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JEE Main 2019 (10 Apr, Shift 2) · Wire Combinations & Special Cases

Wire Combinations & Special Cases question from JEE Main 2019 (10 Apr, Shift 2)

Q24JEE Main 2019 · 10 APR · SHIFT 2Medium

Wire Combinations & Special Cases

In an experiment, brass and steel wires of length 1 m each with areas of cross section 1mm2 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 0.2 mm is, [Given, the Young's Modulus for steel and brass are, respectively, 120 × 109 N/m2 and 60 × 109 N/m2]

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Answer: (a)

  1. Series wires with equal area have the same stress σ\sigma; Δl=σ(LsYs+LbYb)\Delta l = \sigma\left(\frac{L_s}{Y_s} + \frac{L_b}{Y_b}\right)
  2. 0.2×10−3=σ(1120×109+160×109)=σ40×1090.2 \times 10^{-3} = \sigma\left(\frac{1}{120 \times 10^{9}} + \frac{1}{60 \times 10^{9}}\right) = \frac{\sigma}{40 \times 10^{9}}
  3. σ=8×106 N/m2\sigma = 8 \times 10^{6}\text{ N/m}^2, so the answer is (a).
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