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MHT CET 2026 (17 Apr, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2026 (17 Apr, Shift 2)

Q15MHT CET 2026 · 17 APR · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

Radius of gyration of a thin uniform circular disc about the axis passing through its centre and perpendicular to its plane is KcK_c. Radius of gyration of the same disc about a diamter of the disc is KdK_d. the ratio Kd:KcK_d : K_c is

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Answer: (b)

  1. Perpendicular axis: Ic=12MR2⇒Kc2=R22I_c = \frac{1}{2}MR^2 \Rightarrow K_c^2 = \frac{R^2}{2}.
  2. Diameter: Id=14MR2⇒Kd2=R24I_d = \frac{1}{4}MR^2 \Rightarrow K_d^2 = \frac{R^2}{4}.
  3. KdKc=1/41/2=12\frac{K_d}{K_c} = \sqrt{\frac{1/4}{1/2}} = \frac{1}{\sqrt{2}}
  4. So the answer is (b) 1:21 : \sqrt{2}.
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