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Moment of Inertia & Radius of Gyration question from JEE Main 2018

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Moment of Inertia & Radius of Gyration

From a uniform circular disc of radius R and mass 9M, a small disc of radius R/3 is removed as shown in the figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc and passing through centre of disc is:

Figure

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Answer: (b)

  1. Small disc mass =9M×19=M= 9M \times \frac{1}{9} = M, centre at 2R3\frac{2R}{3}.
  2. Ismall=M2R29+M4R29=MR22I_{small} = \frac{M}{2}\frac{R^2}{9} + M\frac{4R^2}{9} = \frac{MR^2}{2}
  3. I=9MR22−MR22=4MR2I = \frac{9MR^2}{2} - \frac{MR^2}{2} = 4MR^2
  4. So the answer is (b).
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