If a capillary of radius r is dipped in water the height of water that rises in it is h and its mass is M . If the radius of the capillary is doubled the mass of water that rises in the capillary will be

Options

A. 4M
B. 2M
C. M
D. `(M)/(2)`

Answer

To solve the problem, we need to analyze how the mass of water that rises in a capillary tube changes when the radius of the capillary is doubled.

### Step-by-Step Solution:

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

2. **Initial Conditions**:
For the initial radius \( R \), the mass of water that rises in the capillary is:
\[ M = \text{Volume} \times \rho = \pi R^2 h \rho \]

3. **Doubling the Radius**:
When the radius is doubled to \( 2R \), we need to find the new height of the liquid (\( h' \)):
\[ h' = \frac{2T \cos \theta}{\rho g (2R)} = \frac{h}{2} \]

4. **Calculating New Mass**:
The new mass of water that rises in the capillary with the new height \( h' \) is:
\[ M' = \text{Volume} \times \rho = \pi (2R)^2 h' \rho = \pi (4R^2) \left(\frac{h}{2}\right) \rho \]
Simplifying this gives:
\[ M' = \pi (4R^2) \left(\frac{h}{2}\right) \rho = 2 \pi R^2 h \rho \]

5. **Relating New Mass to Initial Mass**:
We know from our initial conditions that:
\[ M = \pi R^2 h \rho \]
Thus, we can express \( M' \) in terms of \( M \):
\[ M' = 2M \]

### Conclusion:
The mass of water that rises in the capillary when the radius is doubled is \( 2M \).