A circular disc of radius R is removed from a bigger circular disc of radius 2R such that the circumference of the discs coincoid . The centre of mass of the new disc is `alphaR` from the centre of the bigger disc . the value of `alpha ` is

Options

A. `1//3`
B. `1//2`
C. `1//6`
D. `1//4`

Explanation

To solve the problem of finding the value of \( \alpha \) for the center of mass of a new disc formed by removing a smaller disc from a larger disc, we can follow these steps:

### Step-by-Step Solution:

1. **Define the Problem**:
We have a larger circular disc with radius \( 2R \) and a smaller circular disc with radius \( R \). The smaller disc is removed such that their circumferences coincide.

2. **Determine Areas and Masses**:
- The mass per unit area of the discs is denoted as \( \sigma \).
- The area of the larger disc (mass \( M_1 \)) is given by:
\[ M_1 = \sigma \times \text{Area of larger disc} = \sigma \times \pi (2R)^2 = 4\sigma \pi R^2 \]
- The area of the smaller disc (mass \( M_2 \)) is given by:
\[ M_2 = -\sigma \times \text{Area of smaller disc} = -\sigma \times \pi R^2 \]
(We use negative mass to represent the removal of the smaller disc.)

3. **Set Up the Center of Mass Equation**:
The center of mass \( x_{cm} \) of the system can be calculated using the formula:
\[ x_{cm} = \frac{M_1 x_1 + M_2 x_2}{M_1 + M_2} \]
- Here, \( x_1 = 0 \) (the center of the larger disc) and \( x_2 = R \) (the center of the smaller disc).

4. **Substitute Values into the Equation**:
Substitute the values of \( M_1 \), \( M_2 \), \( x_1 \), and \( x_2 \):
\[ x_{cm} = \frac{(4\sigma \pi R^2)(0) + (-\sigma \pi R^2)(R)}{4\sigma \pi R^2 - \sigma \pi R^2} \]
Simplifying this gives:
\[ x_{cm} = \frac{-\sigma \pi R^3}{3\sigma \pi R^2} = -\frac{R}{3} \]

5. **Interpret the Result**:
The negative sign indicates that the center of mass has shifted towards the larger disc, and its distance from the center of the larger disc is \( \frac{R}{3} \).

6. **Express in Terms of \( \alpha \)**:
Since we defined the distance from the center of the larger disc as \( \alpha R \), we have:
\[ \alpha R = \frac{R}{3} \]
Therefore, we can conclude:
\[ \alpha = \frac{1}{3} \]

### Final Answer:
The value of \( \alpha \) is \( \frac{1}{3} \).