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JEE Main 2022 (25 Jul, Shift 2) · Centre of Mass of Discrete Particles

Centre of Mass of Discrete Particles question from JEE Main 2022 (25 Jul, Shift 2)

Q13JEE Main 2022 · 25 JUL · SHIFT 2Easy

Centre of Mass of Discrete Particles

Three identical spheres each of mass M are placed at the corners of a right angled triangle with mutually perpendicular sides equal to 3 m each. Taking point of intersection of mutually perpendicular sides as origin, the magnitude of position vector of centre of mass of the system will be x\sqrt{x} m. The value of x is ____________.

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

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Answer: 2

  1. Masses MM at (0,0)(0,0), (3,0)(3,0) and (0,3)(0,3) m.
  2. xcm=33=1x_{cm} = \frac{3}{3} = 1 m, ycm=1y_{cm} = 1 m
  3. ∣r⃗cm∣=12+12=2|\vec{r}_{cm}| = \sqrt{1^2 + 1^2} = \sqrt{2} m
  4. So x=2x = 2.
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