times 10^{-3} \, \text{m}, \quad h = 0.05 \, \text{m} \]
- Thus:
\[ \cos \theta = \frac{1000 \times (0.2 \times 10^{-3}) \times 0.05 \times 10}{2 \times 0.072} \]

7. **Calculating**:
- Calculate the right-hand side:
\[ \cos \theta = \frac{1000 \times 0.2 \times 10^{-3} \times 0.05 \times 10}{0.144} = \frac{0.1}{0.144} \approx 0.6944 \]
- Therefore:
\[ \theta = \cos^{-1}(0.6944) \]

8. **Final Calculation**:
- Using a calculator to find the angle:
\[ \theta \approx 45.57^\circ \]

### Final Answer:
The angle of contact between the liquid and glass surface is approximately \( \theta \approx 45.57^\circ \).