A capillary tube of radius 'r' is immersed in water and water rises in it to a height H. Mass of water in the capillary tube is m. If the Capillary of radius 2r is taken and dipped in water, the mass of water that will rise in the capillary tube will be

Options

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

Explanation

To solve the problem, we will use the principles of capillarity and the relationship between the mass of water in a capillary tube and its radius.

### Step-by-Step Solution:

1. **Understanding the Capillary Rise Formula**:
The height \( H \) to which a liquid 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. **Finding the Mass of Water in the First Capillary Tube**:
The mass of water \( m \) in the capillary tube can be expressed as:
\[ m = \rho V \]
where \( V \) is the volume of water in the tube. The volume of water in the capillary tube is given by:
\[ V = \pi r^2 H \]
Substituting \( H \) from the first formula, we get:
\[ V = \pi r^2 \left(\frac{2T \cos \theta}{\rho g r}\right) = \frac{2\pi T \cos \theta}{g \rho} r \]
Thus, the mass \( m \) becomes:
\[ m = \rho \cdot \frac{2\pi T \cos \theta}{g \rho} r = \frac{2\pi T \cos \theta}{g} r \]

3. **Considering the Second Capillary Tube with Radius \( 2r \)**:
Now, if we take a capillary tube of radius \( 2r \), we can find the new mass \( m' \) of water that will rise in it:
\[ H' = \frac{2T \cos \theta}{\rho g (2r)} = \frac{2T \cos \theta}{2\rho g r} = \frac{T \cos \theta}{\rho g r} \]
The volume of water in the second capillary tube is:
\[ V' = \pi (2r)^2 H' = \pi (4r^2) \left(\frac{T \cos \theta}{\rho g r}\right) = \frac{4\pi T \cos \theta}{g \rho} r \]
Therefore, the mass \( m' \) becomes:
\[ m' = \rho \cdot \frac{4\pi T \cos \theta}{g \rho} r = \frac{4\pi T \cos \theta}{g} \]

4. **Finding the Relationship Between \( m' \) and \( m \)**:
From our earlier expression for \( m \):
\[ m = \frac{2\pi T \cos \theta}{g} \]
Therefore, we can express \( m' \) in terms of \( m \):
\[ m' = 2m \]

### Final Answer:
The mass of water that will rise in the capillary tube of radius \( 2r \) is \( 2m \).