The ring of radius 1 m is lying on the surface of liquid. It is lifted. It is lifted from the liquid surface by a force of 4 newtons in such a way that the liquid film in it remains intact. The surface tension of liquid will be

Options

A. `(1)/(2pi)N//m`
B. `(1)/(pi)N//m`
C. `(1)/(3pi)N//m`
D. `(1)/(4pi)N//m`

Answer

To solve the problem, we need to find the surface tension of the liquid when a ring of radius 1 meter is lifted from the surface of the liquid by applying a force of 4 Newtons.

### Step-by-Step Solution:

1. **Identify the Given Data**:
- Radius of the ring, \( r = 1 \, \text{m} \)
- Force applied, \( F = 4 \, \text{N} \)

2. **Calculate the Area of the Ring**:
- The area of the ring in contact with the liquid can be considered as the circumference of the ring multiplied by 2 (as there are two sides of the ring in contact with the liquid).
- The circumference of a circle is given by the formula \( C = 2 \pi r \).
- Therefore, the area \( A \) of the ring is:
\[ A = 2 \times (2 \pi r) = 4 \pi r \]
- Substituting the value of \( r \):
\[ A = 4 \pi (1) = 4 \pi \, \text{m}^2 \]

3. **Relate Force to Surface Tension**:
- The force exerted to lift the ring is equal to the surface tension \( T \) multiplied by the area \( A \):
\[ F = T \times A \]

4. **Rearranging the Formula**:
- To find the surface tension \( T \), we rearrange the formula:
\[ T = \frac{F}{A} \]

5. **Substituting the Values**:
- Now substituting the known values of force and area:
\[ T = \frac{4 \, \text{N}}{4 \pi \, \text{m}^2} \]
- Simplifying this gives:
\[ T = \frac{1}{\pi} \, \text{N/m} \]

6. **Final Answer**:
- The surface tension of the liquid is:
\[ T = \frac{1}{\pi} \, \text{N/m} \]

### Conclusion:
The surface tension of the liquid is \( \frac{1}{\pi} \, \text{N/m} \).

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