The moment of inertia of ring about an axis passing through its diameter is `I`. Then moment of inertia of that ring about an axis passing through its centre and perpendicular to its plane is

Options

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

Explanation

To solve the problem, we need to find the moment of inertia of a ring about an axis that passes through its center and is perpendicular to its plane, given that the moment of inertia about an axis passing through its diameter is \( I \).

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- The moment of inertia of the ring about an axis passing through its diameter is given as \( I \).
- We need to find the moment of inertia about an axis that is perpendicular to the plane of the ring and passes through its center.

2. **Applying the Perpendicular Axis Theorem**:
- The Perpendicular Axis Theorem states that for a planar object, the moment of inertia about an axis perpendicular to the plane (let's call it \( I_{zz} \)) is equal to the sum of the moments of inertia about two perpendicular axes in the plane of the object (let's call them \( I_{xx} \) and \( I_{yy} \)).
- Mathematically, this can be expressed as:
\[ I_{zz} = I_{xx} + I_{yy} \]

3. **Identifying the Axes**:
- In our case, the axes \( I_{xx} \) and \( I_{yy} \) are the axes passing through the diameter of the ring.
- Since the ring is symmetric, we can say:
\[ I_{xx} = I_{yy} = I \]

4. **Substituting into the Perpendicular Axis Theorem**:
- Now substituting \( I_{xx} \) and \( I_{yy} \) into the equation from the Perpendicular Axis Theorem:
\[ I_{zz} = I + I = 2I \]

5. **Conclusion**:
- Therefore, the moment of inertia of the ring about an axis passing through its center and perpendicular to its plane is:
\[ I_{zz} = 2I \]

### Final Answer:
The moment of inertia of the ring about an axis passing through its center and perpendicular to its plane is \( 2I \).