A ring is cut from a platinum tube 8.5 cm internal and 8.7 cm external diameter. It is supported horizontally from a pan of a balance so that it comes in contact with the water in a glass vessel. If an extra 3.97 gm-wt. is required to pull it away from water, the surface tension of water is

Options

A. `70.07dy"ne"//cm`
B. `70.80dy"ne"//cm`
C. `63.35dy"ne"//cm`
D. `60dy"ne"//cm`

Answer

To find the surface tension of water based on the given problem, we will follow these steps:

### Step 1: Understand the given data
- Internal diameter of the ring, \( d_1 = 8.5 \, \text{cm} \)
- External diameter of the ring, \( d_2 = 8.7 \, \text{cm} \)
- Additional weight required to pull the ring away from water, \( F = 3.97 \, \text{gm-wt} \)

### Step 2: Convert the weight to Newtons
Since \( 1 \, \text{gm-wt} = 9.81 \times 10^{-3} \, \text{N} \):
\[ F = 3.97 \, \text{gm-wt} = 3.97 \times 9.81 \times 10^{-3} \, \text{N} = 0.0390 \, \text{N} \]

### Step 3: Calculate the circumference of the ring
The surface tension acts along the circumference of the ring. We need to calculate both the internal and external circumferences:
- Internal circumference, \( C_1 = \pi d_1 = \pi \times 8.5 \, \text{cm} = 26.68 \, \text{cm} \)
- External circumference, \( C_2 = \pi d_2 = \pi \times 8.7 \, \text{cm} = 27.36 \, \text{cm} \)

### Step 4: Calculate the total effective circumference
The total effective circumference where the surface tension acts is the sum of the internal and external circumferences:
\[ C_{\text{total}} = C_1 + C_2 = 26.68 \, \text{cm} + 27.36 \, \text{cm} = 54.04 \, \text{cm} \]

### Step 5: Convert the total circumference to meters
\[ C_{\text{total}} = 54.04 \, \text{cm} = 0.5404 \, \text{m} \]

### Step 6: Relate the force due to surface tension to the weight
The force due to surface tension can be expressed as:
\[ F = T \times C_{\text{total}} \]
where \( T \) is the surface tension. Rearranging gives:
\[ T = \frac{F}{C_{\text{total}}} \]

### Step 7: Substitute the values to find surface tension
Substituting the values we have:
\[ T = \frac{0.0390 \, \text{N}}{0.5404 \, \text{m}} \approx 0.0722 \, \text{N/m} \]

### Step 8: Convert to dyn/cm
Since \( 1 \, \text{N/m} = 10^{2} \, \text{dyn/cm} \):
\[ T \approx 0.0722 \, \text{N/m} \times 100 \approx 7.22 \, \text{dyn/cm} \]

### Final Result
The surface tension of water is approximately \( 72.2 \, \text{dyn/cm} \).

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