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

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

Four identical hollow cylindrical columns of mild steel support a...

Four identical hollow cylindrical columns of mild steel support a big structure of mass 50×103kg.50 \times {10}^{3}kg . The inner and outer radii of each column are 50cm50cm and 100cm100cm respectively. Assuming uniform local distribution, calculate the compression strain of each column.
[use Y=2.0×1011Pa,g=9.8ms−2Y = 2 . 0 \times {10}^{11}Pa, g = 9 . 8m{s}^{- 2}]

Options

  1. A1.87×10−31 . 87 \times {10}^{- 3}
  2. B2.60×10−72 . 60 \times {10}^{- 7}
  3. C3.60×10−83 . 60 \times {10}^{- 8}
  4. D7.07×10−47 . 07 \times {10}^{- 4}

Correct answer

B. 2.60×10−72 . 60 \times {10}^{- 7}

Step-by-step solution

\DeltaLL=mg4AY\dfrac{\DeltaL}{L} = \dfrac{mg}{4AY}
=50×103×9.84π(1002−502)×10−4×2×1011=5×1054×π×150×50×10−4×2×1011= \dfrac{50 \times {10}^{3} \times 9 . 8}{4\pi\left({100}^{2} - {50}^{2}\right) \times {10}^{- 4} \times 2 \times {10}^{11}} = \dfrac{5 \times {10}^{5}}{4 \times \pi \times 150 \times 50 \times {10}^{- 4} \times 2 \times {10}^{11}}
ΔLL=112π×10−5=2.60×10−7\dfrac{\Delta L}{L} = \dfrac{1}{12\pi} \times {10}^{- 5} = 2 . 60 \times {10}^{- 7}
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