Radius of gyration of a ring about a transverse axis passing through its centre is

Options

A. `0.5xx` diameter of ring
B. diameter of ring
C. `2xx` diameter of ring
D. `"(diameter of ring)"^(2)`

Answer

To find the radius of gyration of a ring about a transverse axis passing through its center, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**: We need to find the radius of gyration (k) of a ring when it rotates about a transverse axis that passes through its center.

2. **Define the Ring**: Let's denote the mass of the ring as \( m \) and the radius of the ring as \( r \).

3. **Moment of Inertia**: The moment of inertia \( I \) of a ring about an axis passing through its center and perpendicular to its plane (transverse axis) is given by the formula:
\[ I = m r^2 \]

4. **Radius of Gyration**: The radius of gyration \( k \) is related to the moment of inertia by the equation:
\[ I = m k^2 \]

5. **Equating the Two Expressions**: Since we have two expressions for the moment of inertia, we can set them equal to each other:
\[ m r^2 = m k^2 \]

6. **Simplifying the Equation**: We can cancel \( m \) from both sides (assuming \( m \neq 0 \)):
\[ r^2 = k^2 \]

7. **Finding the Radius of Gyration**: Taking the square root of both sides gives us:
\[ k = r \]

8. **Interpreting the Result**: The radius of gyration of the ring about the transverse axis passing through its center is equal to the radius of the ring.

9. **Answering the Question**: If the question asks for the radius of gyration in terms of the diameter of the ring, we can express it as:
\[ k = \frac{1}{2} \times \text{diameter of the ring} \]

### Final Answer:
The radius of gyration of a ring about a transverse axis passing through its center is \( \frac{1}{2} \) times the diameter of the ring.

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