A capillary tube of radius r can support a liquid of weight `6.28xx10^(-4)N`. If the surface tension of the liquid is `5xx10^(-2)N//m`. The radius of capillary must be

Options

A. `2xx10^(-3)m`
B. `2xx10^(-4)m`
C. `1.5xx10^(-3)`
D. `12.5xx10^(-4)m`

Answer

To solve the problem, we need to find the radius of the capillary tube that can support a given weight of liquid using the surface tension of that liquid. Let's go through the steps systematically.

### Step-by-Step Solution:

1. **Understand the Given Data**:
- Weight of the liquid (W) = \(6.28 \times 10^{-4} \, \text{N}\)
- Surface tension of the liquid (T) = \(5 \times 10^{-2} \, \text{N/m}\)

2. **Formula for Force due to Surface Tension**:
The force due to surface tension (F) can be expressed as:
\[ F = T \times L \]
where \(L\) is the length of the liquid column in contact with the surface.

3. **Determine the Length of Contact**:
For a capillary tube, the length of the liquid column in contact with the surface is the circumference of the tube, which can be calculated as:
\[ L = 2 \pi r \]
where \(r\) is the radius of the capillary tube.

4. **Set Up the Equation**:
The force due to surface tension must equal the weight of the liquid:
\[ T \times L = W \]
Substituting the expression for \(L\):
\[ T \times (2 \pi r) = W \]

5. **Substituting the Known Values**:
Plugging in the values for \(T\) and \(W\):
\[ 5 \times 10^{-2} \times (2 \pi r) = 6.28 \times 10^{-4} \]

6. **Simplifying the Equation**:
We can simplify this equation:
\[ 5 \times 10^{-2} \times 2 \times 3.14 \times r = 6.28 \times 10^{-4} \]
This simplifies to:
\[ 6.28 \times 10^{-2} \times r = 6.28 \times 10^{-4} \]

7. **Solving for Radius (r)**:
Dividing both sides by \(6.28 \times 10^{-2}\):
\[ r = \frac{6.28 \times 10^{-4}}{6.28 \times 10^{-2}} = \frac{1}{100} = 0.01 \, \text{m} \]
Converting to millimeters:
\[ r = 0.01 \, \text{m} = 2 \, \text{mm} \]

### Final Answer:
The radius of the capillary tube must be \(2 \times 10^{-3} \, \text{m}\) or \(2 \, \text{mm}\).

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