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JEE Main 2016 · 10 Apr · Shift 1, Physics question 26

A thin 1 m long rod has a radius of 5 mm. A force of 50 πkN 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 ?

Choose one option
Answer and worked solution

Official answer: B

Y=FLπr2ΔLY=\frac{FL}{\pi r^2\Delta L} and ΔYY=δLL+2δrr+δ(ΔL)ΔL\frac{\Delta Y}Y=\frac{\delta L}L+\frac{2\delta r}r+\frac{\delta(\Delta L)}{\Delta L}.

  • The length L = 1 m has the smallest relative error (10−510^{-5}), so (1) is true.
  • The extension (strain) has by far the largest relative error, so (4) is true.
  • The stress is 50π×103π(5×10−3)2=2×109\frac{50\pi\times10^3}{\pi(5\times10^{-3})^2}=2\times10^9 Pa; with the smallest measurable extension 10−510^{-5} m, Ymax∼2×1014Y_{max}\sim2\times10^{14} N/m², of the order of (3).

Statement (2) is taken as the false one in this derived key. This is a judgement call: the wording "figure of merit" is not standard here.

Topics:Errors in experimentsHooke's law and elastic moduli
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