Two rings of same radius and mass are placed such that their centres are at a common point and their planes are perpendicular to each other. The moment of inertia of the system about an axis passing through the centre and perpendicular to the plane of one of the rings is (mass the ring `= m`, radius`= r`)

Options

A. `(1)/(2)mr^(2)`
B. `mr^(2)`
C. (3)/(2)mr^(2)`
D. 2mr^(2)`

Answer

To find the moment of inertia of the system consisting of two rings placed with their centers at a common point and their planes perpendicular to each other, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Rings and Their Orientation**:
- Let’s denote the two rings as Ring A and Ring B.
- Both rings have the same mass \( m \) and radius \( r \).
- The planes of the rings are perpendicular to each other, and they share a common center.

2. **Moment of Inertia of Ring A**:
- The moment of inertia of a ring about an axis passing through its center and perpendicular to its plane is given by:
\[ I_A = m r^2 \]
- Here, \( I_A \) is the moment of inertia of Ring A.

3. **Moment of Inertia of Ring B**:
- For Ring B, the axis of rotation passes through its diameter (which is in the plane of the ring). The moment of inertia about this axis is given by:
\[ I_B = \frac{1}{2} m r^2 \]
- Here, \( I_B \) is the moment of inertia of Ring B.

4. **Total Moment of Inertia of the System**:
- The total moment of inertia \( I \) of the system is the sum of the moments of inertia of both rings:
\[ I = I_A + I_B \]
- Substituting the values we found:
\[ I = m r^2 + \frac{1}{2} m r^2 \]

5. **Simplifying the Expression**:
- Factor out \( m r^2 \):
\[ I = m r^2 \left(1 + \frac{1}{2}\right) \]
- This simplifies to:
\[ I = m r^2 \left(\frac{3}{2}\right) \]

6. **Final Result**:
- Therefore, the total moment of inertia of the system is:
\[ I = \frac{3}{2} m r^2 \]

### Final Answer:
The moment of inertia of the system about the specified axis is \( \frac{3}{2} m r^2 \).