Question

What force is required to stretch a copper wire $1c{m}^2$ in cross-section to double its length? $Y$ for copper is $1.26\times {10}^{12}dynec{m}^{-2}$

Answer 1

Given that,

Young modulus $Y=1.26\times {10}^{12}dyne/cm$

Area $A=1c{m}^2$

As the length of the wire is doubled, the change in length is equal to its
original length.
If L is the original length

Then, $\Delta l=l$

We know that,

$Y=\frac{F\times L}{A\times \Delta L}$

$F=Y\times A$

$F=0.126\times {10}^{12}\times 1\times {10}^{-4}$

$F=1.2\times {10}^{7}N$

Hence, the force is $1.2\times {10}^{7}N$