A thin metal ring of internal radius 8 cm and external radius 9 cm is supported horizontally from the pan of a balance so that it comes in contact with water in a glass vessel. If is found that an extrea weight of `7.48g` is required to pull the ring out of water. The surface tension of water is `(g=10m//s^(2))`

Options

A. `80xx10^(-3)N//m`
B. `25xx10^(-3)N//m`
C. `45xx10^(-3)N//m`
D. `70xx10^(-3)N//m`

Explanation

To solve the problem, we need to find the surface tension of water based on the extra weight required to pull the ring out of the water. Here are the steps:

### Step 1: Understand the forces acting on the ring
When the ring is submerged in water, it experiences an upward force due to surface tension. This upward force must balance the weight of the ring plus the extra weight required to pull it out of the water.

### Step 2: Convert the extra weight to Newtons
The extra weight given is 7.48 g. We need to convert this mass into force (weight) using the formula:
\[ \text{Weight} = \text{mass} \times g \]
where \( g = 10 \, \text{m/s}^2 \).

Calculating the weight:
\[ \text{Weight} = 7.48 \, \text{g} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} \times 10 \, \text{m/s}^2 = 7.48 \times 10^{-3} \, \text{kg} \times 10 \, \text{m/s}^2 = 7.48 \times 10^{-2} \, \text{N} \]

### Step 3: Calculate the effective radius for surface tension
The surface tension acts along the circumference of the ring. The effective radius for the calculation of surface tension is the average of the internal and external radii:
\[ r_1 = 8 \, \text{cm} = 0.08 \, \text{m} \]
\[ r_2 = 9 \, \text{cm} = 0.09 \, \text{m} \]
The effective radius \( r \) is:
\[ r = \frac{r_1 + r_2}{2} = \frac{0.08 + 0.09}{2} = 0.085 \, \text{m} \]

### Step 4: Calculate the circumference of the ring
The circumference \( C \) of the ring can be calculated using the average radius:
\[ C = 2 \pi r = 2 \pi \times 0.085 \, \text{m} \]

### Step 5: Relate the weight to surface tension
The upward force due to surface tension \( F_T \) can be expressed as:
\[ F_T = \text{Surface Tension} \times \text{Circumference} \]
Thus, we have:
\[ 7.48 \times 10^{-2} \, \text{N} = T \times C \]

### Step 6: Solve for surface tension \( T \)
Substituting the value of circumference:
\[ T = \frac{7.48 \times 10^{-2}}{C} \]
Calculating \( C \):
\[ C = 2 \pi \times 0.085 \approx 0.534 \, \text{m} \]
So,
\[ T = \frac{7.48 \times 10^{-2}}{0.534} \approx 0.140 \, \text{N/m} \]

### Final Answer
The surface tension of water is approximately:
\[ T \approx 0.140 \, \text{N/m} \text{ or } 70 \times 10^{-3} \, \text{N/m} \]