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JEE Mains · Physics · STD 11 - Units,dimensions and measurement
  • JEE Mains 2022
  • Hard
  • Numerical

In an experiment of determine the Young's modulus of wire of a length exactly \(1\; m\), the extension in the length of the wire is measured as \(0.4\,mm\) with an uncertainty of \(\pm 0.02\,mm\) when a load of \(1\,kg\) is applied. The diameter of the wire is measured as \(0.4\,mm\) with an uncertainty of \(\pm 0.01\,mm\). The error in the measurement of Young's modulus \((\Delta Y)\) is found to be \(x \times 10^{10}\,Nm ^{-2}\). The value of \(x\) is \(\left[\right.\) Take \(\left.g =10\,m / s ^{2}\right]\)

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Correct Answer

(C) \(2\)

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Detailed explanation

\(L =1\,m\)

\(\Delta L =0.4 \times 10^{-3} m\)

\(m =1\,kg\)

\(d =0.4 \times 10^{-3}\,m\)

\(\frac{ F }{ A }= Y \frac{\Delta L }{ L }\)

\(Y =\frac{ FL }{ A \Delta L }=\frac{( mg ) \cdot(1)}{\left(\frac{\pi d ^{2}}{4}\right) 0.4 \times 10^{-3}}\)

\(\Rightarrow \frac{10 \times 4}{\pi\left(0.4 \times 10^{-3}\right)^{2} \times 0.4 \times 10^{-3}}\)

\(Y =\frac{40}{\pi\left(0.4 \times 10^{-3}\right)^{3}}\)

\(Y =\frac{40 \times 7}{22 \times 64 \times 10^{-3} \times 10^{-9}}\)

\(Y =0.199 \times 10^{-12} N / m ^{2}\)

\(\frac{\Delta Y }{ Y }=\frac{\Delta F }{ F }+\frac{\Delta L }{ L }+\frac{\Delta A }{ A }+\frac{\Delta(\Delta L )}{(\Delta L )}\)

\(=\frac{0.02}{0.4}+2 \frac{\Delta d }{ d }=\frac{0.2}{4}+2 \times \frac{0.01}{0.4}\)

\(=\frac{0.1}{2}+\frac{0.1}{2}=0.1\)

\(\Rightarrow \Delta Y =0.1 \times Y\)

\(=0.199 \times 10^{11}=1.99 \times 10^{10}\)

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