Moment of inertia of a thin circular plate of mass `M`, radius `R` about an axis passing through its diameter is `I`. The moment of inertia of a circular ring of mass `M`, radius `R` about an axis perpendicular to its plane and passing through its centre is

Options

A. `2I`
B. `I/2`
C. `4I`
D. `I/4`

Explanation

To find the moment of inertia of a circular ring of mass \( M \) and radius \( R \) about an axis perpendicular to its plane and passing through its center, we can follow these steps:

### Step 1: Understand the Moment of Inertia of the Circular Plate
The moment of inertia \( I \) of a thin circular plate of mass \( M \) and radius \( R \) about an axis passing through its diameter is given by the formula:
\[ I = \frac{1}{4} M R^2 \]

### Step 2: Identify the Moment of Inertia of the Circular Ring
The moment of inertia \( I' \) of a circular ring of mass \( M \) and radius \( R \) about an axis perpendicular to its plane and passing through its center is given by the formula:
\[ I' = M R^2 \]

### Step 3: Relate the Moment of Inertia of the Ring to the Plate
From the above two equations, we can relate the moment of inertia of the circular ring to that of the circular plate. We know:
\[ I = \frac{1}{4} M R^2 \]
Thus, we can express \( M R^2 \) in terms of \( I \):
\[ M R^2 = 4I \]

### Step 4: Substitute to Find \( I' \)
Now, substituting \( M R^2 \) into the equation for \( I' \):
\[ I' = M R^2 = 4I \]

### Conclusion
Thus, the moment of inertia of the circular ring about the specified axis is:
\[ I' = 4I \]

### Final Answer
The moment of inertia of the circular ring is \( 4I \).

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