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NEET 2020 (Phase 2) · Centre of Mass of Discrete Particles

Centre of Mass of Discrete Particles question from NEET 2020 (Phase 2)

Q16NEET 2020 · PHASE 2Easy

Centre of Mass of Discrete Particles

Three identical spheres, each of mass MM, are placed at the corners of a right angle triangle with mutually perpendicular sides equal to 2 m (see figure). Taking the point of intersection of the two mutually perpendicular sides as the origin, find the position vector of centre of mass.

Figure

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Answer: (d)

  1. Masses MM at (0,0)(0,0), (2,0)(2,0) and (0,2)(0,2) m.
  2. xcm=M(0+2+0)3M=23x_{cm} = \frac{M(0 + 2 + 0)}{3M} = \frac{2}{3} m, ycm=M(0+0+2)3M=23y_{cm} = \frac{M(0 + 0 + 2)}{3M} = \frac{2}{3} m
  3. So the answer is (d) 23(i^+j^)\frac{2}{3}(\hat{i} + \hat{j}).
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