The moment of inertia (MI) of a disc of radius \( R \) and mass \( M \) about its central axis is:
Show Hint
- \( \frac{MR^2}{4} \)
- \( \frac{MR^2}{2} \)
- \( MR^2 \)
- \( \frac{3MR^2}{2} \)
The Correct Option is B
Solution and Explanation
Step 1: The moment of inertia for a solid disc rotating about its central axis is expressed as: \[ I = \frac{1}{2} MR^2. \] Step 2: By comparing with the provided options, the correct answer is option (b).
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