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MHT CET 2025 (19 Apr, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2025 (19 Apr, Shift 1)

Q41MHT CET 2025 · 19 APR · SHIFT 1Easy

Moment of Inertia & Radius of Gyration

Four particles each of mass M are placed at the corners of a square of side LL. The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is

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Answer: (d)

  1. Each corner is at r=L2r = \frac{L}{\sqrt{2}} from the centre (half the diagonal).
  2. I=4M⋅L22=2ML2I = 4M \cdot \frac{L^2}{2} = 2ML^2
  3. K=I4M=L22=L2K = \sqrt{\frac{I}{4M}} = \sqrt{\frac{L^2}{2}} = \frac{L}{\sqrt{2}}
  4. So the answer is (d) L2\frac{L}{\sqrt{2}}.
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