Liquid rises to a height of 2 cm in a capillary tube and the angle of contact between the solid and the liquid is zero. If the tube is depressed more now so that top of capillary is only 1 cm above the liquid, then the apparent angle of contact between the solid and the liquid is

Options

A. `0^(@)`
B. `30^(@)`
C. `60^(@)`
D. `90^(@)`

Explanation

To solve the problem, we will analyze the situation step by step using the principles of capillarity and the relationship between the height of liquid in the capillary tube and the angle of contact.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- The liquid rises to a height \( h = 2 \) cm in the capillary tube when the angle of contact \( \theta = 0^\circ \).
- The tube is then depressed such that the top of the capillary is only \( h' = 1 \) cm above the liquid.

2. **Use the Capillary Rise Formula**:
- The height to which a liquid rises in a capillary tube is given by the formula:
\[ h = \frac{2 \gamma \cos \theta}{\rho g r} \]
where \( \gamma \) is the surface tension of the liquid, \( \rho \) is the density of the liquid, \( g \) is the acceleration due to gravity, and \( r \) is the radius of the capillary tube.

3. **Set Up the Relationship for the Two Heights**:
- When the height of the liquid is \( h = 2 \) cm, we have:
\[ 2 = \frac{2 \gamma \cos 0^\circ}{\rho g r} \]
- When the height is \( h' = 1 \) cm, we have:
\[ 1 = \frac{2 \gamma \cos \theta'}{\rho g r} \]

4. **Relate the Two Heights**:
- From the two equations, we can set up a ratio:
\[ \frac{h}{h'} = \frac{\cos 0^\circ}{\cos \theta'} \]
- Substituting the known values:
\[ \frac{2}{1} = \frac{1}{\cos \theta'} \]

5. **Solve for \( \cos \theta' \)**:
- Rearranging gives:
\[ \cos \theta' = \frac{1}{2} \]

6. **Determine the Angle \( \theta' \)**:
- The angle whose cosine is \( \frac{1}{2} \) is:
\[ \theta' = 60^\circ \]

### Final Answer:
The apparent angle of contact between the solid and the liquid when the tube is depressed to 1 cm above the liquid is \( 60^\circ \).