Three identical uniform thin metal rods form the three sides of an equilateral triangle. If the moment of inertia of the system of these three rods about an axis passing through the centroid of the triangle and perpendicular to the plane of the triangle is 'n' times the moment of inertia of one rod separately about an axis passing through the centre of the rod and perpendicular to its length, the value of 'n' is

Options

A. 3
B. 6
C. 9
D. 12

Answer

To solve the problem, we need to calculate the moment of inertia of three identical rods forming an equilateral triangle about an axis passing through the centroid of the triangle. We will also compare this with the moment of inertia of a single rod about its own center.

### Step-by-Step Solution:

1. **Identify the Mass and Length of One Rod:**
Let the mass of each rod be \( m \) and the length of each rod be \( a \).

2. **Moment of Inertia of One Rod:**
The moment of inertia \( I \) of a single rod about an axis passing through its center and perpendicular to its length is given by the formula:
\[ I_{\text{rod}} = \frac{1}{12} m a^2 \]

3. **Determine the Position of the Centroid:**
For an equilateral triangle formed by three rods, the centroid is located at an equal distance from all three vertices. The distance from the centroid to the midpoint of each rod can be calculated.

4. **Calculate the Distance from the Centroid to the Midpoint of a Rod:**
The distance \( d \) from the centroid to the midpoint of a rod can be derived using geometry. For an equilateral triangle, this distance is:
\[ d = \frac{a}{2\sqrt{3}} \]

5. **Use the Parallel Axis Theorem:**
The moment of inertia of each rod about the centroid can be calculated using the parallel axis theorem:
\[ I_{\text{centroid}} = I_{\text{rod}} + m d^2 \]
Substituting the values:
\[ I_{\text{centroid}} = \frac{1}{12} m a^2 + m \left(\frac{a}{2\sqrt{3}}\right)^2 \]
Simplifying this:
\[ I_{\text{centroid}} = \frac{1}{12} m a^2 + m \cdot \frac{a^2}{12} = \frac{1}{12} m a^2 + \frac{1}{12} m a^2 = \frac{1}{6} m a^2 \]

6. **Total Moment of Inertia for Three Rods:**
Since there are three identical rods, the total moment of inertia \( I_{\text{total}} \) about the centroid is:
\[ I_{\text{total}} = 3 \cdot I_{\text{centroid}} = 3 \cdot \frac{1}{6} m a^2 = \frac{1}{2} m a^2 \]

7. **Relate Total Moment of Inertia to One Rod's Moment of Inertia:**
Now, we need to express \( I_{\text{total}} \) in terms of \( n \) times the moment of inertia of one rod:
\[ I_{\text{total}} = n \cdot I_{\text{rod}} = n \cdot \frac{1}{12} m a^2 \]
Setting the two expressions for \( I_{\text{total}} \) equal gives:
\[ \frac{1}{2} m a^2 = n \cdot \frac{1}{12} m a^2 \]

8. **Solve for \( n \):**
Dividing both sides by \( m a^2 \) (assuming \( m \) and \( a \) are not zero):
\[ \frac{1}{2} = n \cdot \frac{1}{12} \]
Multiplying both sides by 12:
\[ n = 6 \]

### Final Answer:
The value of \( n \) is \( 6 \).