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MHT CET202611 April 2026Evening ShiftPhysicsRotational MotionActual

Let M and L be the mass and length of thin uniform rod respectively. In first case, axis of rotation is passing through centre and perpendicular to length of rod. In second case axis of rotation is passing through one end and perpendicular to length of rod. The ratio of radius of gyration in first case to second case is

Options

  1. A1 4
  2. B1 2
  3. C1 8
  4. D1 6

Correct answer

B. 1 2

Step-by-step solution

Moment of inertia of a thin uniform rod about an axis passing through its centre and perpendicular to its length is I₁ = ML^2 12 . Radius of gyration in the first case is K₁ = I₁ M = L 12 . Moment of inertia of the rod about an axis passing through one end and perpendicular to its length is I₂ = ML^2 3 . Radius of gyration in the second case is K₂ = I₂ M = L 3 . The ratio of radius of gyration in the first case to the second case is K₁ K₂ = L 12 L 3 = 3 12 = 1 4 = 1 2 .

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