Four identical thin rods each of mass M and length `l_3`, form 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. `1/3 ml^2`
B. `4/3 Ml^2`
C. `2/3 Ml^2`
D. `(13)/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. **Identify the Components**:
- Each rod has a mass \( M \) and length \( l \).
- There are four rods forming a square frame.

2. **Moment of Inertia of a Single Rod**:
- The moment of inertia \( I \) of a thin rod about an axis through its center and perpendicular to its length is given by:
\[ I_{\text{center}} = \frac{1}{12} M l^2 \]

3. **Using the Perpendicular Axis Theorem**:
- For a rod positioned horizontally, we want to find the moment of inertia about an axis through the center of the square and perpendicular to its plane.
- The distance from the center of the square to the center of the rod is \( \frac{l}{2} \).
- We can use the Parallel Axis Theorem to find the moment of inertia about this new axis:
\[ I = I_{\text{center}} + Md^2 \]
- Here, \( d = \frac{l}{2} \), so:
\[ I = \frac{1}{12} M l^2 + M \left(\frac{l}{2}\right)^2 \]
- Simplifying this gives:
\[ I = \frac{1}{12} M l^2 + M \frac{l^2}{4} = \frac{1}{12} M l^2 + \frac{3}{12} M l^2 = \frac{4}{12} M l^2 = \frac{1}{3} M l^2 \]

4. **Total Moment of Inertia for the Square Frame**:
- Since there are four identical rods, the total moment of inertia \( I_{\text{total}} \) is:
\[ I_{\text{total}} = 4 \cdot \frac{1}{3} M l^2 = \frac{4}{3} M l^2 \]

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