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JEE Main 2026 (22 Jan, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2026 (22 Jan, Shift 1)

Q33JEE Main 2026 · 22 JAN · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

A solid sphere of mass 5 kg and radius 10 cm is kept in contact with another solid sphere of mass 10 kg and radius 20 cm. The moment of inertia of this pair of spheres about the tangent passing through the point of contact is ‾ kg⋅m2\underline{\qquad}\ \text{kg}\cdot\text{m}^2.

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Answer: (d)

  1. The tangent is perpendicular to the line of centres, at distances 0.1 m0.1\text{ m} and 0.2 m0.2\text{ m} from the centres.
  2. Small: 25(5)(0.1)2+5(0.1)2=0.02+0.05=0.07\frac{2}{5}(5)(0.1)^2 + 5(0.1)^2 = 0.02 + 0.05 = 0.07
  3. Large: 25(10)(0.2)2+10(0.2)2=0.16+0.40=0.56\frac{2}{5}(10)(0.2)^2 + 10(0.2)^2 = 0.16 + 0.40 = 0.56
  4. Total =0.63 kg m2= 0.63\text{ kg m}^2, so the answer is (d).
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