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The maximum elongation of a steel wire of 1 m length if the elastic limit of steel and its Young's modulus, respectively, are $8 \times 10^8 \mathrm{~N} \mathrm{~m}^{-2}$ and $2 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$, is:
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Verified Answer
The correct answer is:
4 mm
In the case for maximum elongation,
Stress $=$ Elastic limit
$\begin{aligned}
\delta_{\max }=\frac{\sigma_{\text {elastic }} \times L}{\text { Young's modulus }}=\frac{8 \times 10^8 \times 1}{2 \times 10^{11}} & =4 \times 10^{-3} \\
& =4 \mathrm{~mm}
\end{aligned}$
i.e. maximum elongation is 4 mm
Stress $=$ Elastic limit
$\begin{aligned}
\delta_{\max }=\frac{\sigma_{\text {elastic }} \times L}{\text { Young's modulus }}=\frac{8 \times 10^8 \times 1}{2 \times 10^{11}} & =4 \times 10^{-3} \\
& =4 \mathrm{~mm}
\end{aligned}$
i.e. maximum elongation is 4 mm
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