A ring of internal and external diameters `8.5 xx 10^(-2)m` and `8.7 xx 10^(-2)m` is supported horizontally from the pan of a physical balance such that it comes in contact with a liquid. An extra force of 40N is required to pull it away from the liquid . Determine the surface tension of the liquid?

Explanation

To determine the surface tension of the liquid in contact with the ring, we can follow these steps:

### Step 1: Identify the given values
- Internal diameter of the ring, \( D_1 = 8.5 \times 10^{-2} \, \text{m} \)
- External diameter of the ring, \( D_2 = 8.7 \times 10^{-2} \, \text{m} \)
- Force required to pull the ring away from the liquid, \( F = 40 \, \text{N} \)

### Step 2: Calculate the average diameter of the ring
The average diameter \( D \) can be calculated using the formula:
\[ D = \frac{D_1 + D_2}{2} \]
Substituting the values:
\[ D = \frac{8.5 \times 10^{-2} + 8.7 \times 10^{-2}}{2} = \frac{17.2 \times 10^{-2}}{2} = 8.6 \times 10^{-2} \, \text{m} \]

### Step 3: Calculate the radius of the ring
The radius \( r \) is half of the average diameter:
\[ r = \frac{D}{2} = \frac{8.6 \times 10^{-2}}{2} = 4.3 \times 10^{-2} \, \text{m} \]

### Step 4: Use the formula for surface tension
The surface tension \( T \) can be calculated using the formula:
\[ T = \frac{F}{2 \pi r} \]
Substituting the values:
\[ T = \frac{40}{2 \pi (4.3 \times 10^{-2})} \]

### Step 5: Calculate the surface tension
First, calculate \( 2 \pi r \):
\[ 2 \pi r = 2 \times 3.14159 \times 4.3 \times 10^{-2} \approx 0.2707 \, \text{m} \]
Now substituting this back into the surface tension formula:
\[ T = \frac{40}{0.2707} \approx 147.5 \, \text{N/m} \]

### Final Result
Thus, the surface tension of the liquid is approximately:
\[ T \approx 147.5 \, \text{N/m} \]

---