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JEE Main 2023 (24 Jan, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2023 (24 Jan, Shift 2)

Q110JEE Main 2023 · 24 JAN · SHIFT 2Difficult

Moment of Inertia & Radius of Gyration

A uniform solid cylinder with radius R and length L has moment of inertia I 11, about the axis of the cylinder. A concentric solid cylinder of radius R′=R2R' = \frac{R}{2} and length L′=L2L' = \frac{L}{2} is carved out of the original cylinder. If I22 is the moment of inertia of the carved out portion of the cylinder then I1I2=\frac{I_1}{I_2} = __________. (Both I11 and I22 are about the axis of the cylinder)

Numerical answer type. Work it out, then open the answer.

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Answer: 32

  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
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