Form a circular disc of radius R and mass 9M , a small disc of mass M and radius R/3 is removed concentrically . The moment of inertia of the remaining disc about an axis perpendicular to the the plane of the disc and passing its centre is :-

Options

A. `MR^(2)`
B. `4MR^(2)`
C. `4/9 MR^(2)`
D. `40/9MR^(2)`

Explanation

To find the moment of inertia of the remaining disc after removing a smaller disc, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Moment of Inertia of the Original Disc:**
The moment of inertia \( I_1 \) of a solid disc about an axis perpendicular to its plane and passing through its center is given by the formula:
\[ I_1 = \frac{1}{2} M R^2 \]
Here, the mass of the original disc is \( 9M \) and its radius is \( R \). Thus,
\[ I_1 = \frac{1}{2} \times 9M \times R^2 = \frac{9}{2} M R^2 \]

2. **Identify the Moment of Inertia of the Removed Disc:**
The moment of inertia \( I_2 \) of the smaller disc that is removed has a mass \( M \) and a radius \( \frac{R}{3} \). Using the same formula for the moment of inertia:
\[ I_2 = \frac{1}{2} m r^2 \]
Substituting \( m = M \) and \( r = \frac{R}{3} \):
\[ I_2 = \frac{1}{2} M \left(\frac{R}{3}\right)^2 = \frac{1}{2} M \frac{R^2}{9} = \frac{M R^2}{18} \]

3. **Calculate the Moment of Inertia of the Remaining Disc:**
The moment of inertia of the remaining disc \( I \) is given by subtracting the moment of inertia of the smaller disc from that of the original disc:
\[ I = I_1 - I_2 \]
Substituting the values we calculated:
\[ I = \frac{9}{2} M R^2 - \frac{M R^2}{18} \]

4. **Finding a Common Denominator:**
To subtract these fractions, we need a common denominator. The least common multiple of 2 and 18 is 18. Thus, we rewrite \( I_1 \):
\[ I_1 = \frac{9}{2} M R^2 = \frac{81}{18} M R^2 \]
Now we can perform the subtraction:
\[ I = \frac{81}{18} M R^2 - \frac{1}{18} M R^2 = \frac{80}{18} M R^2 \]

5. **Simplifying the Result:**
Simplifying \( \frac{80}{18} \):
\[ I = \frac{40}{9} M R^2 \]

### Final Result:
The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through its center is:
\[ I = \frac{40}{9} M R^2 \]