Physics - Elasticity Question with Solution | TestHub

The wire is stretched to double its length. Therefore, if the length of the wore was $l$, change in the length $∆l=l$.

Applying Hooke's law,

$$\begin{gathered} Y=\frac{\text{stress}}{\text{strain}}=\frac{Fl}{A∆l} \\ \Rightarrow F=\frac{YA∆l}{l}=YA \\ \Rightarrow F=2\times {10}^{11}\times 1\times {10}^{-4}=2\times {10}^7 \mathrm{N} \end{gathered}$$

$=2\times {10}^{11}\times {10}^{-4}\left(\frac{2l-l}{l}\right)$

$=2\times {10}^7 \mathrm{N}$