A very light, rectangular wire-frame of dimensions `7 cm xx 5 cm` hangs just above the free surface of a liquid of surface tension T, with its plane parallel to the free surface. The wire -frame is just brought in contact with the liquid surface and then, lifted up. If the force required to lift the wire-frame is 3.36 N, then what is the value of T in `(N m^(-1))`?

Answer

To solve the problem, we need to determine the surface tension \( T \) of the liquid based on the force required to lift the wire-frame.

### Step-by-Step Solution:

1. **Identify the dimensions of the wire-frame**:
The dimensions of the wire-frame are given as \( 7 \, \text{cm} \times 5 \, \text{cm} \).

2. **Calculate the perimeter of the wire-frame**:
The perimeter \( P \) of a rectangle is calculated using the formula:
\[ P = 2 \times (\text{length} + \text{width}) \]
Substituting the dimensions:
\[ P = 2 \times (7 \, \text{cm} + 5 \, \text{cm}) = 2 \times 12 \, \text{cm} = 24 \, \text{cm} \]
Converting this to meters:
\[ P = 24 \, \text{cm} = 0.24 \, \text{m} \]

3. **Understand the force due to surface tension**:
When the wire-frame is lifted, the force \( F \) required to lift it is related to the surface tension \( T \) of the liquid. The force due to surface tension acting on the wire-frame is given by:
\[ F = 2 \times T \times P \]
Here, the factor of 2 accounts for the two sides of the wire-frame that are in contact with the liquid.

4. **Substitute the known values**:
We know that the force \( F \) required to lift the wire-frame is \( 3.36 \, \text{N} \). Substituting the values into the equation:
\[ 3.36 = 2 \times T \times 0.24 \]

5. **Solve for \( T \)**:
Rearranging the equation to solve for \( T \):
\[ T = \frac{3.36}{2 \times 0.24} \]
Simplifying the denominator:
\[ T = \frac{3.36}{0.48} \]
Performing the division:
\[ T = 7 \, \text{N/m} \]

### Final Answer:
The value of the surface tension \( T \) is \( 7 \, \text{N/m} \).

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