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MHT CET 2026 (13 Apr, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2026 (13 Apr, Shift 2)

Q8MHT CET 2026 · 13 APR · SHIFT 2Easy

Moment of Inertia & Radius of Gyration

Moment of inertia of a disc of mass MM and radius 'R' about any of its diameter is MR2/4MR^2/4. The moment of inertia of this disc about an axis normal to the disc and passing through a point on its edge will be, (x/2)MR2(x/2)MR^2. The value of xx is

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Answer: (b)

  1. Perpendicular axis theorem: Icm=2×MR24=MR22I_{cm} = 2 \times \frac{MR^2}{4} = \frac{MR^2}{2} (normal to disc through centre).
  2. Parallel axis through a point on the edge: I=MR22+MR2=32MR2I = \frac{MR^2}{2} + MR^2 = \frac{3}{2}MR^2
  3. Comparing with x2MR2\frac{x}{2}MR^2: x=3x = 3.
  4. So the answer is (b) 33.
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