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MHT CET 2021 (20 Sep, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2021 (20 Sep, Shift 1)

Q135MHT CET 2021 · 20 SEP · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

A solid sphere of mass MM, radius RR has moment of inertia 'II' about its diameter. It is recast into a disc of thickness 't' whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains 'I'. Radius of the disc will be

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Answer: (c)

  1. Sphere: I=25MR2I = \frac{2}{5}MR^2; disc about its edge, perpendicular to plane: I=32Mr2I = \frac{3}{2}Mr^2.
  2. 32Mr2=25MR2⇒r2=4R215\frac{3}{2}Mr^2 = \frac{2}{5}MR^2 \Rightarrow r^2 = \frac{4R^2}{15}
  3. So the answer is (c) 2R15\frac{2R}{\sqrt{15}}.
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