The force required to stretch a wire of cross-section 1 cm ^ 2 to...

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,

$Y = \dfrac{\text{stress}}{\text{strain}} = \dfrac{Fl}{A ∆ l} ⇒ F = \dfrac{YA ∆ l}{l} = YA ⇒ F = 2 \times {10}^{11} \times 1 \times {10}^{- 4} = 2 \times {10}^{7}N$

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

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