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 the remaining part of the disc about a perpendicular axis, passing through the centre

Options

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

Explanation

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: Understand the Problem
We have a disc of radius \( R \) and mass \( M \). A circular hole of diameter \( R \) (which means the radius of the hole is \( R/2 \)) is cut from the disc, with the rim of the hole passing through the center of the disc.

### Step 2: 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:
\[ I_{\text{disc}} = \frac{1}{2} M R^2 \]

### Step 3: Calculate the Mass of the Circular Hole
The area of the full disc is:
\[ A_{\text{disc}} = \pi R^2 \]
The area of the hole (which is a circle with radius \( R/2 \)) is:
\[ A_{\text{hole}} = \pi \left(\frac{R}{2}\right)^2 = \frac{\pi R^2}{4} \]
The mass of the hole can be found using the ratio of the areas, since the mass is uniformly distributed:
\[ M_{\text{hole}} = M \cdot \frac{A_{\text{hole}}}{A_{\text{disc}}} = M \cdot \frac{\frac{\pi R^2}{4}}{\pi R^2} = \frac{M}{4} \]

### Step 4: Calculate the Moment of Inertia of the Hole
The moment of inertia of the hole about the same axis (using the formula for a disc) is:
\[ I_{\text{hole}} = \frac{1}{2} M_{\text{hole}} \left(\frac{R}{2}\right)^2 = \frac{1}{2} \cdot \frac{M}{4} \cdot \frac{R^2}{4} = \frac{MR^2}{32} \]

### Step 5: Use the Parallel Axis Theorem
Since the hole is not centered at the axis of the full disc, we need to use the parallel axis theorem to find the moment of inertia of the hole about the center of the disc:
\[ I_{\text{hole, center}} = I_{\text{hole}} + M_{\text{hole}} \cdot d^2 \]
where \( d \) is the distance from the center of the disc to the center of the hole. Since the hole's center is at the edge of the disc, \( d = R/2 \):
\[ I_{\text{hole, center}} = \frac{MR^2}{32} + \frac{M}{4} \cdot \left(\frac{R}{2}\right)^2 \]
Calculating the second term:
\[ \frac{M}{4} \cdot \frac{R^2}{4} = \frac{MR^2}{16} \]
Thus,
\[ I_{\text{hole, center}} = \frac{MR^2}{32} + \frac{MR^2}{16} = \frac{MR^2}{32} + \frac{2MR^2}{32} = \frac{3MR^2}{32} \]

### Step 6: Calculate the Moment of Inertia of the Remaining Part
Now we can find the moment of inertia of the remaining part of the disc:
\[ I_{\text{remaining}} = I_{\text{disc}} - I_{\text{hole, center}} \]
Substituting the values:
\[ I_{\text{remaining}} = \frac{1}{2} MR^2 - \frac{3MR^2}{32} \]
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 part of the disc about the perpendicular axis passing through the center is:
\[ \boxed{\frac{13MR^2}{32}} \]