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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

Options

  1. A7 3 m
  2. B12 7 m
  3. C10 7 m
  4. D9 7 m

Correct answer

B. 12 7 m

Step-by-step solution

Let rod is placed along x -axis. Mass of element P Q of length d x situated at x = x is d m = λ d x = 2 + x d x The CM of the element has coordinates ( x , 0, 0). Therefore, x -coordinates of CM of the rod will be x C M = ∫ 0 3 x d m ∫ 0 3 d m = ∫ 0 3 x 2 + x d x ∫ 0 3 2 + x d x = ∫ 0 3 2 x + x 2 d x ∫ 0 3 2 + x d x = 2 x 2 2 + x 3 3 0 3 2 x + x 2 2 0 3 = 3 2 + 3 3 3 2 × 3 + 3 2 2 = 9 + 9 6 + 9 / 2 = 18 × 2 21 = 12 7 m

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