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MHT CET 2026 (13 May, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2026 (13 May, Shift 1)

Q2MHT CET 2026 · 13 MAY · SHIFT 1Easy

Moment of Inertia & Radius of Gyration

The moment of inertia of two equal masses each of mass m at separation L connected by a rod of mass M about an axis passing through centre and perpendicular to the length of rod is

Figure

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Answer: (b)

  1. Rod about its centre: Irod=ML212I_{rod} = \frac{ML^2}{12}.
  2. Each mass is at L2\frac{L}{2} from the axis: Imasses=2m(L2)2=mL22=6mL212I_{masses} = 2m\left(\frac{L}{2}\right)^2 = \frac{mL^2}{2} = \frac{6mL^2}{12}
  3. I=ML212+6mL212=(M+6m)L212I = \frac{ML^2}{12} + \frac{6mL^2}{12} = \frac{(M+6m)L^2}{12}
  4. So the answer is (b) (M+6m)L212\frac{(M+6m)L^2}{12}.
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