A concentric hole of radius R/2 is cut from a thin circular plate of mass M and radius R. The moment of inertia of the remaining plate about its axis will be

Options

A. (a)`13/24MR^(2)`
B. (b)`11/24MR^(2)`
C. (c)`13/32MR^(2)`
D. (d)`15/32MR^(2)`

Explanation

To find the moment of inertia of a thin circular plate with a concentric hole cut out, we can follow these steps:

### Step 1: Define the mass and radius of the original plate
Let the mass of the original circular plate be \( M \) and the radius be \( R \).

### Step 2: Calculate the mass of the hole
The radius of the hole is \( \frac{R}{2} \). The area of the original plate is \( \pi R^2 \) and the area of the hole is \( \pi \left(\frac{R}{2}\right)^2 = \frac{\pi R^2}{4} \).

The mass of the hole can be calculated using the ratio of the areas, assuming uniform density:
\[ \text{Mass of the hole} = M \times \frac{\text{Area of the hole}}{\text{Area of the original plate}} = M \times \frac{\frac{\pi R^2}{4}}{\pi R^2} = M \times \frac{1}{4} = \frac{M}{4} \]

### Step 3: Calculate the mass of the remaining plate
The mass of the remaining plate after the hole is cut out is:
\[ \text{Mass of remaining plate} = M - \frac{M}{4} = \frac{3M}{4} \]

### Step 4: Calculate the moment of inertia of the original plate
The moment of inertia \( I \) of a thin circular plate about its central axis is given by:
\[ I_{\text{original}} = \frac{1}{2} M R^2 \]

### Step 5: Calculate the moment of inertia of the hole
The moment of inertia of the hole about the same axis can be calculated as:
\[ I_{\text{hole}} = \frac{1}{2} \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2 = \frac{1}{2} \cdot \frac{M}{4} \cdot \frac{R^2}{4} = \frac{M R^2}{32} \]

### Step 6: Use the parallel axis theorem
Since the hole is concentric, we do not need to use the parallel axis theorem. The moment of inertia of the remaining plate is simply the moment of inertia of the original plate minus the moment of inertia of the hole:
\[ I_{\text{remaining}} = I_{\text{original}} - I_{\text{hole}} = \frac{1}{2} M R^2 - \frac{M R^2}{32} \]

### Step 7: Simplify the expression
To simplify:
\[ I_{\text{remaining}} = \frac{16M R^2}{32} - \frac{M R^2}{32} = \frac{(16M - M) R^2}{32} = \frac{15M R^2}{32} \]

### Conclusion
The moment of inertia of the remaining plate about its axis is:
\[ I_{\text{remaining}} = \frac{15M R^2}{32} \]