Water rises to a height of `10cm` in a glass capillary tube. If the area of cross section of the tube is reduced to one fourth of the former value what is the height of water rise now?

Options

A. 20 cm
B. 5 cm
C. 2.5 cm
D. 7 cm

Answer

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

### Step-by-Step Solution:

1. **Understand the Initial Condition**:
- The initial height of water rise in the capillary tube is given as \( h_1 = 10 \, \text{cm} \).

2. **Identify the Relationship**:
- The height of liquid rise in a capillary tube is given by the formula:
\[ h = \frac{2T \cos \theta}{\rho g r} \]
where:
- \( T \) = surface tension of the liquid,
- \( \theta \) = angle of contact,
- \( \rho \) = density of the liquid,
- \( g \) = acceleration due to gravity,
- \( r \) = radius of the capillary tube.

3. **Change in Cross-Sectional Area**:
- The problem states that the area of cross-section is reduced to one-fourth of the original value.
- The area \( A \) of a circular tube is given by \( A = \pi r^2 \). If the area is reduced to one-fourth, we have:
\[ A' = \frac{1}{4} A = \frac{1}{4} \pi r^2 \]
- This implies:
\[ r' = \sqrt{\frac{A'}{\pi}} = \sqrt{\frac{1}{4} \pi r^2} = \frac{1}{2} r \]
- Thus, the new radius \( r' \) is half of the original radius \( r \).

4. **Effect on Height of Water Rise**:
- Since the height of the liquid rise \( h \) is inversely proportional to the radius \( r \) (from the formula), we can write:
\[ h \propto \frac{1}{r} \]
- If the radius is halved, the new height \( h_2 \) can be calculated as:
\[ h_2 = \frac{2T \cos \theta}{\rho g r'} = \frac{2T \cos \theta}{\rho g \left(\frac{1}{2} r\right)} = 2 \cdot \frac{2T \cos \theta}{\rho g r} = 2h_1 \]
- Therefore, the new height of water rise is:
\[ h_2 = 2 \times 10 \, \text{cm} = 20 \, \text{cm} \]

5. **Final Answer**:
- The height of water rise in the capillary tube after the area of cross-section is reduced to one-fourth is \( \boxed{20 \, \text{cm}} \).