On dipping one end of a capiilary in liquid and inclining the capillary at an angles `30^(@)` and `60^(@)` with the vertical, the lengths of liquid columns in it are found to be `l_(1)` and `l_(2)` respectively. The ratio of `l_(1)` and `l_(2)` is

Options

A. `1:sqrt(3)`
B. `1:sqrt(2)`
C. `sqrt(2):1`
D. `sqrt(3):1`

Explanation

To solve the problem, we need to find the ratio of the lengths of the liquid columns \( l_1 \) and \( l_2 \) in a capillary tube inclined at angles \( 30^\circ \) and \( 60^\circ \) with the vertical.

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The length of the liquid column in a capillary tube can be expressed as:
\[ l = \frac{H}{\cos(\alpha)} \]
where \( H \) is the height of the liquid column when the tube is vertical, and \( \alpha \) is the angle of inclination with the vertical.

2. **Applying the Formula for Each Angle**:
- For \( l_1 \) (when the angle is \( 30^\circ \)):
\[ l_1 = \frac{H}{\cos(30^\circ)} \]
- For \( l_2 \) (when the angle is \( 60^\circ \)):
\[ l_2 = \frac{H}{\cos(60^\circ)} \]

3. **Finding the Cosine Values**:
- We know that:
\[ \cos(30^\circ) = \frac{\sqrt{3}}{2} \]
\[ \cos(60^\circ) = \frac{1}{2} \]

4. **Substituting the Cosine Values**:
- Substitute the cosine values into the expressions for \( l_1 \) and \( l_2 \):
\[ l_1 = \frac{H}{\frac{\sqrt{3}}{2}} = \frac{2H}{\sqrt{3}} \]
\[ l_2 = \frac{H}{\frac{1}{2}} = 2H \]

5. **Finding the Ratio \( \frac{l_1}{l_2} \)**:
- Now, we can find the ratio:
\[ \frac{l_1}{l_2} = \frac{\frac{2H}{\sqrt{3}}}{2H} \]
- The \( 2H \) terms cancel out:
\[ \frac{l_1}{l_2} = \frac{1}{\sqrt{3}} \]

6. **Final Ratio**:
- Thus, the ratio \( l_1 : l_2 \) is:
\[ l_1 : l_2 = 1 : \sqrt{3} \]

### Conclusion:
The ratio of the lengths of the liquid columns \( l_1 \) and \( l_2 \) is \( 1 : \sqrt{3} \).