From a disc of radius `R and mass M`, a circular hole of diameter `R`, whose rim passes through the centre is cut. What is the moment of inertia of remaining part of the disc about a perependicular axis, passing through the centre ?

Options

A. `13 MR^(2)//32`
B. `11 MR^(2)//32`
C. `9 MR^(2)//32`
D. `15 MR^(2)//32`

Answer

To find the moment of inertia of the remaining part of the disc after cutting a circular hole, we can follow these steps:

### Step 1: Calculate the Moment of Inertia of the Full Disc
The moment of inertia \( I \) of a full disc about an axis perpendicular to its plane and passing through its center is given by the formula:
\[ I_{\text{total}} = \frac{1}{2} M R^2 \]
where \( M \) is the mass of the disc and \( R \) is its radius.

### Step 2: Determine the Mass of the Removed Portion
The diameter of the hole is \( R \), which means the radius of the hole \( r \) is:
\[ r = \frac{R}{2} \]
The area of the full disc is:
\[ A_{\text{disc}} = \pi R^2 \]
The area of the hole is:
\[ A_{\text{hole}} = \pi \left(\frac{R}{2}\right)^2 = \frac{\pi R^2}{4} \]
The mass of the removed portion (the hole) can be calculated using the ratio of areas:
\[ M_{\text{removed}} = M \cdot \frac{A_{\text{hole}}}{A_{\text{disc}}} = M \cdot \frac{\frac{\pi R^2}{4}}{\pi R^2} = \frac{M}{4} \]

### Step 3: Calculate the Moment of Inertia of the Removed Portion
Using the parallel axis theorem, the moment of inertia of the removed portion about the same axis can be calculated as:
\[ I_{\text{removed}} = I_{\text{CM}} + M_{\text{removed}} d^2 \]
where \( I_{\text{CM}} \) is the moment of inertia of the hole about its own center of mass, and \( d \) is the distance from the center of the disc to the center of the hole.

The moment of inertia of the hole about its center is:
\[ I_{\text{CM}} = \frac{1}{2} M_{\text{removed}} r^2 = \frac{1}{2} \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2 = \frac{1}{2} \cdot \frac{M}{4} \cdot \frac{R^2}{4} = \frac{MR^2}{32} \]
The distance \( d \) from the center of the disc to the center of the hole is \( \frac{R}{2} \). Therefore, we have:
\[ I_{\text{removed}} = \frac{MR^2}{32} + \left(\frac{M}{4}\right) \left(\frac{R}{2}\right)^2 = \frac{MR^2}{32} + \frac{M}{4} \cdot \frac{R^2}{4} = \frac{MR^2}{32} + \frac{MR^2}{16} \]
Converting \( \frac{MR^2}{16} \) to a common denominator:
\[ \frac{MR^2}{16} = \frac{2MR^2}{32} \]
Thus,
\[ I_{\text{removed}} = \frac{MR^2}{32} + \frac{2MR^2}{32} = \frac{3MR^2}{32} \]

### Step 4: Calculate the Moment of Inertia of the Remaining Part
Now, we can find the moment of inertia of the remaining part of the disc:
\[ I_{\text{remaining}} = I_{\text{total}} - I_{\text{removed}} = \frac{1}{2} MR^2 - \frac{3MR^2}{32} \]
Finding a common denominator (which is 32):
\[ I_{\text{remaining}} = \frac{16MR^2}{32} - \frac{3MR^2}{32} = \frac{13MR^2}{32} \]

### Final Answer
The moment of inertia of the remaining part of the disc about a perpendicular axis passing through the center is:
\[ \boxed{\frac{13MR^2}{32}} \]