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JEE Main 2022 (27 Jul, Shift 2) · Wire Combinations & Special Cases

Wire Combinations & Special Cases question from JEE Main 2022 (27 Jul, Shift 2)

Q13JEE Main 2022 · 27 JUL · SHIFT 2Medium

Wire Combinations & Special Cases

A steel wire of length 3.2 m(Ys=2.0×1011 N m−2)3.2\text{ m} \left(Y_s = 2.0 \times 10^{11}\text{ N m}^{-2}\right) and a copper wire of length 4.4 m(Yc=1.1×1011 N m−2)4.4\text{ m} \left(Y_c = 1.1 \times 10^{11}\text{ N m}^{-2}\right), both of radius 1.4 mm1.4\text{ mm} are connected end to end. When stretched by a load, the net elongation is found to be 1.4 mm1.4\text{ mm}. The load applied, in Newton, will be: (( Given π=227\pi = \frac{22}{7})

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Answer: (d)

  1. Same load FF and area A=227×(1.4×10−3)2=6.16×10−6 m2A = \frac{22}{7} \times (1.4 \times 10^{-3})^2 = 6.16 \times 10^{-6}\text{ m}^2
  2. Δl=FA(LsYs+LcYc)=FA(1.6+4.0)×10−11=FA×5.6×10−11\Delta l = \frac{F}{A}\left(\frac{L_s}{Y_s} + \frac{L_c}{Y_c}\right) = \frac{F}{A}(1.6 + 4.0) \times 10^{-11} = \frac{F}{A} \times 5.6 \times 10^{-11}
  3. F=1.4×10−3×6.16×10−65.6×10−11=154 NF = \frac{1.4 \times 10^{-3} \times 6.16 \times 10^{-6}}{5.6 \times 10^{-11}} = 154\text{ N}
  4. So the answer is (d) 154.
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