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JEE Main 2025 (2 Apr, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2025 (2 Apr, Shift 1)

Q56JEE Main 2025 · 2 APR · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

Moment of inertia of a rod of mass ' M ' and length ' L ' about an axis passing through its center and normal to its length is ' α\alpha '. Now the rod is cut into two equal parts and these parts are joined symmetrically to form a cross shape. Moment of inertia of cross about an axis passing through its center and normal to plane containing cross is:

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Answer: (a)

  1. Original rod: α=ML212\alpha = \frac{ML^2}{12}.
  2. Each half (mass M2\frac{M}{2}, length L2\frac{L}{2}) is centred on the axis: I=112⋅M2⋅L24=ML296I = \frac{1}{12}\cdot\frac{M}{2}\cdot\frac{L^2}{4} = \frac{ML^2}{96}.
  3. Cross: I=2×ML296=ML248=α4I = 2\times\frac{ML^2}{96} = \frac{ML^2}{48} = \frac{\alpha}{4}
  4. So the answer is (a) α/4\alpha/4.
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