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MHT CET 2025 (5 May, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2025 (5 May, Shift 2)

Q43MHT CET 2025 · 5 MAY · SHIFT 2Easy

Moment of Inertia & Radius of Gyration

The moment of inertia of a thin uniform rod of mass ' MM ' and length ' LL ', about an axis perpendicular to length of the rod and at a distance ' L/4L/4 ' from one end is

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Answer: (d)

  1. The point at L4\frac{L}{4} from one end is at d=L2−L4=L4d = \frac{L}{2} - \frac{L}{4} = \frac{L}{4} from the centre.
  2. I=ML212+ML216=4ML2+3ML248I = \frac{ML^2}{12} + M\frac{L^2}{16} = \frac{4ML^2 + 3ML^2}{48}
  3. I=7ML248I = \frac{7ML^2}{48}
  4. So the answer is (d) 7ML248\frac{7ML^2}{48}.
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