An iron rod of mass M and length L is cut into n equal parts by cutting it perpendicular to its length. If I is the M.I. of the rod, about an axis passing through its centre and perpendicular to its axis, then the moment of interia of each part about the similar axis

Options

A. `(l)/(n)`
B. `(l)/(n^(2))`
C. `(l)/(n^(3))`
D. `(l)/(n^(2))`

Explanation

To solve the problem, we need to find the moment of inertia of each part of the iron rod after it has been cut into \( n \) equal parts.

### Step-by-Step Solution:

1. **Understand the Initial Conditions**:
- We have an iron rod of mass \( M \) and length \( L \).
- The moment of inertia \( I \) of the rod about an axis passing through its center and perpendicular to its length is given by the formula:
\[ I = \frac{ML^2}{12} \]

2. **Cutting the Rod**:
- The rod is cut into \( n \) equal parts.
- Therefore, the length of each part \( L' \) will be:
\[ L' = \frac{L}{n} \]
- The mass of each part \( M' \) will be:
\[ M' = \frac{M}{n} \]

3. **Moment of Inertia of Each Part**:
- We need to find the moment of inertia \( I' \) of one part about the same axis (through its center and perpendicular to its length).
- The formula for the moment of inertia for a rod about its center is:
\[ I' = \frac{M'L'^2}{12} \]
- Substituting \( M' \) and \( L' \):
\[ I' = \frac{\left(\frac{M}{n}\right)\left(\frac{L}{n}\right)^2}{12} \]

4. **Simplifying the Expression**:
- Now, substituting \( L' = \frac{L}{n} \):
\[ I' = \frac{\frac{M}{n} \cdot \frac{L^2}{n^2}}{12} \]
- This simplifies to:
\[ I' = \frac{ML^2}{12n^3} \]

5. **Final Result**:
- Thus, the moment of inertia of each part about the same axis is:
\[ I' = \frac{ML^2}{12n^3} \]

### Conclusion:
The moment of inertia of each part of the rod about the axis passing through its center and perpendicular to its length is given by:
\[ I' = \frac{ML^2}{12n^3} \]