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MHT CET 2020 (16 Oct, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2020 (16 Oct, Shift 2)

Q142MHT CET 2020 · 16 OCT · SHIFT 2Easy

Moment of Inertia & Radius of Gyration

A thin uniform rod has mass MM and length LL The moment of inertia about an axis perpendicular to it and passing through the point at a distance L3\frac{L}{3} from one of its ends, will be

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Answer: (c)

  1. Distance from the centre: d=L2−L3=L6d = \frac{L}{2} - \frac{L}{3} = \frac{L}{6}.
  2. I=ML212+ML236=3ML2+ML236I = \frac{ML^2}{12} + \frac{ML^2}{36} = \frac{3ML^2 + ML^2}{36}
  3. I=ML29I = \frac{ML^2}{9}
  4. So the answer is (c) ML29\frac{ML^2}{9}.
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