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. `(13)/( 24) MR^2`
B. `(11)/( 24) MR^2`
C. `(13)/(32) MR^2`
D. `(15)/(32) MR^2`

Explanation

To find the moment of inertia of the remaining plate after a concentric hole of radius \( R/2 \) is cut from a thin circular plate of mass \( M \) and radius \( R \), we can follow these steps:

### Step 1: Calculate the moment of inertia of the complete circular plate
The moment of inertia \( I \) of a solid circular plate about its axis is given by the formula:

\[ I_{\text{complete}} = \frac{1}{2} M R^2 \]

### Step 2: Calculate the area of the complete plate and the hole
The area \( A \) of the complete plate is:

\[ A_{\text{complete}} = \pi R^2 \]

The area \( A \) of the hole (which has radius \( R/2 \)) is:

\[ A_{\text{hole}} = \pi \left(\frac{R}{2}\right)^2 = \pi \frac{R^2}{4} \]

### Step 3: Calculate the area of the remaining portion
The area of the remaining portion after cutting out the hole is:

\[ A_{\text{remaining}} = A_{\text{complete}} - A_{\text{hole}} = \pi R^2 - \pi \frac{R^2}{4} = \pi R^2 \left(1 - \frac{1}{4}\right) = \pi R^2 \cdot \frac{3}{4} = \frac{3\pi R^2}{4} \]

### Step 4: Calculate the mass density of the plate
The mass density \( \rho \) of the plate can be defined as:

\[ \rho = \frac{M}{A_{\text{complete}}} = \frac{M}{\pi R^2} \]

### Step 5: Calculate the mass of the hole
The mass \( M_{\text{hole}} \) of the removed portion (the hole) can be calculated using its area and the mass density:

\[ M_{\text{hole}} = \rho \cdot A_{\text{hole}} = \frac{M}{\pi R^2} \cdot \pi \frac{R^2}{4} = \frac{M}{4} \]

### Step 6: Calculate the mass of the remaining portion
The mass \( M_{\text{remaining}} \) of the remaining portion is:

\[ M_{\text{remaining}} = M - M_{\text{hole}} = M - \frac{M}{4} = \frac{3M}{4} \]

### Step 7: Calculate the moment of inertia of the hole
To find the moment of inertia of the hole about the same axis, we use the parallel axis theorem. The moment of inertia of the hole about its own center is:

\[ I_{\text{hole, center}} = \frac{1}{2} M_{\text{hole}} \left(\frac{R}{2}\right)^2 = \frac{1}{2} \cdot \frac{M}{4} \cdot \frac{R^2}{4} = \frac{MR^2}{32} \]

Now, we need to apply the parallel axis theorem to find the moment of inertia about the main axis:

\[ I_{\text{hole}} = I_{\text{hole, center}} + M_{\text{hole}} \cdot d^2 \]
where \( d = \frac{R}{2} \):

\[ I_{\text{hole}} = \frac{MR^2}{32} + \frac{M}{4} \cdot \left(\frac{R}{2}\right)^2 = \frac{MR^2}{32} + \frac{M}{4} \cdot \frac{R^2}{4} = \frac{MR^2}{32} + \frac{MR^2}{16} \]

Converting \( \frac{MR^2}{16} \) to a common denominator:

\[ \frac{MR^2}{16} = \frac{2MR^2}{32} \]

Thus,

\[ I_{\text{hole}} = \frac{MR^2}{32} + \frac{2MR^2}{32} = \frac{3MR^2}{32} \]

### Step 8: Calculate the moment of inertia of the remaining plate
Now we can find the moment of inertia of the remaining plate:

\[ I_{\text{remaining}} = I_{\text{complete}} - I_{\text{hole}} = \frac{1}{2} M R^2 - \frac{3}{32} M R^2 \]

Finding a common denominator (32):

\[ I_{\text{remaining}} = \frac{16MR^2}{32} - \frac{3MR^2}{32} = \frac{(16 - 3)MR^2}{32} = \frac{13MR^2}{32} \]

### Final Answer
The moment of inertia of the remaining plate about its axis is:

\[ \boxed{\frac{13MR^2}{32}} \]