Out of a disc of mass M and radius R a concentric disc of mass m and radius r is removed. The M.I. of the remaining part about the symmetric axis will be :

Options

A. `(M-m)(R+r)^(2)//2`
B. `(M-m)(R^(2)-r^(2))//2`
C. `(M-m)(R-r)^(2)//2`
D. `(M-m)(R^(2)+r^(2))//2`

Answer

To solve the problem of finding the moment of inertia (M.I.) of the remaining part of a disc after removing a concentric disc, we can follow these steps:

### Step 1: Understand the Problem
We have a larger disc of mass \( M \) and radius \( R \). From this disc, we remove a smaller concentric disc of mass \( m \) and radius \( r \). We need to find the moment of inertia of the remaining part about the symmetric axis.

### Step 2: Moment of Inertia of the Whole Disc
The moment of inertia \( I \) of a solid disc about its central axis is given by the formula:
\[ I = \frac{1}{2} M R^2 \]
For the larger disc, this will be:
\[ I_1 = \frac{1}{2} M R^2 \]

### Step 3: Moment of Inertia of the Removed Disc
For the smaller disc that is removed, its moment of inertia about the same axis is:
\[ I_2 = \frac{1}{2} m r^2 \]

### Step 4: Calculate the Moment of Inertia of the Remaining Part
The moment of inertia of the remaining part after removing the smaller disc can be calculated using the principle of subtraction:
\[ I = I_1 - I_2 \]
Substituting the values we found:
\[ I = \frac{1}{2} M R^2 - \frac{1}{2} m r^2 \]

### Step 5: Factor Out Common Terms
We can factor out \( \frac{1}{2} \):
\[ I = \frac{1}{2} \left( M R^2 - m r^2 \right) \]

### Step 6: Relate Masses and Densities
Since all discs are made of the same material, we can relate the masses and areas using density \( \sigma \). The mass of the larger disc is:
\[ M = \sigma \cdot \pi R^2 \]
And for the smaller disc:
\[ m = \sigma \cdot \pi r^2 \]
Thus, the remaining mass after removing the smaller disc is:
\[ M - m = \sigma \cdot \pi (R^2 - r^2) \]

### Step 7: Substitute Back into the Moment of Inertia Equation
Now we can substitute \( M \) and \( m \) back into our equation for \( I \):
\[ I = \frac{1}{2} \left( \sigma \cdot \pi R^2 R^2 - \sigma \cdot \pi r^2 r^2 \right) \]
This simplifies to:
\[ I = \frac{\sigma \cdot \pi}{2} \left( R^4 - r^4 \right) \]

### Final Result
Thus, the moment of inertia of the remaining part about the symmetric axis is:
\[ I = \frac{1}{2} \left( M R^2 - m r^2 \right) \]