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JEE Main 2022 (26 Jul, Shift 1) · Wire Combinations & Special Cases

Wire Combinations & Special Cases question from JEE Main 2022 (26 Jul, Shift 1)

Q17JEE Main 2022 · 26 JUL · SHIFT 1Difficult

Wire Combinations & Special Cases

In an experiment to determine the Young's modulus of wire of a length exactly 1 m1\text{ m}, the extension in the length of the wire is measured as 0.4 mm0.4\text{ mm} with an uncertainty of ±0.02 mm\pm 0.02\text{ mm} when a load of 1 kg1\text{ kg} is applied. The diameter of the wire is measured as 0.4 mm0.4\text{ mm} with an uncertainty of ±0.01 mm\pm 0.01\text{ mm}. The error in the measurement of Young's modulus (ΔY)(\Delta \text{Y}) is found to be x×1010 Nm−2x \times 10^{10}\text{ Nm}^{-2}. The value of xx is _________________. ((take g=10 ms−2\text{g} = 10\text{ ms}^{-2})

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

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Answer: 2

  1. Y=4MgLπd2l=4×10×1π(0.4×10−3)2(0.4×10−3)≈2×1011 N m−2Y = \dfrac{4MgL}{\pi d^2 l} = \dfrac{4 \times 10 \times 1}{\pi (0.4 \times 10^{-3})^2 (0.4 \times 10^{-3})} \approx 2 \times 10^{11}\text{ N m}^{-2}
  2. ΔYY=2Δdd+Δll=2×0.010.4+0.020.4=0.1\dfrac{\Delta Y}{Y} = 2\dfrac{\Delta d}{d} + \dfrac{\Delta l}{l} = 2 \times \dfrac{0.01}{0.4} + \dfrac{0.02}{0.4} = 0.1
  3. ΔY=0.1×2×1011=2×1010 N m−2\Delta Y = 0.1 \times 2 \times 10^{11} = 2 \times 10^{10}\text{ N m}^{-2}
  4. So x=2x = 2.
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