Water rises to a height h in a capillary tube of radius r. The mass of water in the capillary tube is 10 g. The mass of water rising in another capillary tube of radius 4r will be

Options

A. 30g
B. 40g
C. 20g
D. 10g

Explanation

To solve the problem, we need to understand the relationship between the mass of water that rises in a capillary tube and the radius of the tube. The mass of water that rises in a capillary tube is directly proportional to the radius of the tube when other factors remain constant.

### Step-by-Step Solution:

1. **Identify the Given Information**:
- Mass of water in the first capillary tube (radius \( r \)): \( m_1 = 10 \, \text{g} \)
- Radius of the first capillary tube: \( r_1 = r \)
- Radius of the second capillary tube: \( r_2 = 4r \)

2. **Understand the Relationship**:
The mass of water that rises in a capillary tube is given by the formula:
\[ m \propto r \]
This means that the mass of water (\( m \)) is directly proportional to the radius (\( r \)) of the capillary tube.

3. **Set Up the Proportionality**:
From the relationship, we can express the masses in terms of their respective radii:
\[ m_1 = k \cdot r_1 \quad \text{and} \quad m_2 = k \cdot r_2 \]
where \( k \) is a constant that depends on other factors like the height of the liquid column, surface tension, etc.

4. **Write the Ratios**:
Taking the ratio of the two masses:
\[ \frac{m_1}{m_2} = \frac{r_1}{r_2} \]

5. **Substitute the Known Values**:
Substitute \( m_1 = 10 \, \text{g} \), \( r_1 = r \), and \( r_2 = 4r \):
\[ \frac{10 \, \text{g}}{m_2} = \frac{r}{4r} \]
Simplifying the right side gives:
\[ \frac{10 \, \text{g}}{m_2} = \frac{1}{4} \]

6. **Solve for \( m_2 \)**:
Cross-multiplying gives:
\[ 10 \, \text{g} = \frac{1}{4} m_2 \]
Therefore:
\[ m_2 = 10 \, \text{g} \times 4 = 40 \, \text{g} \]

### Final Answer:
The mass of water rising in the capillary tube of radius \( 4r \) will be \( 40 \, \text{g} \).