The moment of inertia of a disc about an axis passing through its center and normal to its plane is I. The disc is now folded along a diameter such that the two halves are mutually perpendicular. Its moment of inertia about this diameter will now be:
Correct answer: C. I/2
- A. I
- B. I/√2
- C. I/2
- D. I/4
Explanation
The moment of inertia of a disc about its central axis perpendicular to its plane is I. When the disc is folded along a diameter to form two perpendicular halves, each half retains a symmetrical distribution about the diameter. The moment of inertia of each half-disc about the new axis is half of the original disc's moment of inertia, resulting in a total moment of inertia of I/2 for the folded configuration. Options A and B are incorrect because they do not consider the symmetrical folding and resulting mass distribution. Option D incorrectly assumes a greater reduction in moment of inertia than what occurs.I = ½ mR² For a diameter; I' = ½ I
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