- 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 \(\left.g =10\,m / s ^{2}\right]\)
Answer & Solution
(C) \(2\)
Step-by-step Solution
Detailed explanation
\(\Delta L =0.4 \times 10^{-3} m\)
\(m =1\,kg\)
\(d =0.4 \times 10^{-3}\,m\)
\(\frac{ F }{ A }= Y \frac{\Delta L }{ L }\)
\(Y =\frac{ FL }{ A \Delta L }=\frac{( mg ) \cdot(1)}{\left(\frac{\pi d ^{2}}{4}\right) 0.4 \times 10^{-3}}\)
\(\Rightarrow \frac{10 \times 4}{\pi\left(0.4 \times 10^{-3}\right)^{2} \times 0.4 \times 10^{-3}}\)
\(Y =\frac{40}{\pi\left(0.4 \times 10^{-3}\right)^{3}}\)
\(Y =\frac{40 \times 7}{22 \times 64 \times 10^{-3} \times 10^{-9}}\)
\(Y =0.199 \times 10^{-12} N / m ^{2}\)
\(\frac{\Delta Y }{ Y }=\frac{\Delta F }{ F }+\frac{\Delta L }{ L }+\frac{\Delta A }{ A }+\frac{\Delta(\Delta L )}{(\Delta L )}\)
\(=\frac{0.02}{0.4}+2 \frac{\Delta d }{ d }=\frac{0.2}{4}+2 \times \frac{0.01}{0.4}\)
\(=\frac{0.1}{2}+\frac{0.1}{2}=0.1\)
\(\Rightarrow \Delta Y =0.1 \times Y\)
\(=0.199 \times 10^{11}=1.99 \times 10^{10}\)
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