Two rings of same mass and radius R are placed with their planes perpendicular to each other and centre at a common point. The radius of gyration of the system about an axis passing through the centre and perpendicular to the plane of one ring is

Options

A. 2R
B. `(R )/(sqrt(2))`
C. `sqrt((3)/(2))R`
D. `(sqrt(3)R)/(2)`

Answer

To find the radius of gyration of the system of two rings placed with their planes perpendicular to each other, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have two rings of the same mass \( m \) and radius \( R \).
- The rings are positioned such that their planes are perpendicular to each other, with their centers coinciding at a common point.

2. **Identifying the Axis of Rotation**:
- We need to find the radius of gyration about an axis that passes through the center and is perpendicular to the plane of one of the rings.

3. **Moment of Inertia of Each Ring**:
- For a ring, the moment of inertia about an axis perpendicular to its plane and through its center is given by:
\[ I = mR^2 \]
- However, since we are considering the axis perpendicular to the plane of one ring, we need to use the perpendicular axis theorem for the second ring.

4. **Applying the Perpendicular Axis Theorem**:
- According to the perpendicular axis theorem, for a planar object:
\[ I_z = I_x + I_y \]
- Here, \( I_z \) is the moment of inertia about the axis perpendicular to the plane of the ring, and \( I_x \) and \( I_y \) are the moments of inertia about the two axes in the plane of the ring.
- For the second ring (which is in the plane of the first ring), we can express its moment of inertia about the axis perpendicular to its plane as:
\[ I_{ring2} = I_{ring2,x} + I_{ring2,y} = mR^2 + mR^2 = 2mR^2 \]

5. **Total Moment of Inertia of the System**:
- The total moment of inertia \( I_{total} \) about the chosen axis is the sum of the moment of inertia of both rings:
\[ I_{total} = I_{ring1} + I_{ring2} = mR^2 + 2mR^2 = 3mR^2 \]

6. **Finding the Radius of Gyration**:
- The radius of gyration \( k \) is defined by the equation:
\[ I_{total} = mk^2 \]
- Substituting the total moment of inertia:
\[ 3mR^2 = mk^2 \]
- Dividing both sides by \( m \):
\[ k^2 = 3R^2 \]
- Taking the square root gives:
\[ k = \sqrt{3}R \]

7. **Final Result**:
- The radius of gyration of the system about the specified axis is:
\[ k = \frac{\sqrt{3}}{2}R \]

### Conclusion:
The radius of gyration of the system about the axis passing through the center and perpendicular to the plane of one ring is \( \frac{\sqrt{3}}{2}R \).