The M.I. of a thin rod of length 1 about the perpendicular axis through its centre is I. The M.I. of square structure made by four such rods a perpendicular axis to the plane and through the centre will be:

Options

A. 4I
B. 8I
C. 12I
D. 16I

Answer

To solve the problem, we need to find the moment of inertia (M.I.) of a square structure made by four thin rods, each of length 1 meter, about a perpendicular axis through the center of the square.

### Step-by-Step Solution:

1. **Understand the Moment of Inertia of a Single Rod:**
The moment of inertia \( I \) of a thin rod of length \( L \) about an axis perpendicular to its length and through its center is given by the formula:
\[ I = \frac{mL^2}{12} \]
Here, \( L = 1 \) meter, so we can denote the moment of inertia of one rod as:
\[ I = \frac{m(1^2)}{12} = \frac{m}{12} \]

2. **Determine the Configuration of the Square:**
When we arrange four such rods to form a square, each rod will be positioned at the sides of the square. The length of each side of the square is 1 meter.

3. **Calculate the Moment of Inertia about the Center:**
To find the moment of inertia of the square structure about an axis perpendicular to the plane of the square and through its center, we can use the perpendicular axis theorem. According to this theorem:
\[ I_{z} = I_{x} + I_{y} \]
where \( I_{z} \) is the moment of inertia about the perpendicular axis, and \( I_{x} \) and \( I_{y} \) are the moments of inertia about the two axes in the plane of the square.

4. **Calculate \( I_{x} \) and \( I_{y} \):**
Each side of the square contributes to the moment of inertia. For each rod, the moment of inertia about its own center is \( \frac{m}{12} \), but we need to use the parallel axis theorem to find the moment of inertia about the center of the square.
The distance from the center of the square to the center of each rod (which is at a distance of \( \frac{1}{2} \) meter) is:
\[ d = \frac{1}{2} \]
Thus, the moment of inertia for each rod about the center of the square is:
\[ I = \frac{m}{12} + m \left(\frac{1}{2}\right)^2 = \frac{m}{12} + \frac{m}{4} = \frac{m}{12} + \frac{3m}{12} = \frac{4m}{12} = \frac{m}{3} \]

5. **Total Moment of Inertia for the Square:**
Since there are four rods, the total moment of inertia \( I_{total} \) for the square structure is:
\[ I_{total} = 4 \times \frac{m}{3} = \frac{4m}{3} \]

6. **Relate to Given Moment of Inertia \( I \):**
From the earlier calculation, we know that \( I = \frac{m}{12} \). Therefore, we can express \( m \) in terms of \( I \):
\[ m = 12I \]
Substituting this back into the total moment of inertia:
\[ I_{total} = \frac{4(12I)}{3} = 16I \]

### Final Answer:
The moment of inertia of the square structure about the perpendicular axis through its center is:
\[ \boxed{16I} \]