AIIMS 2006 Physics Moment of Inertia MCQ Question

To solve the problem of finding the moment of inertia of a bent rod about an axis through its center, we need to first understand the effect of bending the rod.

Step 1: Understanding the Original Moment of Inertia

For a straight rod of mass $M$ and length $L$, the moment of inertia $I$ about an axis through its center and perpendicular to its length is given by:

$$I = \frac{1}{12} ML^2$$

This formula is derived from the integration of mass distribution along the length of the rod.

Step 2: Analyzing the Bent Rod

When the rod is bent in the middle to form an angle of $60^\circ$, we can visualize it as two straight segments, each of length $\frac{L}{2}$, making an angle of $60^\circ$ with respect to each other. Let's denote the two halves of the rod as $AB$ and $BC$.

Step 3: Moment of Inertia of Each Segment

Since the two halves are symmetric and contribute equally to the moment of inertia about the center axis, we can calculate the moment of inertia for one half and then double it.

The moment of inertia for each half (straight segment) about its own center is:

$$I_{\text{half}} = \frac{1}{12} \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{1}{12} \frac{M}{2} \frac{L^2}{4} = \frac{ML^2}{96}$$

However, because these halves are not rotating around their own centers but about the center of the original rod, we need to use the parallel axis theorem to find the moment of inertia about the center.

Step 4: Applying the Parallel Axis Theorem

The parallel axis theorem states:

$$I = I_{cm} + Md^2$$

Where:

• $I_{cm}$ is the moment of inertia about the center of mass,
• $M$ is the mass,
• $d$ is the distance from the center of mass to the new axis.

For each half of the rod, the distance $d$ from the axis through the center to the center of each half is $\frac{L}{4}$. So, we need to add $M/2 \cdot (d^2)$ for each half.

Calculating the parallel axis adjustment for one half:

$$d = \frac{L}{4} \quad \text{and} \quad M = \frac{M}{2}$$

Thus, the additional moment of inertia for one half becomes:

$$I_{\text{shift}} = \frac{M}{2} \left(\frac{L}{4}\right)^2 = \frac{M}{2} \cdot \frac{L^2}{16} = \frac{ML^2}{32}$$

Step 5: Total Moment of Inertia for the Bent Rod

Now combining these results, the total moment of inertia of the bent rod is:

$$I_{\text{total}} = 2 \left( I_{\text{half}} + I_{\text{shift}} \right) = 2 \left( \frac{ML^2}{96} + \frac{ML^2}{32} \right)$$

Finding a common denominator (which is 96):

1. Convert $\frac{ML^2}{32}$ to a fraction with 96 as a denominator:

$$\frac{ML^2}{32} = \frac{3ML^2}{96}$$

1. Now we can add:

$$I_{\text{total}} = 2 \left( \frac{ML^2}{96} + \frac{3ML^2}{96} \right) = 2 \cdot \frac{4ML^2}{96} = \frac{8ML^2}{96} = \frac{ML^2}{12}$$

Conclusion

Thus, the moment of inertia of the bent rod about the same axis is:

$$I = \frac{1}{12} ML^2$$

This leads us to conclude that the correct answer to the question is option B) $\frac{1}{12} ML^2$.

Why Other Options Are Incorrect

• A) $\frac{1}{48} ML^2$: This value is too low and does not account for the correct calculation of the moment of inertia after bending.
• **C) $ \frac{1}{24} ML^