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JEE Main 2024 (8 Apr, Shift 1) · Wire Combinations & Special Cases

Wire Combinations & Special Cases question from JEE Main 2024 (8 Apr, Shift 1)

Q9JEE Main 2024 · 8 APR · SHIFT 1Difficult

Wire Combinations & Special Cases

Young's modulus is determined by the equation given by Y=49000ml dynecm2\text{Y} = 49000\frac{\text{m}}{\text{l}}\,\frac{\text{dyne}}{\text{cm}^2} where MM is the mass and ll is the extension of wire used in the experiment. Now error in Young modules (Y)(Y) is estimated by taking data from M−lM - l plot in graph paper. The smallest scale divisions are 5 g5\text{ g} and 0.02 cm0.02\text{ cm} along load axis and extension axis respectively. If the value of MM and ll are 500 g500\text{ g} and 2 cm2\text{ cm} respectively then percentage error of YY is:

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Answer: (a)

  1. Y∝MlY \propto \dfrac{M}{l}, so ΔYY=ΔMM+Δll\dfrac{\Delta Y}{Y} = \dfrac{\Delta M}{M} + \dfrac{\Delta l}{l}.
  2. ΔMM=5500=1%\dfrac{\Delta M}{M} = \dfrac{5}{500} = 1\%, Δll=0.022=1%\dfrac{\Delta l}{l} = \dfrac{0.02}{2} = 1\%
  3. So the answer is (a) 2%.
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