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MHT CET 2026 (12 May, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2026 (12 May, Shift 1)

Q11MHT CET 2026 · 12 MAY · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

Four solid spheres, each of diameter 5\sqrt{5} cm and mass 0.50.5 kg are placed with their centres at the corners of a square of side 44 cm. The M.I. of the system about the diagonal of the square is N×10−4N \times 10^{-4} kg-m22, then NN is

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Answer: (a)

  1. r=52r = \frac{\sqrt5}{2} cm, so 25mr2=25(0.5)(1.25×10−4)=0.25×10−4\frac{2}{5}mr^2 = \frac{2}{5}(0.5)(1.25\times10^{-4}) = 0.25\times10^{-4} kg m2^2.
  2. Two spheres on the diagonal: 2×0.25×10−4=0.5×10−42\times0.25\times10^{-4} = 0.5\times10^{-4}.
  3. Other two are at half-diagonal 222\sqrt2 cm: each 0.25×10−4+0.5×8×10−4=4.25×10−40.25\times10^{-4} + 0.5\times8\times10^{-4} = 4.25\times10^{-4}.
  4. Total =0.5+8.5=9×10−4= 0.5 + 8.5 = 9\times10^{-4} kg m2^2, so N=9N = 9, answer (a).
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