A circular disc of radius R is removed from a bigger circular dise of radius such that the circumferences of the discs coincide. The centre of mass of the new disc is `alphaR` from the centre of the bigger disc. The value of a is:

Options

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

Explanation

To solve the problem, we need to find the value of \( \alpha \) when a circular disc of radius \( r \) is removed from a larger circular disc of radius \( R \) such that their circumferences coincide.

### Step-by-Step Solution:

1. **Understand the Geometry**:
- The circumference of the smaller disc is equal to the circumference of the larger disc. Therefore, we have:
\[ 2\pi r = 2\pi R \]
- This implies that the radius of the larger disc \( R \) is equal to the diameter of the smaller disc \( r \):
\[ R = 2r \]

2. **Determine the Areas**:
- The area of the smaller disc \( A_1 \) is:
\[ A_1 = \pi r^2 \]
- The area of the larger disc \( A_2 \) is:
\[ A_2 = \pi R^2 = \pi (2r)^2 = 4\pi r^2 \]
- The area of the remaining part after removing the smaller disc is:
\[ A_{remaining} = A_2 - A_1 = 4\pi r^2 - \pi r^2 = 3\pi r^2 \]

3. **Calculate the Masses**:
- Assuming uniform density, the mass of the smaller disc \( m_1 \) is proportional to its area:
\[ m_1 = k \cdot A_1 = k \cdot \pi r^2 \]
- The mass of the larger disc \( m_2 \) is:
\[ m_2 = k \cdot A_2 = k \cdot 4\pi r^2 \]
- The mass of the remaining part after removing the smaller disc is:
\[ m_{remaining} = m_2 - m_1 = k \cdot 4\pi r^2 - k \cdot \pi r^2 = 3k \cdot \pi r^2 \]

4. **Set Up the Center of Mass Equation**:
- Let the center of mass of the remaining disc be at a distance \( \alpha R \) from the center of the larger disc.
- The center of mass of the smaller disc is at a distance \( R = 2r \) from the center of the larger disc.
- Using the formula for the center of mass:
\[ x_{cm} = \frac{m_1 \cdot x_1 + m_{remaining} \cdot x_{remaining}}{m_1 + m_{remaining}} \]
- Here, \( x_1 = R = 2r \) and \( x_{remaining} = 0 \) (as it is at the center of the larger disc):
\[ x_{cm} = \frac{(k \cdot \pi r^2)(2r) + (3k \cdot \pi r^2)(0)}{k \cdot \pi r^2 + 3k \cdot \pi r^2} \]
- Simplifying this gives:
\[ x_{cm} = \frac{2kr^3\pi}{4kr^2\pi} = \frac{2r}{4} = \frac{r}{2} \]

5. **Relate to \( \alpha R \)**:
- Since \( R = 2r \), we can express \( \frac{r}{2} \) in terms of \( R \):
\[ x_{cm} = \frac{R}{4} \]
- Therefore, we have:
\[ \alpha R = \frac{R}{4} \implies \alpha = \frac{1}{4} \]

### Final Answer:
Thus, the value of \( \alpha \) is:
\[ \alpha = \frac{1}{3} \]