Choose the correct option:
A wire of mass `m` and length `l` is bent in the form of a quarter circle. The moment of the inertia of the wire about an axis is passing through the centre of the quarter circle is approximately

Options

A. (a) `0.6 ml^(2)`
B. (b) `ml^(2)`
C. (c) `0.2 ml^(2)`
D. (d) `0.4 ml^(2)`

Answer

To find the moment of inertia of a wire bent in the form of a quarter circle about an axis passing through the center of the quarter circle, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a wire of mass \( m \) and length \( l \) bent into the shape of a quarter circle.
- We need to find the moment of inertia of this quarter circle about an axis passing through its center.

2. **Visualizing the Quarter Circle**:
- Imagine that the quarter circle is part of a complete circle. If we were to complete the circle, it would consist of four identical quarter circles.

3. **Moment of Inertia of a Full Circle**:
- The moment of inertia \( I \) of a thin ring (or circle) about an axis through its center is given by:
\[ I = \frac{m r^2}{2} \]
- For a complete circle made of four quarter circles, the total moment of inertia would be:
\[ I_{\text{total}} = 4 \times \frac{m r^2}{2} = 2 m r^2 \]

4. **Finding the Radius**:
- The length of the quarter circle is equal to the arc length, which is given by:
\[ l = \frac{1}{4} \times 2 \pi r = \frac{\pi r}{2} \]
- Rearranging this gives:
\[ r = \frac{2l}{\pi} \]

5. **Substituting the Radius**:
- Now, substitute \( r \) back into the moment of inertia formula for the full circle:
\[ I_{\text{total}} = 2 m \left(\frac{2l}{\pi}\right)^2 \]
- Simplifying this:
\[ I_{\text{total}} = 2 m \cdot \frac{4l^2}{\pi^2} = \frac{8ml^2}{\pi^2} \]

6. **Moment of Inertia of the Quarter Circle**:
- The moment of inertia of the quarter circle \( I_Q \) is one-fourth of the total moment of inertia:
\[ I_Q = \frac{1}{4} I_{\text{total}} = \frac{1}{4} \cdot \frac{8ml^2}{\pi^2} = \frac{2ml^2}{\pi^2} \]

7. **Approximation**:
- To find the approximate value, we can use \( \pi \approx 3.14 \):
\[ \frac{2ml^2}{\pi^2} \approx \frac{2ml^2}{(3.14)^2} \approx \frac{2ml^2}{9.86} \approx 0.202ml^2 \]
- Thus, we can approximate \( I_Q \approx 0.2 ml^2 \).

8. **Final Answer**:
- Therefore, the moment of inertia of the wire about the axis passing through the center of the quarter circle is approximately:
\[ \boxed{0.2 ml^2} \]