A thin wire of length `L` and uniform linear mass density `rho` is bent into a circular loop with centre at O as shown. The moment of inertia of the loop about the axis `XX'` is

Options

A. `(piL^(3))/(8pi^(2))`
B. `(piL^(3))/(16pi^(2))`
C. `(5piL^(3))/(16pi^(2))`
D. `(3piL^(3))/(8pi^(2))`

Explanation

To find the moment of inertia of a thin wire bent into a circular loop about a specified axis, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have a thin wire of length \( L \) and uniform linear mass density \( \rho \) bent into a circular loop. We need to find the moment of inertia of this loop about the axis \( XX' \).

2. **Identify the Moment of Inertia Formula**: The moment of inertia \( I \) of a ring about an axis through its center and perpendicular to its plane is given by:
\[ I_{cm} = m r^2 \]
where \( m \) is the mass of the ring and \( r \) is the radius.

3. **Use the Parallel Axis Theorem**: The moment of inertia about an axis parallel to the axis through the center of mass (CM) can be calculated using:
\[ I = I_{cm} + md^2 \]
where \( d \) is the distance from the center of mass axis to the new axis. In this case, since the axis \( XX' \) is in the plane of the loop and passing through the tangent, we will use:
\[ I = I_{cm} + m \left( \frac{L}{2\pi} \right)^2 \]

4. **Calculate Mass \( m \)**: The mass of the wire can be calculated using its linear mass density:
\[ m = \rho L \]

5. **Calculate the Radius \( r \)**: The radius of the circular loop can be found from the relationship between the length of the wire and the circumference of the circle:
\[ L = 2\pi r \implies r = \frac{L}{2\pi} \]

6. **Substitute Values into the Moment of Inertia Formula**:
- First, calculate \( I_{cm} \):
\[ I_{cm} = m r^2 = (\rho L) \left( \frac{L}{2\pi} \right)^2 = \frac{\rho L^3}{4\pi^2} \]
- Now, substitute \( I_{cm} \) and \( m \) into the moment of inertia formula:
\[ I = I_{cm} + m \left( \frac{L}{2\pi} \right)^2 \]
\[ I = \frac{\rho L^3}{4\pi^2} + \rho L \left( \frac{L}{2\pi} \right)^2 \]
\[ = \frac{\rho L^3}{4\pi^2} + \frac{\rho L^3}{4\pi^2} = \frac{3\rho L^3}{4\pi^2} \]

7. **Final Result**: Thus, the moment of inertia of the loop about the axis \( XX' \) is:
\[ I = \frac{3\rho L^3}{8\pi^2} \]