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JEE Main 2021 (27 Jul, Shift 1) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2021 (27 Jul, Shift 1)

Q136JEE Main 2021 · 27 JUL · SHIFT 1Medium

Moment of Inertia & Radius of Gyration

The figure shows two solid discs with radius R and r respectively. If mass per unit area is same for both, what is the ratio of MI of bigger disc around axis AB (Which is ⊥\perp to the plane of the disc and passing through its centre) to MI of smaller disc around one of its diameters lying on its plane? Given 'M' is the mass of the larger disc. (MI stands for moment of inertia)

Figure

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Answer: (d)

  1. Bigger disc about AB (perpendicular, through centre): I1=MR22I_1 = \frac{MR^2}{2}.
  2. Same mass per area: smaller disc mass m=Mr2R2m = M\frac{r^2}{R^2}; about a diameter I2=mr24=Mr44R2I_2 = \frac{mr^2}{4} = \frac{Mr^4}{4R^2}.
  3. I1I2=MR2/2Mr4/4R2=2R4r4\frac{I_1}{I_2} = \frac{MR^2/2}{Mr^4/4R^2} = \frac{2R^4}{r^4}
  4. So the answer is (d) 2R4:r42R^4 : r^4.
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