The maximum force, in addition to the weight required to pull a wire of `5.0cm` long from the surface of water at temperature `20^@C`, is 728 dynes. The surface tension of water is

Options

A. `7.28(N)/(cm)`
B. `7.28("dyne")/(cm)`
C. `72.8("dyne")/(cm)`
D. `7.28xx10^2("dyne")/(cm)`

Explanation

To find the surface tension of water given the maximum force required to pull a wire from its surface, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We need to calculate the surface tension of water using the maximum force needed to pull a wire out of the water. The length of the wire is given as 5.0 cm, and the maximum force required is 728 dynes.

2. **Identify the Formula for Surface Tension**: The surface tension (S) can be calculated using the formula:
\[ S = \frac{F}{L} \]
where \( F \) is the force acting on the wire and \( L \) is the length of the wire in contact with the water.

3. **Consider the Forces Acting on the Wire**: When the wire is pulled from the water, the surface tension acts along the length of the wire. Since the wire is in contact with the water on both sides, the effective length contributing to the surface tension is \( 2L \).

4. **Calculate the Effective Length**: Given that the length of the wire \( L = 5.0 \, \text{cm} \), the effective length for surface tension is:
\[ 2L = 2 \times 5.0 \, \text{cm} = 10.0 \, \text{cm} \]

5. **Substitute Values into the Surface Tension Formula**: Now we can substitute the values into the surface tension formula:
\[ S = \frac{F}{2L} = \frac{728 \, \text{dynes}}{10.0 \, \text{cm}} \]

6. **Perform the Calculation**:
\[ S = \frac{728}{10} = 72.8 \, \text{dynes/cm} \]

7. **Conclusion**: The surface tension of water at 20°C is \( 72.8 \, \text{dynes/cm} \).

### Final Answer:
The surface tension of water is **72.8 dynes/cm**.

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