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JEE Main 2016 (10 Apr, Shift 1) · Wire Combinations & Special Cases

Wire Combinations & Special Cases question from JEE Main 2016 (10 Apr, Shift 1)

Q27JEE Main 2016 · 10 APR · SHIFT 1Difficult

Wire Combinations & Special Cases

A thin 1 m long rod has a radius of 5 mm. A force of 50 π\pikN is applied at one end to determine its Young’s modulus. Assume that the force is exactly known. If the least count in the measurement of all lengths is 0.01 mm, which of the following statements is false?

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Answer: (c)

  1. Stress =Fπr2=50π×103π(5×10−3)2=2×109 N/m2= \dfrac{F}{\pi r^2} = \dfrac{50\pi \times 10^3}{\pi (5 \times 10^{-3})^2} = 2 \times 10^9\text{ N/m}^2.
  2. Smallest measurable extension is the least count 0.01 mm=10−5 m0.01\text{ mm} = 10^{-5}\text{ m}, so minimum strain =10−5/1=10−5= 10^{-5}/1 = 10^{-5}.
  3. Maximum Y=2×10910−5=2×1014 N/m2Y = \dfrac{2 \times 10^9}{10^{-5}} = 2 \times 10^{14}\text{ N/m}^2, not 101410^{14}, so (c) is false.
  4. Fractional errors: length 10−510^{-5} (smallest, so largest figure of merit), radius 2×0.015=4×10−32 \times \dfrac{0.01}{5} = 4 \times 10^{-3}, extension largest. So (a), (b), (d) are true.
  5. So the answer is (c).
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