Water rises in a capillary upto a height `h`. If now this capillary is tilted by an angle of `45^(@)` then the length of the water column in the capillary becomes-

Options

A. `2h`
B. `(h)/(2)`
C. `(h)/(sqrt(2))`
D. `hsqrt(2)`

Answer

To solve the problem of how the length of the water column in a capillary tube changes when the tube is tilted at an angle of 45 degrees, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Initial Condition**:
- Initially, the capillary tube is vertical, and the water rises to a height \( h \).

2. **Identify the Change in Orientation**:
- The capillary tube is then tilted at an angle of \( 45^\circ \).

3. **Analyze the Vertical Height**:
- When the capillary is tilted, the vertical height of the water column remains the same at \( h \). This is because the pressure difference that causes the rise of water in the capillary is unchanged.

4. **Relate the Length of the Water Column to the Angle**:
- Let \( h' \) be the length of the water column along the inclined capillary. The vertical height can be related to the inclined length using trigonometry:
\[ h = h' \cdot \cos(45^\circ) \]

5. **Calculate \( h' \)**:
- We know that \( \cos(45^\circ) = \frac{1}{\sqrt{2}} \). Therefore, we can rearrange the equation:
\[ h' = \frac{h}{\cos(45^\circ)} = h \cdot \sqrt{2} \]

6. **Final Result**:
- Thus, the length of the water column in the capillary when tilted at \( 45^\circ \) becomes:
\[ h' = h \sqrt{2} \]

### Conclusion:
The length of the water column in the capillary when tilted at an angle of \( 45^\circ \) is \( h \sqrt{2} \).