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JEE Main 2020 (4 Sep, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2020 (4 Sep, Shift 2)

Q149JEE Main 2020 · 4 SEP · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

Consider two uniform discs of the same thickness and different radii R1 = R and R2 = α\alphaR made of the same material. If the ratio of their moments of inertia I1 and I2, respectively, about their axes is I1: I2 = 1: 16 then the value of α\alpha is:

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Answer: (b)

  1. Same material and thickness: M∝R2M \propto R^2, so I=MR22∝R4I = \frac{MR^2}{2} \propto R^4.
  2. α4=16⇒α=2\alpha^4 = 16 \Rightarrow \alpha = 2
  3. So the answer is (b) 2.
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