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MHT CET 2026 (16 Apr, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from MHT CET 2026 (16 Apr, Shift 2)

Q36MHT CET 2026 · 16 APR · SHIFT 2Difficult

Moment of Inertia & Radius of Gyration

A uniform solid cylinder with radius R and length L has moment of inertia I1I_1 about the axis of the cylinder. A concentric solid cylinder of radius R/2R/2 and length L/2L/2 is carved out of the original cylinder. If I2I_2 is the moment of inertia of the carved out portion of the cylinder then I1/I2I_1/I_2 is (Both I1I_1 and I2I_2 are about the axis of the cylinder)

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Answer: (d)

  1. I1=MR22I_1 = \frac{MR^2}{2}
  2. Carved mass =M×14×12=M8= M \times \frac{1}{4} \times \frac{1}{2} = \frac{M}{8}: I2=12M8R24=MR264I_2 = \frac{1}{2}\frac{M}{8}\frac{R^2}{4} = \frac{MR^2}{64}
  3. I1I2=32\frac{I_1}{I_2} = 32, so the answer is (d) 32:132:1.
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