Four identical thin rods each of mass `M` and length `l`, from a square frame. Moment of inertia of this frame about an axis through the centre of the square and perpendicular to its plane is

Options

A. `4/3 Ml^(2)`
B. `2/ Ml^(2)`
C. `13/3 Ml^(2)`
D. `1/3 Ml^(2)`

Answer

To find the moment of inertia of a square frame made of four identical thin rods, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have four identical thin rods, each of mass \( M \) and length \( l \), forming a square frame. The rods are positioned at the edges of the square.

2. **Identify the Axis of Rotation**:
- The axis of rotation is through the center of the square and perpendicular to its plane.

3. **Calculate the Moment of Inertia for One Rod**:
- For a thin rod rotating about its midpoint, the moment of inertia \( I_{\text{cm}} \) is given by the formula:
\[ I_{\text{cm}} = \frac{1}{12} M l^2 \]

4. **Apply the Parallel Axis Theorem**:
- To find the moment of inertia about the center of the square, we use the parallel axis theorem. The theorem states:
\[ I = I_{\text{cm}} + Md^2 \]
- Here, \( d \) is the distance from the center of the rod to the center of the square. Since the square has a side length \( l \), the distance \( d \) from the center of the rod to the center of the square is \( \frac{l}{2} \).

5. **Calculate the Moment of Inertia for One Rod about the Center of the Square**:
- Substitute \( d = \frac{l}{2} \) into the parallel axis theorem:
\[ I = \frac{1}{12} M l^2 + M \left(\frac{l}{2}\right)^2 \]
- This simplifies to:
\[ I = \frac{1}{12} M l^2 + M \cdot \frac{l^2}{4} = \frac{1}{12} M l^2 + \frac{3}{12} M l^2 = \frac{4}{12} M l^2 = \frac{1}{3} M l^2 \]

6. **Calculate the Total Moment of Inertia for All Four Rods**:
- Since there are four identical rods, the total moment of inertia \( I_{\text{total}} \) is:
\[ I_{\text{total}} = 4 \times I = 4 \times \frac{1}{3} M l^2 = \frac{4}{3} M l^2 \]

### Final Answer:
The moment of inertia of the square frame about an axis through the center of the square and perpendicular to its plane is:
\[ \frac{4}{3} M l^2 \]