A capillary tube of radius `r` is immersed in a liquid. The liquid rises to a height `h`. The corresponding mass is `m`. What mass of water shall rise in the capillary if the radius of the tube is doubled?

Options

A. `m`
B. `2m`
C. `3m`
D. `4m`

Explanation

To solve the problem, we need to understand the relationship between the height of the liquid column in a capillary tube and the radius of the tube. The height to which a liquid rises in a capillary tube is given by the formula:

\[ h = \frac{2s \cos \theta}{\rho g r} \]

where:
- \( h \) is the height of the liquid column,
- \( s \) is the surface tension of the liquid,
- \( \theta \) is the contact angle,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity,
- \( r \) is the radius of the capillary tube.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- The initial radius of the capillary tube is \( r \).
- The height of the liquid column is \( h \).
- The mass of the liquid is \( m \).

2. **Determine the volume of the liquid in the capillary tube:**
- The volume \( V \) of the liquid that rises to height \( h \) in a tube of radius \( r \) is given by:
\[ V = \pi r^2 h \]

3. **Calculate the mass of the liquid using its density:**
- The mass \( m \) of the liquid can be expressed as:
\[ m = \rho V = \rho (\pi r^2 h) \]

4. **Now consider the case when the radius is doubled:**
- If the radius is doubled, the new radius \( r' = 2r \).
- We need to find the new height \( h' \) of the liquid column when the radius is \( 2r \).

5. **Substituting the new radius into the height formula:**
- The new height \( h' \) can be calculated as:
\[ h' = \frac{2s \cos \theta}{\rho g (2r)} = \frac{2s \cos \theta}{2\rho g r} = \frac{s \cos \theta}{\rho g r} \]

6. **Calculate the new volume of the liquid:**
- The new volume \( V' \) of the liquid that rises to height \( h' \) in a tube of radius \( 2r \) is:
\[ V' = \pi (2r)^2 h' = \pi (4r^2) \left(\frac{s \cos \theta}{\rho g r}\right) = \frac{4\pi r s \cos \theta}{\rho g} \]

7. **Calculate the new mass of the liquid:**
- The new mass \( m' \) of the liquid can be expressed as:
\[ m' = \rho V' = \rho \left(\frac{4\pi r s \cos \theta}{\rho g}\right) = \frac{4\pi r s \cos \theta}{g} \]

8. **Relate the new mass to the initial mass:**
- From the earlier expressions, we know:
\[ m = \rho (\pi r^2 h) = \frac{2\pi r s \cos \theta}{g} \]
- Therefore, we can see that:
\[ m' = 2m \]

### Conclusion:
Thus, if the radius of the capillary tube is doubled, the mass of the liquid that rises in the capillary tube will also double. Hence, the answer is:

**Final Answer: \( 2m \)**