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JEE Main 2022 (29 Jun, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2022 (29 Jun, Shift 2)

Q117JEE Main 2022 · 29 JUN · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is I1. The same rod is bent into a ring and its moment of inertia about a diameter is I2. If I1I2\frac{I_1}{I_2} is xπ23\frac{x\pi^2}{3}, then the value of x will be ____________.

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

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Answer: 8

  1. Rod about one end: I1=mL23I_1 = \frac{mL^2}{3}.
  2. Ring: 2πR=L⇒R=L2π2\pi R = L \Rightarrow R = \frac{L}{2\pi}; about a diameter I2=mR22=mL28π2I_2 = \frac{mR^2}{2} = \frac{mL^2}{8\pi^2}.
  3. I1I2=8π23\frac{I_1}{I_2} = \frac{8\pi^2}{3}
  4. So x=8x = 8.
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