A capillary tube of radius r is immersed in water and water rises in to a height h. The mass of water in the capillary tube is 5g. Another capillary tube of radius 2 r is immersed in water. The mass of water that will rise in this tube is

Options

A. 2.5 g
B. 5.0 g
C. 10 g
D. 20 g

Explanation

To solve the problem step by step, we can follow these instructions:

### Step 1: Understand the Capillary Rise
The height to which a liquid rises in a capillary tube is inversely proportional to the radius of the tube. This relationship can be expressed as:
\[ r_1 h_1 = r_2 h_2 \]
where \( r_1 \) and \( h_1 \) are the radius and height for the first tube, and \( r_2 \) and \( h_2 \) are the radius and height for the second tube.

### Step 2: Define the Given Data
For the first capillary tube:
- Radius \( r_1 = r \)
- Height \( h_1 = h \)
- Mass of water \( m_1 = 5 \, \text{g} \)

For the second capillary tube:
- Radius \( r_2 = 2r \)

### Step 3: Relate the Heights
Using the relationship from Step 1, we can express the height \( h_2 \) for the second tube:
\[ r_1 h_1 = r_2 h_2 \implies r \cdot h = 2r \cdot h_2 \]
Dividing both sides by \( r \):
\[ h = 2h_2 \implies h_2 = \frac{h}{2} \]

### Step 4: Calculate the Mass of Water in the Second Tube
The mass of water in a capillary tube can be calculated using the formula:
\[ m = \rho \pi r^2 h \]
For the first tube:
\[ m_1 = \rho \pi r^2 h = 5 \, \text{g} \]

For the second tube, substituting \( r_2 = 2r \) and \( h_2 = \frac{h}{2} \):
\[ m_2 = \rho \pi (2r)^2 \left(\frac{h}{2}\right) \]
Calculating this gives:
\[ m_2 = \rho \pi (4r^2) \left(\frac{h}{2}\right) = 2 \rho \pi r^2 h \]

### Step 5: Relate \( m_2 \) to \( m_1 \)
Since \( m_1 = \rho \pi r^2 h \):
\[ m_2 = 2 m_1 \]
Substituting \( m_1 = 5 \, \text{g} \):
\[ m_2 = 2 \times 5 \, \text{g} = 10 \, \text{g} \]

### Final Answer
The mass of water that will rise in the second capillary tube is \( 10 \, \text{g} \).

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