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JEE Main 2020 (2 Sep, Shift 2) · Composite & Cut-Out Bodies

Composite & Cut-Out Bodies question from JEE Main 2020 (2 Sep, Shift 2)

Q9JEE Main 2020 · 2 SEP · SHIFT 2Difficult

Composite & Cut-Out Bodies

A square shaped hole of side l = a2\frac{a}{2} is carved out at a distance d = a2\frac{a}{2} from the centre ‘O’ of a uniform circular disk of radius a. If the distance of the centre of mass of the remaining portion from O is −aX-\frac{a}{X}, value of X (to the nearest integer) is:

Figure

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

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Answer: 23

  1. Mass ∝\propto area. Disc: πa2\pi a^2 at O. Square hole: a24\frac{a^2}{4} at x=a2x = \frac{a}{2}.
  2. xcm=−a24⋅a2πa2−a24=−a8π−2x_{cm} = \frac{-\frac{a^2}{4}\cdot\frac{a}{2}}{\pi a^2 - \frac{a^2}{4}} = -\frac{a}{8\pi - 2}
  3. 8π−2≈23.18\pi - 2 \approx 23.1
  4. So X≈23X \approx 23.
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