In a capillary tube having area of cross - section A, water rises to a height h. If cross-sectional area is reduced to `(A)/(9)`, the rise of water in the capillary tube is

Options

A. 4h
B. 3h
C. 2h
D. h

Answer

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

### Step-by-Step Solution:

1. **Understanding Capillary Rise**:
The height of liquid (h) in a capillary tube is given by the formula:
\[ h = \frac{2T \cos \theta}{\rho g R} \]
where:
- \( T \) 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. **Relating Radius and Area**:
The cross-sectional area \( A \) of the capillary tube is related to the radius \( R \) by:
\[ A = \pi R^2 \]
Therefore, the radius can be expressed as:
\[ R = \sqrt{\frac{A}{\pi}} \]

3. **Inversely Proportional Relationship**:
From the formula for height \( h \), we can see that the height is inversely proportional to the radius \( R \):
\[ h \propto \frac{1}{R} \]
Since \( R \) is related to the area \( A \) (as \( R \propto \sqrt{A} \)), we can say:
\[ h \propto \frac{1}{\sqrt{A}} \]

4. **Comparing Initial and Reduced Area**:
Let the initial area be \( A \) and the new area be \( \frac{A}{9} \). The new height \( h_2 \) can be expressed in terms of the initial height \( h \):
\[ \frac{h_2}{h} = \sqrt{\frac{A}{\frac{A}{9}}} \]
Simplifying this gives:
\[ \frac{h_2}{h} = \sqrt{9} = 3 \]
Therefore, we have:
\[ h_2 = 3h \]

5. **Conclusion**:
When the cross-sectional area of the capillary tube is reduced to \( \frac{A}{9} \), the height of water rises to \( 3h \).

### Final Answer:
The rise of water in the capillary tube when the cross-sectional area is reduced to \( \frac{A}{9} \) is \( 3h \).