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 a paper disc with a hole cut out of it, floating in a liquid with surface tension \( S \), we will follow these steps:

### Step 1: Understand the Geometry of the Problem
The paper disc has a radius \( R \) and a hole with a radius \( r \) cut out from it. The effective area of the disc that is in contact with the liquid is the circumference of the outer circle minus the circumference of the inner circle (the hole).

### Step 2: Calculate the Length in Contact with the Liquid
The length of the circumference of the outer circle (the disc) is given by:
\[ L_{\text{outer}} = 2\pi R \]
The length of the circumference of the inner circle (the hole) is given by:
\[ L_{\text{inner}} = 2\pi r \]
The effective length in contact with the liquid is:
\[ L_{\text{effective}} = L_{\text{outer}} - L_{\text{inner}} = 2\pi R - 2\pi r = 2\pi (R - r) \]

### Step 3: Apply the Formula for Surface Tension Force
The force due to surface tension \( F \) is given by the product of surface tension \( S \) and the length in contact with the liquid:
\[ F = S \cdot L_{\text{effective}} \]
Substituting the effective length we found:
\[ F = S \cdot 2\pi (R - r) \]

### Step 4: Final Expression for the Force
Thus, the force on the disc due to the surface tension is:
\[ F = 2\pi S (R - r) \]