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MHT CET2016PhysicsRotational Motion

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

Options

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

Correct answer

B. 1 2

Step-by-step solution

Moment of inertia of rod whose axis of rotation is passing through center and perpendicular to the plane of rod is I = M L 2 12 a n d I = M K 1 2 (where K 1 radius of gravitation) ⇒ K 1 = L 2 3 ..... (i) When axis of rotation of rod is passing through one end of rod, then I = M K 2 2 = M L 2 3 ⇒ K 2 = L 3 ..... (ii) Taking ratios of (i) and (ii) we get K 1 K 2 = L 2 3 × 3 L = 1 2 ⇒ K 1 K 2 = 1 2

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