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 understand the relationship between the height of capillary rise and the cross-sectional area of the capillary tube.

### Step-by-Step Solution:

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

2. **Relating Cross-Sectional Area to Radius**:
The cross-sectional area \( A \) of a capillary tube is related to its radius \( R \) by the formula:
\[ A = \pi R^2 \]
Therefore, the radius can be expressed in terms of the area:
\[ R = \sqrt{\frac{A}{\pi}} \]

3. **Inversely Proportional Relationship**:
From the capillary rise formula, we see that the height \( h \) is inversely proportional to the radius \( R \). Hence, we can express this relationship as:
\[ h \propto \frac{1}{R} \]

4. **Substituting for Radius**:
Since \( R \) is related to the area \( A \) as \( R \propto A^{1/2} \), we can substitute this into our height equation:
\[ h \propto \frac{1}{A^{1/2}} \]

5. **Comparing Two Scenarios**:
Let’s denote the height of capillary rise in the first tube (cross-sectional area \( A \)) as \( h_1 \) and in the second tube (cross-sectional area \( 4A \)) as \( h_2 \). According to our relationship:
\[ \frac{h_1}{h_2} = \frac{A^{1/2}}{(4A)^{1/2}} = \frac{A^{1/2}}{2A^{1/2}} = \frac{1}{2} \]
This implies:
\[ h_2 = 2h_1 \]

6. **Final Result**:
If the height of water rises to \( h \) in the first capillary tube (cross-sectional area \( A \)), then in the second capillary tube (cross-sectional area \( 4A \)), the height of water will rise to:
\[ h_2 = \frac{h}{2} \]

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