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JEE Main 201910 AprilEvening Shift
Exam
JEE Main
Year
2019
Date
10 April
Shift
Evening Shift
Subject
Physics
Chapter
Mechanical Properties of Solids

JEE Main 2019 · 10 April · Evening Shift · Physics · Mechanical Properties of Solids

In an experiment, brass and steel wires of length 1m each with...

In an experiment, brass and steel wires of length 1m1m each with areas of cross section 1mm21m{m}^{2} are used. The wires are connected in series and one end of the combined wire is connected to a rigid support and other end is subjected to elongation. The stress required to produce a net elongation of 0.2mm0.2mm is,
[Given, the Young's Modulus for steel and brass are, respectively, 120×109N/m2120 \times {10}^{9}N / {m}^{2} and 60×109N/m260 \times {10}^{9}N / {m}^{2} ]

Options

  1. A8.0×106N/m28.0 \times {10}^{6}N / {m}^{2}
  2. B1.2×106N/m21.2 \times {10}^{6}N / {m}^{2}
  3. C0.2×106N/m20.2 \times {10}^{6}N / {m}^{2}
  4. D1.8×106N/m21.8 \times {10}^{6}N / {m}^{2}

Correct answer

A. 8.0×106N/m28.0 \times {10}^{6}N / {m}^{2}

Step-by-step solution

Stress will be same in both rods
Stress =Y1Δl1l=Y2Δl2l= {Y}_{1}\dfrac{\Delta{l}_{1}}{l} = {Y}_{2}\dfrac{\Delta{l}_{2}}{l} … (1)
And
Δl1+Δl2=0.2×10−3m\Delta{l}_{1} + \Delta{l}_{2} = 0.2 \times {10}^{- 3}m … (2)
From equation (1) and (2)
Stress ×(lY1+lY2)=0.2×10−3m\times \left(\dfrac{l}{{Y}_{1}} + \dfrac{l}{{Y}_{2}}\right) = 0 .2 \times {10}^{- 3}m
Stress ×(1120×109+160×109)=0.2×10−3m\times \left(\dfrac{1}{120 \times {10}^{9}} + \dfrac{1}{60 \times {10}^{9}}\right) = 0.2 \times {10}^{- 3}m
Stress =8×106N/m2= 8 \times {10}^{6}N / {m}^{2}
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