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JEE Main 2026 (28 Jan, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2026 (28 Jan, Shift 1)

Q24JEE Main 2026 · 28 JAN · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

A solid sphere of radius 10 cm is rotating about an axis which is at a distance 15 cm from its centre. The radius of gyration about this axis is n cm\sqrt{n}\text{ cm}. The value of nn is

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Answer: 265

  1. Parallel axis theorem: k2=25R2+d2k^2 = \frac{2}{5}R^2 + d^2.
  2. k2=25(10)2+(15)2=40+225=265 cm2k^2 = \frac{2}{5}(10)^2 + (15)^2 = 40 + 225 = 265\ \text{cm}^2
  3. k=265k = \sqrt{265} cm, so n=265n = 265.
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