A 10 cm long wire is placed horizontal on the surface of water and is gently pulled up with a force of `2xx10^(-2)` N to keep the wire in equilibrium. The surface tension, in `Nm^(-1)` of water is

Options

A. 0.1
B. 0.2
C. 0.001
D. 0.002

Explanation

To find the surface tension of water when a 10 cm long wire is placed horizontally on its surface and pulled up with a force of \(2 \times 10^{-2}\) N, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
- We have a wire of length \(L = 10 \text{ cm} = 0.1 \text{ m}\) placed horizontally on the surface of water.
- The wire is being pulled up with a force \(F = 2 \times 10^{-2} \text{ N}\).

2. **Identify Effective Length**:
- When the wire is placed on the surface of the water, the surface tension acts along both sides of the wire.
- Therefore, the effective length \(L_{\text{eff}}\) that contributes to the surface tension is \(2L\).
- Thus, \(L_{\text{eff}} = 2 \times 0.1 \text{ m} = 0.2 \text{ m}\).

3. **Use the Formula for Surface Tension**:
- Surface tension (\(\sigma\)) is defined as the force per unit length. The formula is:
\[ \sigma = \frac{F}{L_{\text{eff}}} \]
- Substituting the values we have:
\[ \sigma = \frac{2 \times 10^{-2} \text{ N}}{0.2 \text{ m}} \]

4. **Calculate Surface Tension**:
- Performing the calculation:
\[ \sigma = \frac{2 \times 10^{-2}}{0.2} = \frac{2 \times 10^{-2}}{2 \times 10^{-1}} = 0.1 \text{ N/m} \]

5. **Final Answer**:
- The surface tension of water is \(0.1 \text{ N/m}\).