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 to the surface tension and the radius of the capillary.

### 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. **Recall the formula for the force due to surface tension:**
The force due to surface tension can be expressed as:
\[ F = T \cdot 2\pi r \]
where \( T \) is the surface tension and \( r \) is the radius of the capillary.

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

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

5. **Calculate \( 2\pi \):**
Using \( \pi \approx 3.14 \):
\[ 2\pi \approx 2 \times 3.14 = 6.28 \]

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

7. **Simplify the equation:**
\[ T = \frac{6.28 \times 10^{-4}}{6.28 \times 2 \times 10^{-3}} = \frac{1}{2} \times 10^{-1} \]

8. **Calculate the final value of surface tension:**
\[ T = 0.5 \times 10^{-1} = 5 \times 10^{-2} \, \text{N/m} \]

### Final Answer:
The surface tension of the liquid is:
\[ T = 5 \times 10^{-2} \, \text{N/m} \]