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JEE Main 202131 AugustMorning Shift
Exam
JEE Main
Year
2021
Date
31 August
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2021 · 31 August · Morning Shift · Physics · Mechanical Properties of Solids

A uniform heavy rod of weight 10kg ms ^ - 2 , cross-sectional area...

A uniform heavy rod of weight 10kgms−2,10kg{ms}^{- 2}, cross-sectional area 100cm2100{cm}^{2} and length 20cm20cm is hanging from a fixed support. Young modulus of the material of the rod is 2×1011Nm−2.2 \times {10}^{11}N{m}^{- 2} . Neglecting the lateral contraction, find the elongation of rod due to its own weight:

Options

  1. A5×10−10m5 \times {10}^{- 10}m
  2. B4×10−8m4 \times {10}^{- 8}m
  3. C5×10−8m5 \times {10}^{- 8}m
  4. D2×10−9m2 \times {10}^{- 9}m

Correct answer

A. 5×10−10m5 \times {10}^{- 10}m

Step-by-step solution

Elongation due to self weight,
Wl2AY=10×0.22×2×1011×100×10−4=5×10−10m\dfrac{Wl}{2AY} = \dfrac{10 \times 0 . 2}{2 \times 2 \times {10}^{11} \times 100 \times {10}^{- 4}} = 5 \times {10}^{- 10}m
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