A uniform metal disc of radius R is taken and out of it a disc of diameter R is cut-off from the end. The center of mass of the remaining part will be

Options

A. `R/4` from the center
B. `R/3` from the center
C. `R/5` from the center
D. `R/6` from the center.

Explanation

To find the center of mass of the remaining part of a uniform metal disc after cutting out a smaller disc, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry**:
- We have a larger disc of radius \( R \).
- We cut out a smaller disc of diameter \( R \), which means the radius of the smaller disc is \( \frac{R}{2} \).

2. **Calculate the Areas**:
- The area of the larger disc \( A_1 \) is given by:
\[ A_1 = \pi R^2 \]
- The area of the smaller disc \( A_2 \) is:
\[ A_2 = \pi \left(\frac{R}{2}\right)^2 = \pi \frac{R^2}{4} \]

3. **Determine the Position of the Centers**:
- The center of mass of the larger disc is at its center, which we can consider as the origin (0, 0).
- The center of the smaller disc, which is cut from the edge of the larger disc, is at the point \( (R, 0) \).

4. **Use the Center of Mass Formula**:
- The center of mass \( x_{cm} \) of the remaining part can be calculated using the formula:
\[ x_{cm} = \frac{A_1 x_1 - A_2 x_2}{A_1 - A_2} \]
- Here, \( x_1 = 0 \) (center of the larger disc), and \( x_2 = R \) (center of the smaller disc).

5. **Substitute the Values**:
- Substitute the areas and positions into the formula:
\[ x_{cm} = \frac{\pi R^2 \cdot 0 - \pi \frac{R^2}{4} \cdot R}{\pi R^2 - \pi \frac{R^2}{4}} \]
- Simplifying this gives:
\[ x_{cm} = \frac{-\frac{\pi R^3}{4}}{\pi R^2 - \frac{\pi R^2}{4}} = \frac{-\frac{R^3}{4}}{R^2 \left(1 - \frac{1}{4}\right)} = \frac{-\frac{R^3}{4}}{\frac{3R^2}{4}} = -\frac{R}{3} \]

6. **Final Position of the Center of Mass**:
- Since we are measuring from the center of the larger disc, the center of mass of the remaining part is at:
\[ x_{cm} = -\frac{R}{3} \]
- This indicates that the center of mass is \( \frac{R}{3} \) to the left of the center of the larger disc.