JEE Mains · Physics · STD 11- Mechanical properties of solids
- JEE Mains 2024
- Hard
- MCQ
Young's modulus is determined by the equation given by \(\mathrm{Y}=49000 \frac{\mathrm{m}}{\ell} \frac{\text { dyne }}{\mathrm{cm}^2}\) where \(\mathrm{M}\) is the mass and \(\ell\) is the extension of wre used in the experiment. Now error in Young modules \((\mathrm{Y})\) is estimated by taking data from \(M-\ell\) plot in graph paper. The smallest scale divisions are \(5 \mathrm{~g}\) and \(0.02\) \(\mathrm{cm}\) along load axis and extension axis respectively. If the value of \(M\) and \(\ell\) are \(500 \mathrm{~g}\) and \(2 \mathrm{~cm}\) respectively then percentage error of \(\mathrm{Y}\) is _______.
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(2 \%\)
Step-by-step Solution
Detailed explanation
\(\frac{\Delta \mathrm{Y}}{\mathrm{Y}} =\frac{\Delta \mathrm{m}}{\mathrm{m}}+\frac{\Delta \ell}{\ell}\)
\(=\frac{5}{500}+\frac{0.02}{2}=0.01+0.01\)
\(\frac{\Delta \mathrm{Y}}{\mathrm{Y}} =0.02 \Rightarrow \% \frac{\Delta \mathrm{Y}}{\mathrm{Y}}=2 \%\)
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