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. `(I)/(2)`
B. `2I`
C. `(I)/(4)`
D. `4I`

Explanation

To find the moment of inertia of a ring about an axis passing through its center and perpendicular to its plane, we can use the perpendicular axis theorem. Here’s a step-by-step solution:

### Step 1: Understand the Given Information
We know that the moment of inertia of the ring about an axis passing through its diameter is given as \( I \). This means that if we take an axis along the diameter of the ring, the moment of inertia about that axis is \( I \).

### Step 2: Identify the Axes
For a ring, we can define two axes:
- Axis \( x \): along the diameter of the ring.
- Axis \( y \): another diameter perpendicular to axis \( x \).

Since the ring is symmetrical, the moment of inertia about both axes \( x \) and \( y \) is the same:
- \( I_x = I \)
- \( I_y = I \)

### Step 3: Apply the Perpendicular Axis Theorem
The perpendicular axis theorem states that for a planar object (like a ring), the moment of inertia about an axis perpendicular to the plane of the object (let's call this axis \( z \)) is equal to the sum of the moments of inertia about two perpendicular axes in the plane of the object (axes \( x \) and \( y \)).

Mathematically, this can be expressed as:
\[ I_z = I_x + I_y \]

### Step 4: Substitute the Values
Since we have established that:
- \( I_x = I \)
- \( I_y = I \)

We can substitute these values into the equation:
\[ I_z = I + I = 2I \]

### Step 5: Conclusion
Thus, the moment of inertia of the ring about an axis passing through its center and perpendicular to its plane is:
\[ I_z = 2I \]

### Final Answer
The correct answer is \( 2I \).

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