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Question: Answered & Verified by Expert
If linear density of a rod of length 3 m varies as λ=2+x, then the position of the centre of gravity of the rod is
PhysicsCenter of Mass Momentum and CollisionJEE Main
Options:
  • A 73m
  • B 127m
  • C 107m
  • D 97m
Solution:
1614 Upvotes Verified Answer
The correct answer is: 127m
Let rod is placed along x-axis. Mass of element PQ of length dx situated at x=x is




dm=λ dx=2+x dx

The CM of the element has coordinates (x, 0, 0).

Therefore, x-coordinates of CM of the rod will be

xCM=03xdm03dm

=03x2+xdx032+xdx

=032x+x2dx032+xdx

=2x22+x33032x+x2203

=32+3332×3+322=9+96+9/2

=18×221=127m

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