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JEE Main 2025 (7 Apr, Shift 2) · Moment of Inertia & Radius of Gyration

Moment of Inertia & Radius of Gyration question from JEE Main 2025 (7 Apr, Shift 2)

Q59JEE Main 2025 · 7 APR · SHIFT 2Medium

Moment of Inertia & Radius of Gyration

Let M and R be the mass and radius of a disc. A small disc of radius R/3R/3 is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis ABAB passing through the centre OO and perpendicular to the plane of disc is 4xMR2\frac{4}{x}MR^2. The value of xx is ____________.

Figure

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

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Answer: 9

  1. Small disc mass M9\frac{M}{9}, centre at 2R3\frac{2R}{3} from O.
  2. Ismall=12M9R29+M94R29=MR218I_{small} = \frac{1}{2}\frac{M}{9}\frac{R^2}{9} + \frac{M}{9}\frac{4R^2}{9} = \frac{MR^2}{18}
  3. I=MR22−MR218=49MR2I = \frac{MR^2}{2} - \frac{MR^2}{18} = \frac{4}{9}MR^2
  4. So x=9x = 9.
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