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

Options

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

Explanation

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

### Step 1: Understand the Problem
We have a uniform disk of radius \( R \) and we cut out a smaller disk of diameter \( R \) (which means its radius is \( \frac{R}{2} \)). We need to find the center of mass of the remaining part.

### Step 2: Calculate the Mass of the Original Disk
The mass of the original disk can be expressed as:
\[ M = \sigma \cdot A \]
where \( \sigma \) is the density and \( A \) is the area of the disk. The area of the original disk is:
\[ A = \pi R^2 \]
Thus, the mass of the original disk is:
\[ M = \sigma \cdot \pi R^2 \]

### Step 3: Calculate the Mass of the Cut-Out Disk
The area of the cut-out disk (with radius \( \frac{R}{2} \)) is:
\[ A_{cut} = \pi \left(\frac{R}{2}\right)^2 = \pi \frac{R^2}{4} \]
The mass of the cut-out disk is:
\[ M_{cut} = \sigma \cdot A_{cut} = \sigma \cdot \pi \frac{R^2}{4} \]

### Step 4: Determine the Center of Mass of the Cut-Out Disk
The center of mass of the cut-out disk is located at a distance \( \frac{R}{2} \) from the center of the original disk.

### Step 5: Apply the Center of Mass Formula
We can treat the remaining part as a combination of the original disk and a negative mass for the cut-out disk. The center of mass \( x_{cm} \) can be calculated using the formula:
\[ x_{cm} = \frac{M \cdot 0 - M_{cut} \cdot \frac{R}{2}}{M - M_{cut}} \]
Substituting the values we have:
\[ x_{cm} = \frac{M \cdot 0 - \left(\sigma \cdot \pi \frac{R^2}{4}\right) \cdot \frac{R}{2}}{M - M_{cut}} \]

### Step 6: Simplify the Expression
Substituting \( M = \sigma \cdot \pi R^2 \) and \( M_{cut} = \sigma \cdot \pi \frac{R^2}{4} \):
\[ x_{cm} = \frac{0 - \left(\sigma \cdot \pi \frac{R^2}{4}\right) \cdot \frac{R}{2}}{\sigma \cdot \pi R^2 - \sigma \cdot \pi \frac{R^2}{4}} \]
This simplifies to:
\[ x_{cm} = \frac{-\frac{\sigma \cdot \pi R^3}{8}}{\sigma \cdot \pi R^2 \left(1 - \frac{1}{4}\right)} = \frac{-\frac{R}{8}}{\frac{3R^2}{4}} = \frac{-R}{6} \]

### Step 7: Conclusion
The center of mass of the remaining part is located at a distance of \( \frac{R}{6} \) from the center of the original disk, in the opposite direction of the cut-out disk.

### Final Answer:
The center of mass of the remaining part will be \( \frac{R}{6} \) away from the center of the original disk, in the direction opposite to the cut-out disk.

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