A wire ring of diameter 0.03m is dipped in a liquid and pulled out gently. If a force of 0.1N is required to break the film, then what is the surface tension of the liquid?

Explanation

To find the surface tension of the liquid, we can follow these steps:

### Step 1: Identify the Given Values
- Diameter of the wire ring (d) = 0.03 m
- Force required to break the film (F) = 0.1 N

### Step 2: Calculate the Radius of the Ring
The radius (r) is half of the diameter:
\[ r = \frac{d}{2} = \frac{0.03 \, \text{m}}{2} = 0.015 \, \text{m} \]

### Step 3: Calculate the Circumference of the Ring
The circumference (C) of the ring is given by the formula:
\[ C = 2 \pi r \]
Substituting the value of r:
\[ C = 2 \pi (0.015 \, \text{m}) = 0.09424778 \, \text{m} \approx 0.0942 \, \text{m} \]

### Step 4: Use the Formula for Surface Tension
The surface tension (T) is defined as the force (F) divided by the length (C):
\[ T = \frac{F}{C} \]
Substituting the values we have:
\[ T = \frac{0.1 \, \text{N}}{0.0942 \, \text{m}} \approx 1.06 \, \text{N/m} \]

### Final Answer
The surface tension of the liquid is approximately:
\[ T \approx 1.06 \, \text{N/m} \]

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