One quarter sector is cut from a uniform circular disc of radius R. The sector has mass M. IT is made to rotate about a line perpendicular to its plane and passing through the centre of the original disc. Its moment of inertia about the axis of rotation is

Options

A. `MR^(2)//2`
B. `MR^(2)//4`
C. `MR^(2)//8`
D. `sqrt(2)MR^(2)`

Explanation

To find the moment of inertia of a quarter sector cut from a uniform circular disc of radius \( R \) and mass \( M \), we can follow these steps:

### Step 1: Determine the mass of the full disc
Since the mass of the quarter sector is given as \( M \), the mass of the complete disc can be calculated as:
\[ \text{Mass of complete disc} = 4 \times M = 4M \]

### Step 2: Calculate the moment of inertia of the complete disc
The moment of inertia \( I \) of a complete disc about an axis perpendicular to its plane and passing through its center is given by the formula:
\[ I_{\text{disc}} = \frac{1}{2} M R^2 \]
Substituting the mass of the complete disc:
\[ I_{\text{disc}} = \frac{1}{2} (4M) R^2 = 2M R^2 \]

### Step 3: Calculate the moment of inertia of the quarter sector
By symmetry, the moment of inertia of the quarter sector can be found by taking one-fourth of the moment of inertia of the complete disc:
\[ I_{\text{quarter sector}} = \frac{1}{4} I_{\text{disc}} = \frac{1}{4} (2M R^2) = \frac{1}{2} M R^2 \]

### Conclusion
The moment of inertia of the quarter sector about the axis of rotation is:
\[ \boxed{\frac{1}{2} M R^2} \]

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