A disc of paper of radius R is floating on the surface of water of surface tension T . If r=20 cm and T=0.070N//m , then the force of surface tension on the disc is

Options

A. `2.2xx10^(-2)N`
B. `4.4xx10^(-2)N`
C. `8.8xx10^(-2)N`
D. `44xx10^(-2)`N

Answer

To find the force of surface tension acting on a disc of paper floating on the surface of water, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Radius of the disc, \( r = 20 \, \text{cm} = 0.20 \, \text{m} \)
- Surface tension, \( T = 0.070 \, \text{N/m} \)

2. **Calculate the circumference of the disc:**
The circumference \( C \) of the disc can be calculated using the formula:
\[ C = 2\pi r \]
Substituting the value of \( r \):
\[ C = 2 \pi (0.20) \, \text{m} = 0.40 \pi \, \text{m} \]

3. **Calculate the force due to surface tension:**
The force \( F \) due to surface tension acting on the perimeter of the disc is given by:
\[ F = T \times C \]
Substituting the values of \( T \) and \( C \):
\[ F = 0.070 \, \text{N/m} \times (0.40 \pi) \, \text{m} \]

4. **Calculate the numerical value:**
Using \( \pi \approx 3.14 \):
\[ F = 0.070 \times 0.40 \times 3.14 \]
\[ F = 0.070 \times 1.256 \approx 0.088 \, \text{N} \]

5. **Convert to scientific notation:**
\[ F \approx 8.8 \times 10^{-2} \, \text{N} \]

Thus, the force of surface tension on the disc is approximately \( 8.8 \times 10^{-2} \, \text{N} \).