The force of surface tension on a ring situated on the surface of water is ` 14 pi xx 10^(-4)` N . Then the diameter of the ring is

Options

A. 5 mm
B. 1 cm
C. 2 cm
D. 4 cm

Explanation

To solve the problem, we need to find the diameter of a ring placed on the surface of water given the force of surface tension acting on it. The force of surface tension is given as \( F = 14 \pi \times 10^{-4} \) N.

### Step-by-Step Solution:

1. **Understanding the Force of Surface Tension**:
The force of surface tension \( F \) acting on the ring can be expressed as:
\[ F = T \times L \]
where \( T \) is the surface tension of water and \( L \) is the total length in contact with the water.

2. **Calculating Total Length in Contact**:
Since the ring is thin, the total length in contact with the water is the circumference of the ring on both the outer and inner surfaces. Therefore, the total length \( L \) is:
\[ L = 2 \times \text{Circumference} = 2 \times (2 \pi r) = 4 \pi r \]

3. **Substituting into the Force Equation**:
Substituting \( L \) into the force equation gives:
\[ F = T \times 4 \pi r \]

4. **Rearranging for Radius**:
We can rearrange this equation to solve for the radius \( r \):
\[ r = \frac{F}{4 \pi T} \]

5. **Substituting the Given Values**:
The force \( F \) is given as \( 14 \pi \times 10^{-4} \) N. We need the value of surface tension \( T \) for water, which is typically around \( 70 \times 10^{-3} \) N/m. Substituting these values into the equation:
\[ r = \frac{14 \pi \times 10^{-4}}{4 \pi \times 70 \times 10^{-3}} \]

6. **Simplifying the Equation**:
The \( \pi \) cancels out:
\[ r = \frac{14 \times 10^{-4}}{4 \times 70 \times 10^{-3}} = \frac{14 \times 10^{-4}}{280 \times 10^{-3}} = \frac{14}{280} \times 10^{-1} = \frac{1}{20} \times 10^{-1} \]
\[ r = \frac{1}{200} \text{ m} = 0.005 \text{ m} \]

7. **Calculating the Diameter**:
The diameter \( d \) is twice the radius:
\[ d = 2r = 2 \times 0.005 = 0.01 \text{ m} = 1 \text{ cm} \]

### Final Answer:
The diameter of the ring is \( 1 \text{ cm} \).