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JEE Main 2007 (AIEEE) · Composite & Cut-Out Bodies

Composite & Cut-Out Bodies question from JEE Main 2007 (AIEEE)

Q13JEE Main 2007 · AIEEEEasy

Composite & Cut-Out Bodies

A circular disc of radius RR is removed from a bigger circular disc of radius 2R2R such that the circumferences of the discs coincide. The center of mass of the new disc is αR\alpha R from the center of the bigger disc. The value of α\alpha is

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Answer: (b)

OholeRCMCM at R/3 left of O
  1. Mass ∝\propto area: big disc 4m4m at O, removed disc mm with centre at distance RR from O.
  2. Treat the hole as mass −m-m: xcm=4m(0)−m(R)4m−m=−R3x_{cm} = \frac{4m(0) - m(R)}{4m - m} = -\frac{R}{3}
  3. The CM is R3\frac{R}{3} from O, on the side away from the hole, so α=13\alpha = \frac{1}{3}.
  4. So the answer is (b) 1/31/3.
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