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 radius of the tube is doubled, mass of water that will rise in the capillary tube will now be

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between mass and volume in a capillary tube
The mass of the water in the capillary tube can be expressed as:
\[ M = \rho \cdot V \]
where \( \rho \) is the density of water and \( V \) is the volume of water in the tube.

### Step 2: Calculate the volume of water in the capillary tube
The volume \( V \) of water in a cylindrical capillary tube is given by:
\[ V = A \cdot H \]
where \( A \) is the cross-sectional area of the tube and \( H \) is the height of the water column.

For a tube of radius \( R \), the cross-sectional area \( A \) is:
\[ A = \pi R^2 \]
Thus, the volume becomes:
\[ V = \pi R^2 H \]

### Step 3: Substitute the volume into the mass equation
Substituting the expression for volume into the mass equation gives:
\[ M = \rho \cdot (\pi R^2 H) \]

### Step 4: Relate height \( H \) to the radius \( R \)
From the capillary rise formula, we know:
\[ H = \frac{2T \cos \theta}{R} \]
where \( T \) is the surface tension and \( \theta \) is the contact angle.

### Step 5: Substitute \( H \) into the mass equation
Now, substituting \( H \) into the mass equation:
\[ M = \rho \cdot \left(\pi R^2 \cdot \frac{2T \cos \theta}{R}\right) \]
This simplifies to:
\[ M = 2 \pi \rho T \cos \theta \cdot R \]

### Step 6: Analyze the effect of doubling the radius
If the radius of the tube is doubled (i.e., \( R \) becomes \( 2R \)), we can find the new mass \( M' \):
\[ M' = 2 \pi \rho T \cos \theta \cdot (2R) \]
This simplifies to:
\[ M' = 4 \pi \rho T \cos \theta \cdot R \]

### Step 7: Relate the new mass to the original mass
From our original mass equation:
\[ M = 2 \pi \rho T \cos \theta \cdot R \]
Thus, we can express \( M' \) in terms of \( M \):
\[ M' = 2M \]

### Conclusion
Therefore, when the radius of the tube is doubled, the mass of water that will rise in the capillary tube will also double.

### Final Answer
The mass of water that will rise in the capillary tube when the radius is doubled is:
\[ M' = 2M \]

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