A film of soap solution is formed on a loop frame loop of `6.28 cm` long thread is gently put on the film and the film is broken with a needle inside the loop. The tread loop takes the shape of a circle. Find the tension in the thread. Surface tension of soap solution `= 0.030 n//m`.

Explanation

To solve the problem, we need to find the tension in the thread loop formed by the soap film. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Length of the thread loop (L) = 6.28 cm = 0.0628 m
- Surface tension of soap solution (S) = 0.030 N/m

### Step 2: Determine the radius of the circular loop
The length of the thread loop is equal to the circumference of the circle formed by the loop.

The formula for the circumference of a circle is:
\[ L = 2\pi r \]

From this, we can solve for the radius \( r \):
\[ r = \frac{L}{2\pi} \]

Substituting the values:
\[ r = \frac{0.0628 \, \text{m}}{2 \times 3.14} = \frac{0.0628}{6.28} \approx 0.01 \, \text{m} \]

### Step 3: Analyze the forces acting on the thread
When the soap film is broken, the tension \( T \) in the thread will provide the necessary force to balance the surface tension acting along the arc of the circle.

### Step 4: Set up the equation for tension
The surface tension acts along the length of the arc. For a small angle \( d\theta \), the inward force due to tension can be expressed as:
\[ \text{Inward force} = 2T \sin\left(\frac{d\theta}{2}\right) \approx 2T \cdot \frac{d\theta}{2} = T d\theta \]

The force due to surface tension acting along the arc length is:
\[ \text{Surface tension force} = S \cdot \text{arc length} = S \cdot (2r \cdot d\theta) \]

Setting these two forces equal gives:
\[ T d\theta = S \cdot (2r \cdot d\theta) \]

### Step 5: Simplify the equation
Since \( d\theta \) is common on both sides, we can cancel it out:
\[ T = S \cdot 2r \]

### Step 6: Substitute the known values
Now substituting the values of surface tension \( S \) and radius \( r \):
\[ T = 0.030 \, \text{N/m} \cdot 2 \cdot 0.01 \, \text{m} \]
\[ T = 0.030 \cdot 0.02 = 0.0006 \, \text{N} \]

### Step 7: Final answer
Thus, the tension in the thread is:
\[ T = 6 \times 10^{-4} \, \text{N} \]