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JEE Main 20198 AprilMorning Shift
Exam
JEE Main
Year
2019
Date
8 April
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2019 · 8 April · Morning Shift · Physics · Mechanical Properties of Solids

A steel wire having a radius of 2.0mm , carrying a load of 4kg, is...

A steel wire having a radius of 2.0mm2.0mm , carrying a load of 4kg,4kg, is hanging from a ceiling. Given that g=3.1\pims−2,g = 3.1\pim{s}^{- 2}, what will be the tensile stress that would be developed in the wire?

Options

  1. A5.2×106Nm−25.2 \times {10}^{6}N{m}^{- 2}
  2. B6.2×106Nm−26.2 \times {10}^{6}N{m}^{- 2}
  3. C4.8×106Nm−24 . 8 \times {10}^{6}N{m}^{- 2}
  4. D3.1×106Nm−23.1 \times {10}^{6}N{m}^{- 2}

Correct answer

D. 3.1×106Nm−23.1 \times {10}^{6}N{m}^{- 2}

Step-by-step solution

Stress =mgπr2=4×3.1×ππ×(2×10−3)2<br/>= \dfrac{mg}{\pi{r}^{2}} = \dfrac{4 \times 3.1 \times \pi}{\pi \times {\left(2 \times {10}^{- 3}\right)}^{2}} <br/>= 3.1 \times {10}^{6}N / {m}^{2}
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