Question

A thin metal rod of mass \(M\) and length \(L\) is cut into 4 equal parts by cutting it perpendicular to its length. If moment of inertia of the rod about an axis passing through its centre and perpendicular to its axis is \(I\), then moment of inertia of each part about the similar axis is

• A. \( \dfrac{I}{16} \)
• B. \( \dfrac{I}{32} \)
• C. \( \dfrac{I}{128} \)
• D. \( \dfrac{I}{64} \)

Hint:

When a rod is cut into equal parts, moment of inertia reduces rapidly because it depends on the square of length.

Answer 1

The Correct Option is D

Step 1: Moment of inertia of the original rod.
For a thin rod of mass \(M\) and length \(L\), the moment of inertia about an axis passing through its centre and perpendicular to its length is \[ I = \frac{1}{12}ML^2. \]
Step 2: Mass and length of each part.
After cutting the rod into 4 equal parts: \[ \text{Mass of each part} = \frac{M}{4}, \quad \text{Length of each part} = \frac{L}{4}. \]
Step 3: Moment of inertia of each part.
Moment of inertia of one part about its centre is \[ I' = \frac{1}{12}\left(\frac{M}{4}\right)\left(\frac{L}{4}\right)^2 = \frac{1}{12}\cdot\frac{M}{4}\cdot\frac{L^2}{16} = \frac{ML^2}{768}. \]
Step 4: Expressing in terms of \(I\).
Since \( I = \frac{ML^2}{12} \), \[ I' = \frac{I}{64}. \]
Step 5: Conclusion.
The moment of inertia of each part about the similar axis is \( \dfrac{I}{64} \).