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MHT CET 2024 (2 May, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2024 (2 May, Shift 2)

Q85MHT CET 2024 · 2 MAY · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

A solid sphere of mass ' mm ', radius ' RR ', having moment of inertia about an axis passing through center of mass as 'I' is recast into a disc of thickness ' tt ' whose moment of inertia about an axis passing through the rim (edge) \& perpendicular to plane remains 'I'. Then the radius of disc is

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Answer: (a)

  1. Sphere about centre: I=25mR2I = \frac{2}{5}mR^2.
  2. Disc about an axis through the rim, perpendicular to plane: I=32mr2I = \frac{3}{2}mr^2.
  3. 32mr2=25mR2⇒r2=4R215\frac{3}{2}mr^2 = \frac{2}{5}mR^2 \Rightarrow r^2 = \frac{4R^2}{15}
  4. So the answer is (a) r=2R15r = \frac{2R}{\sqrt{15}}.
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