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JEE Main202131 Aug 2021Morning ShiftPhysicsMechanical Properties of SolidsActual

A uniform heavy rod of weight 10 kg m s - 2 , cross-sectional area 100 cm 2 and length 20 cm is hanging from a fixed support. Young modulus of the material of the rod is 2 × 10 11 N m - 2 . Neglecting the lateral contraction, find the elongation of rod due to its own weight:

Options

  1. A5 × 10 - 10   m
  2. B4 × 10 - 8   m
  3. C5 × 10 - 8   m
  4. D2 × 10 - 9   m

Correct answer

A. 5 × 10 - 10   m

Step-by-step solution

Elongation due to self weight, W l 2 A Y = 10 × 0 . 2 2 × 2 × 10 11 × 100 × 10 - 4 = 5 × 10 - 10   m

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