A rectangular piece of dimension `lxxb` is cut out of central portion of a uniform circular disc of mass m and radius r. The moment of inertia of the remaining piece about an axis perpendicular to the plane of the disc and passing through its centre is :

Options

A. `m[r^(2)-(lb)/(6pir^(2))(l^(2)+b^(2))`
B. `(m)/(2)[r^(2)-(lb)/(6pir^(2))(l^(2)+b^(2))]`
C. `(m)/(2)[r^(2)-((l^(2)+b^(2)))/(6)]`
D. not determinable as mass of the rectangular piece is not given

Explanation

To find the moment of inertia of the remaining piece after cutting a rectangular section from a uniform circular disc, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**:
We have a uniform circular disc of mass \( m \) and radius \( r \). A rectangular piece of dimensions \( L \times B \) is cut out from the center of the disc. We need to find the moment of inertia of the remaining portion about an axis perpendicular to the plane of the disc and passing through its center.

2. **Calculate the Moment of Inertia of the Original Disc**:
The moment of inertia \( I \) of a uniform circular disc about an axis perpendicular to its plane and through its center is given by:
\[ I_{\text{disc}} = \frac{1}{2} m r^2 \]

3. **Calculate the Mass of the Cut-Out Rectangle**:
To find the mass of the rectangle, we first need to determine the mass per unit area of the disc. The area \( A \) of the disc is:
\[ A = \pi r^2 \]
Thus, the mass per unit area \( \sigma \) is:
\[ \sigma = \frac{m}{\pi r^2} \]
The area of the rectangle is:
\[ A_{\text{rectangle}} = L \times B \]
Therefore, the mass \( M \) of the rectangle is:
\[ M = \sigma \times A_{\text{rectangle}} = \frac{m}{\pi r^2} \times (L \times B) = \frac{m L B}{\pi r^2} \]

4. **Calculate the Moment of Inertia of the Cut-Out Rectangle**:
The moment of inertia of the rectangle about the same axis (using the parallel axis theorem) is:
\[ I_{\text{rectangle}} = \frac{1}{12} M (L^2 + B^2) \]
Substituting \( M \):
\[ I_{\text{rectangle}} = \frac{1}{12} \left(\frac{m L B}{\pi r^2}\right) (L^2 + B^2) \]

5. **Calculate the Moment of Inertia of the Remaining Piece**:
The moment of inertia of the remaining piece is the moment of inertia of the original disc minus the moment of inertia of the rectangle:
\[ I_{\text{remaining}} = I_{\text{disc}} - I_{\text{rectangle}} \]
Substituting the values:
\[ I_{\text{remaining}} = \frac{1}{2} m r^2 - \frac{1}{12} \left(\frac{m L B}{\pi r^2}\right) (L^2 + B^2) \]

6. **Final Expression**:
The final expression for the moment of inertia of the remaining piece is:
\[ I_{\text{remaining}} = \frac{1}{2} m r^2 - \frac{m L B (L^2 + B^2)}{12 \pi r^2} \]