Radius of a capillary is `2xx10^(-3)m`. A liquid of weight `6.28xx10^(-4)N` may remain in the capillary, then the surface tension of liquid will be:

Options

A. `5xx10^(-3)N//m`
B. `5xx10^(-2) N//m`
C. 5N/m
D. 50 N/m

Explanation

To find the surface tension of the liquid in the capillary, we can use the formula that relates the weight of the liquid column in the capillary to the surface tension. The formula is given by:

\[ F = 2 \pi R T \]

Where:
- \( F \) is the weight of the liquid (in Newtons),
- \( R \) is the radius of the capillary (in meters),
- \( T \) is the surface tension of the liquid (in Newtons per meter).

### Step-by-Step Solution:

1. **Identify the given values**:
- Radius of the capillary, \( R = 2 \times 10^{-3} \, \text{m} \)
- Weight of the liquid, \( F = 6.28 \times 10^{-4} \, \text{N} \)

2. **Rearrange the formula to solve for surface tension \( T \)**:
\[ T = \frac{F}{2 \pi R} \]

3. **Substitute the known values into the equation**:
\[ T = \frac{6.28 \times 10^{-4}}{2 \pi (2 \times 10^{-3})} \]

4. **Calculate the denominator**:
- First, calculate \( 2 \times 10^{-3} \):
\[ 2 \times 10^{-3} = 0.002 \, \text{m} \]
- Then calculate \( 2 \pi (2 \times 10^{-3}) \):
\[ 2 \pi (0.002) \approx 0.0125664 \, \text{m} \]

5. **Now substitute this back into the equation for \( T \)**:
\[ T = \frac{6.28 \times 10^{-4}}{0.0125664} \]

6. **Perform the division**:
\[ T \approx 0.05 \, \text{N/m} \]

### Final Answer:
The surface tension of the liquid is approximately \( 0.05 \, \text{N/m} \) or \( 5 \times 10^{-2} \, \text{N/m} \).