A circular disc of radius `R` is removed from a bigger circular disc of radius `2 R` such that the circumferences of the discs coincide. The center of mass of new disc is `alpha R` from the center of the bigger disc. The value of `alpha` is

Options

A. (a)`1//2`
B. (b)`1//6`
C. (c)`1//4`
D. (d)`1//3`

Explanation

To solve the problem, we need to find the value of \(\alpha\) given that a circular disc of radius \(R\) is removed from a larger circular disc of radius \(2R\). The centers of both discs coincide, and we need to determine the center of mass of the remaining shape.

### Step-by-Step Solution:

1. **Define the Masses**:
- Let the mass of the larger disc (radius \(2R\)) be \(M_1\).
- The area of the larger disc is \(A_1 = \pi (2R)^2 = 4\pi R^2\).
- Let the mass of the smaller disc (radius \(R\)) be \(M_2\).
- The area of the smaller disc is \(A_2 = \pi R^2\).
- The mass per unit area (density) for both discs is the same, so we can express the masses as:
\[ M_1 = \sigma \cdot 4\pi R^2 \quad \text{and} \quad M_2 = \sigma \cdot \pi R^2 \]
where \(\sigma\) is the mass per unit area.

2. **Determine the Center of Mass**:
- The center of mass of the larger disc is at the origin (0,0).
- The center of mass of the smaller disc (which is removed) is at the origin as well.
- When the smaller disc is removed, we need to find the new center of mass of the remaining shape.

3. **Calculate the Center of Mass of the Remaining Shape**:
- The new center of mass \(x\) can be calculated using the formula:
\[ x_{\text{cm}} = \frac{M_1 \cdot x_1 - M_2 \cdot x_2}{M_1 - M_2} \]
- Here, \(x_1 = 0\) (center of the larger disc) and \(x_2 = 0\) (center of the smaller disc).
- Substituting the values:
\[ x_{\text{cm}} = \frac{M_1 \cdot 0 - M_2 \cdot 0}{M_1 - M_2} = 0 \]
- However, since we are removing the smaller disc, we need to consider the effective shift in the center of mass due to the removal of mass.

4. **Effective Mass Calculation**:
- The effective mass of the remaining shape is \(M_1 - M_2\).
- The center of mass of the remaining shape can be calculated as:
\[ x_{\text{cm}} = \frac{M_1 \cdot 0 - M_2 \cdot R}{M_1 - M_2} \]
- Substituting \(M_1\) and \(M_2\):
\[ x_{\text{cm}} = \frac{0 - \sigma \cdot \pi R^2 \cdot R}{\sigma \cdot 4\pi R^2 - \sigma \cdot \pi R^2} \]
- Simplifying gives:
\[ x_{\text{cm}} = \frac{-\sigma \cdot \pi R^3}{\sigma \cdot 3\pi R^2} = -\frac{R}{3} \]

5. **Relate to \(\alpha\)**:
- We have \(x_{\text{cm}} = -\frac{R}{3}\).
- Since the problem states that the center of mass is \(\alpha R\) from the center of the larger disc, we equate:
\[ \alpha R = -\frac{R}{3} \]
- Thus, \(\alpha = -\frac{1}{3}\).

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