From a circular disc of radius R and 9M , a small disc of mass M and radius `(R )/(3)` is removed concentrically .The moment of inertia of the remaining disc about and axis perpendicular to the plane of the disc and passing through its centre is

Options

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

Explanation

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

### Step 1: Understand the Problem
We have a larger disc of radius \( R \) and mass \( 9M \). A smaller disc of radius \( \frac{R}{3} \) and mass \( M \) is removed from the center of the larger disc. We need to calculate the moment of inertia of the remaining part of the disc about an axis perpendicular to the plane of the disc and passing through its center.

### Step 2: Calculate the Moment of Inertia of the Larger 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:
\[ I = \frac{1}{2} M R^2 \]
For the larger disc:
\[ I_1 = \frac{1}{2} \times 9M \times R^2 = \frac{9MR^2}{2} \]

### Step 3: Calculate the Moment of Inertia of the Smaller Disc
The moment of inertia \( I_2 \) of the smaller disc (which we will treat as a negative mass since it is removed) is calculated using the same formula, but with its radius \( \frac{R}{3} \) and mass \( M \):
\[ I_2 = \frac{1}{2} M \left(\frac{R}{3}\right)^2 = \frac{1}{2} M \frac{R^2}{9} = \frac{MR^2}{18} \]

### Step 4: Adjust for the Removal of the Smaller Disc
Since we are removing the smaller disc, we treat its moment of inertia as negative:
\[ I_{\text{remaining}} = I_1 - I_2 = \frac{9MR^2}{2} - \frac{MR^2}{18} \]

### Step 5: Find a Common Denominator and Simplify
To combine these fractions, we need a common denominator. The least common multiple of 2 and 18 is 18:
\[ I_{\text{remaining}} = \frac{9MR^2 \cdot 9}{18} - \frac{MR^2}{18} = \frac{81MR^2}{18} - \frac{MR^2}{18} = \frac{80MR^2}{18} \]

### Step 6: Simplify the Result
Now we can simplify \( \frac{80MR^2}{18} \):
\[ I_{\text{remaining}} = \frac{40MR^2}{9} \]

### Final Answer
The moment of inertia of the remaining disc about the specified axis is:
\[ \boxed{\frac{40MR^2}{9}} \]

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