Given a uniform disc of mass `M` and radius `R`. A small disc of radius `R//2` is cut from this disc in such a way that the distance between the centres of the two discs is `R//2`. Find the moment of inertia of the remaining disc about a diameter of the original disc perpendicular to the line connecting the centres of the two discs

Options

A. `3MR2//32`
B. `5MR2//16`
C. `11MR2//64`
D. none of these

Explanation

To find the moment of inertia of the remaining disc after cutting a smaller disc from a larger uniform disc, we can follow these steps:

### Step 1: Determine the mass per unit area of the original disc
The mass per unit area (σ) of the original disc can be calculated using the formula:
\[ \sigma = \frac{M}{\text{Area of the original disc}} = \frac{M}{\pi R^2} \]

### Step 2: Calculate the mass of the smaller disc
The area of the smaller disc (radius \( R/2 \)) is:
\[ \text{Area of smaller disc} = \pi \left(\frac{R}{2}\right)^2 = \frac{\pi R^2}{4} \]
Now, using the mass per unit area, the mass of the smaller disc (\( M' \)) is:
\[ M' = \sigma \times \text{Area of smaller disc} = \frac{M}{\pi R^2} \times \frac{\pi R^2}{4} = \frac{M}{4} \]

### Step 3: Calculate the moment of inertia of the original disc about its diameter
The moment of inertia (\( I_1 \)) of the original disc about an axis through its center and along its diameter is given by:
\[ I_1 = \frac{1}{4} M R^2 \]

### Step 4: Calculate the moment of inertia of the smaller disc about its center
The moment of inertia (\( I_{cm} \)) of the smaller disc about its own center is:
\[ I_{cm} = \frac{1}{4} M' \left(\frac{R}{2}\right)^2 = \frac{1}{4} \left(\frac{M}{4}\right) \left(\frac{R^2}{4}\right) = \frac{MR^2}{64} \]

### Step 5: Use the parallel axis theorem to find the moment of inertia of the smaller disc about the axis of the original disc
The distance \( d \) between the center of the smaller disc and the center of the original disc is \( R/2 \). According to the parallel axis theorem:
\[ I_2 = I_{cm} + M' d^2 = \frac{MR^2}{64} + \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2 \]
Calculating the second term:
\[ \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2 = \frac{M}{4} \cdot \frac{R^2}{4} = \frac{MR^2}{16} \]
Thus, we have:
\[ I_2 = \frac{MR^2}{64} + \frac{MR^2}{16} = \frac{MR^2}{64} + \frac{4MR^2}{64} = \frac{5MR^2}{64} \]

### Step 6: Calculate the moment of inertia of the remaining disc
The moment of inertia of the remaining disc (\( I_{\text{remaining}} \)) is given by:
\[ I_{\text{remaining}} = I_1 - I_2 = \frac{1}{4} MR^2 - \frac{5MR^2}{64} \]
Converting \( \frac{1}{4} MR^2 \) to a fraction with a denominator of 64:
\[ \frac{1}{4} MR^2 = \frac{16MR^2}{64} \]
Now substituting:
\[ I_{\text{remaining}} = \frac{16MR^2}{64} - \frac{5MR^2}{64} = \frac{11MR^2}{64} \]

### Final Answer
The moment of inertia of the remaining disc about the specified axis is:
\[ \boxed{\frac{11MR^2}{64}} \]