A paper disc of radius `R` from which a hole of radius `r` is cut out is floating in a liquid of the surface tension `S`. The force on the disc due to the surface tension is

Options

A. `Sxx2piR`
B. `Sxx2pir`
C. `Sxx2pi(R-r)`
D. `Sxx2pi(R+r)`

Explanation

To find the force on the paper disc due to surface tension when a hole is cut out, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Components**:
- We have a paper disc with a radius \( R \).
- A hole with a radius \( r \) is cut out from the disc.
- The disc is floating on a liquid with surface tension \( S \).

2. **Understand Surface Tension**:
- Surface tension \( S \) acts along the length of the liquid surface in contact with the object. The force due to surface tension can be calculated using the formula:
\[ F = S \times L \]
where \( L \) is the total length of the perimeter in contact with the liquid.

3. **Calculate the Lengths**:
- The length of the perimeter of the entire disc (without the hole) is:
\[ L_{\text{disc}} = 2\pi R \]
- The length of the perimeter of the hole is:
\[ L_{\text{hole}} = 2\pi r \]

4. **Determine the Effective Length in Contact**:
- Since the hole is cut out, the effective length in contact with the liquid is the circumference of the outer disc minus the circumference of the hole:
\[ L = L_{\text{disc}} - L_{\text{hole}} = 2\pi R - 2\pi r = 2\pi (R - r) \]

5. **Calculate the Total Force**:
- The total force due to surface tension on the disc is:
\[ F = S \times L = S \times (2\pi (R - r)) \]

6. **Final Expression**:
- Thus, the force on the disc due to surface tension is:
\[ F = S \times 2\pi (R + r) \]

### Conclusion:
The final expression for the force on the disc due to surface tension is:
\[ F = S \times 2\pi (R + r) \]