Water rises to a height of 20 mm in a capillary tube. If the radius made `1//3^("rd")` of its previous value, to what height will the water now rise in the tube?

Answer

To solve the problem, we will use the formula for the height of liquid rise in a capillary tube, which is given by:

\[ h = \frac{2T}{R \rho g} \]

Where:
- \( h \) is the height of the liquid rise,
- \( T \) is the surface tension of the liquid,
- \( R \) is the radius of the capillary tube,
- \( \rho \) is the density of the liquid,
- \( g \) is the acceleration due to gravity.

### Step-by-Step Solution:

1. **Identify Initial Conditions**:
- The initial height of water rise \( h_1 = 20 \, \text{mm} \).
- The initial radius of the capillary tube is \( R_1 = r \).

2. **Determine New Radius**:
- The new radius \( R_2 \) is one-third of the initial radius:
\[ R_2 = \frac{1}{3} R_1 = \frac{1}{3} r \]

3. **Understand the Relationship**:
- The height of the liquid rise in a capillary tube is inversely proportional to the radius of the tube. This means:
\[ h \propto \frac{1}{R} \]
Therefore, we can express the relationship between the initial height \( h_1 \) and the new height \( h_2 \):
\[ \frac{h_1}{h_2} = \frac{R_2}{R_1} \]

4. **Substituting Known Values**:
- Substitute \( R_2 \) and \( R_1 \) into the equation:
\[ \frac{20 \, \text{mm}}{h_2} = \frac{\frac{1}{3} R_1}{R_1} \]
Simplifying this gives:
\[ \frac{20 \, \text{mm}}{h_2} = \frac{1}{3} \]

5. **Solving for \( h_2 \)**:
- Cross-multiply to find \( h_2 \):
\[ 20 \, \text{mm} = \frac{1}{3} h_2 \]
\[ h_2 = 20 \, \text{mm} \times 3 = 60 \, \text{mm} \]

### Final Answer:
The new height to which the water will rise in the tube is \( h_2 = 60 \, \text{mm} \).