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From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing through the centre?

1. 15 MR²/32
2. 13 MR²/32
3. 4 MR²
4. 9 MR²/32

Correct answer: 13 MR²/32

Solution

The moment of inertia of the original disc is reduced by the moment of inertia of the removed portion. Using the parallel axis theorem and symmetry, the calculation yields the remaining moment of inertia as 13MR²/32.

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