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MHT CET 2020 (16 Oct, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2020 (16 Oct, Shift 2)

Q146MHT CET 2020 · 16 OCT · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

The ratio of radii of gyration of a ring to a disc (both circular) of same radii and mass, about a tangential axis perpendicular to the plane is

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Answer: (d)

  1. Ring, tangent perpendicular to plane: MR2+MR2=2MR2⇒Kr=2RMR^2 + MR^2 = 2MR^2 \Rightarrow K_r = \sqrt{2}R.
  2. Disc, tangent perpendicular to plane: 12MR2+MR2=32MR2⇒Kd=32R\frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2 \Rightarrow K_d = \sqrt{\frac{3}{2}}R.
  3. KrKd=23/2=23\frac{K_r}{K_d} = \sqrt{\frac{2}{3/2}} = \frac{2}{\sqrt{3}}
  4. So the answer is (d) 23\frac{2}{\sqrt{3}}.
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