Water rises to a height `h` in a capillary tube of cross-sectional area A. the height to which water will rise in a capillary tube of cross-sectional area `4A` will be

Options

A. ` h `
B. ` h//2 `
C. ` h//4 `
D. ` 4h `

Explanation

To solve the problem, we need to determine how the height of water rising in a capillary tube changes when the cross-sectional area of the tube is altered.

### Step-by-Step Solution:

1. **Understanding Capillary Action**:
The height to which a liquid rises in a capillary tube is inversely proportional to the radius of the tube. This relationship can be derived from the balance of forces acting on the liquid column.

2. **Initial Conditions**:
Let the height of water rise in the first capillary tube (cross-sectional area A) be \( h \). The radius \( r_1 \) of this tube can be expressed in terms of area:
\[ A = \pi r_1^2 \implies r_1 = \sqrt{\frac{A}{\pi}} \]

3. **New Conditions**:
For the second capillary tube, the cross-sectional area is \( 4A \). The radius \( r_2 \) of this tube can be expressed as:
\[ A_2 = 4A = \pi r_2^2 \implies r_2 = \sqrt{\frac{4A}{\pi}} = 2\sqrt{\frac{A}{\pi}} = 2r_1 \]

4. **Height Relationship**:
Since the height \( h \) is inversely proportional to the radius, we can express this relationship as:
\[ h \propto \frac{1}{r} \]
Therefore, we can write the ratio of the heights in the two tubes as:
\[ \frac{h_1}{h_2} = \frac{r_2}{r_1} \]

5. **Substituting Values**:
Substituting \( r_2 = 2r_1 \) into the height ratio gives:
\[ \frac{h}{h_2} = \frac{2r_1}{r_1} = 2 \]
This leads to:
\[ h_2 = \frac{h}{2} \]

6. **Final Result**:
The height to which water will rise in the capillary tube of cross-sectional area \( 4A \) is:
\[ h_2 = \frac{h}{2} \]

### Conclusion:
The height to which water will rise in a capillary tube of cross-sectional area \( 4A \) is \( \frac{h}{2} \).