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JEE Main 2022 (26 Jul, Shift 2) · Wire Combinations & Special Cases

Wire Combinations & Special Cases question from JEE Main 2022 (26 Jul, Shift 2)

Q14JEE Main 2022 · 26 JUL · SHIFT 2Medium

Wire Combinations & Special Cases

A uniform heavy rod of mass 20 kg20\text{ kg}, cross sectional area 0.4 m20.4\text{ m}^2 and length 20 m20\text{ m} is hanging from a fixed support. Neglecting the lateral contraction, the elongation in the rod due to its own weight is x×10−9 mx \times 10^{-9}\text{ m}. The value of xx is _______________.

(Given, young modulus Y = 2 ×\times 1011 Nm−2 and g = 10 ms−2)

Numerical answer type. Work it out, then open the answer.

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Answer: 25

  1. Own-weight elongation: Δl=WL2AY\Delta l = \frac{WL}{2AY} with W=20×10=200 NW = 20 \times 10 = 200\text{ N}
  2. Δl=200×202×0.4×2×1011=40001.6×1011=2.5×10−8 m\Delta l = \frac{200 \times 20}{2 \times 0.4 \times 2 \times 10^{11}} = \frac{4000}{1.6 \times 10^{11}} = 2.5 \times 10^{-8}\text{ m}
  3. Δl=25×10−9 m\Delta l = 25 \times 10^{-9}\text{ m}, so x=25x = 25.
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