If a capillary tube is `45^(@) and 60^(@)` from the vertical then the ratoio of lengths `l_(1) and l_(2)` liquid columns in it will be

Options

A. ` 1: sqrt(2)`
B. ` sqrt(2):1`
C. `2:1`
D. `1:4`

Explanation

To solve the problem of finding the ratio of lengths \( l_1 \) and \( l_2 \) of liquid columns in a capillary tube at angles of contact \( 45^\circ \) and \( 60^\circ \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Formula for Capillary Rise**:
The height of the liquid column in a capillary tube is given by the formula:
\[ h = \frac{2T \cos \theta}{\rho g r} \]
where:
- \( h \) is the height of the liquid column,
- \( T \) is the surface tension of the liquid,
- \( \theta \) is the angle of contact,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( r \) is the radius of the capillary tube.

2. **Set Up the Equations for Both Angles**:
For the angle \( 45^\circ \):
\[ l_1 = \frac{2T \cos 45^\circ}{\rho g r} \]
For the angle \( 60^\circ \):
\[ l_2 = \frac{2T \cos 60^\circ}{\rho g r} \]

3. **Substitute the Values of Cosine**:
We know that:
- \( \cos 45^\circ = \frac{1}{\sqrt{2}} \)
- \( \cos 60^\circ = \frac{1}{2} \)

Therefore, we can rewrite the equations:
\[ l_1 = \frac{2T \cdot \frac{1}{\sqrt{2}}}{\rho g r} = \frac{2T}{\sqrt{2} \rho g r} \]
\[ l_2 = \frac{2T \cdot \frac{1}{2}}{\rho g r} = \frac{T}{\rho g r} \]

4. **Find the Ratio \( \frac{l_1}{l_2} \)**:
Now, we can find the ratio of the lengths:
\[ \frac{l_1}{l_2} = \frac{\frac{2T}{\sqrt{2} \rho g r}}{\frac{T}{\rho g r}} \]

Simplifying this gives:
\[ \frac{l_1}{l_2} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

5. **Final Answer**:
Thus, the ratio of lengths \( l_1 \) and \( l_2 \) is:
\[ \frac{l_1}{l_2} = \sqrt{2} \]