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MHT CET 2023 (11 May, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2023 (11 May, Shift 2)

Q108MHT CET 2023 · 11 MAY · SHIFT 2Difficult

Moment of Inertia & Radius of Gyration

From a disc of mass 'MM' and radius 'RR', a circular hole of diameter 'RR' is cut whose rim passes through the centre. The moment of inertia of the remaining part of the disc about perpendicular axis passing through the centre is

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Answer: (a)

  1. Hole radius R2\frac{R}{2}, mass M4\frac{M}{4}, centre at R2\frac{R}{2} from O.
  2. Ihole=12M4R24+M4R24=3MR232I_{hole} = \frac{1}{2}\frac{M}{4}\frac{R^2}{4} + \frac{M}{4}\frac{R^2}{4} = \frac{3MR^2}{32}
  3. I=MR22−3MR232=13MR232I = \frac{MR^2}{2} - \frac{3MR^2}{32} = \frac{13MR^2}{32}
  4. So the answer is (a).
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