Four identical rods are joined end to end to form a square. The mass of each rod is `M`. The moment of inertia of the square about the median line is

Options

A. `(Ml^(2))/3`
B. `(Ml^(2))/4`
C. `(Ml^(2))/6`
D. none of these

Explanation

To find the moment of inertia of a square formed by four identical rods about the median line, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Configuration**:
- We have a square formed by four identical rods, each of mass \( M \) and length \( L \).

2. **Determine the Axis of Rotation**:
- The median line in this case is the line that bisects the square horizontally (or vertically). This line runs through the center of the square and is perpendicular to the length of the rods.

3. **Calculate the Moment of Inertia for Each Rod**:
- For the two horizontal rods (let's call them Rod 1 and Rod 2), the moment of inertia about the median line can be calculated using the formula for a rod about an axis through its center:
\[ I = \frac{1}{12} M L^2 \]
- Since there are two horizontal rods, the total moment of inertia for these rods is:
\[ I_{horizontal} = 2 \times \frac{1}{12} M L^2 = \frac{1}{6} M L^2 \]

4. **Calculate the Moment of Inertia for the Vertical Rods**:
- For the two vertical rods (Rod 3 and Rod 4), we need to use the parallel axis theorem because their axis of rotation is not through their center. The distance from the median line to the center of each vertical rod is \( \frac{L}{2} \).
- The moment of inertia for each vertical rod about its own center is:
\[ I_{center} = \frac{1}{12} M L^2 \]
- Using the parallel axis theorem:
\[ I_{vertical} = I_{center} + M d^2 \]
where \( d = \frac{L}{2} \):
\[ I_{vertical} = \frac{1}{12} M L^2 + M \left(\frac{L}{2}\right)^2 = \frac{1}{12} M L^2 + M \frac{L^2}{4} = \frac{1}{12} M L^2 + \frac{3}{12} M L^2 = \frac{4}{12} M L^2 = \frac{1}{3} M L^2 \]
- Since there are two vertical rods, the total moment of inertia for these rods is:
\[ I_{vertical} = 2 \times \frac{1}{3} M L^2 = \frac{2}{3} M L^2 \]

5. **Combine the Moments of Inertia**:
- The total moment of inertia of the square about the median line is the sum of the moments of inertia of all four rods:
\[ I_{total} = I_{horizontal} + I_{vertical} = \frac{1}{6} M L^2 + \frac{2}{3} M L^2 \]
- To combine these, convert \( \frac{2}{3} \) to sixths:
\[ \frac{2}{3} = \frac{4}{6} \]
- Thus,
\[ I_{total} = \frac{1}{6} M L^2 + \frac{4}{6} M L^2 = \frac{5}{6} M L^2 \]

### Final Result:
The moment of inertia of the square about the median line is:
\[ I = \frac{5}{6} M L^2 \]