Physics - Elasticity Question with Solution | TestHub

The cross-sectional area of a wire of radius $r=1\mathrm{mm}=10^{-3}\mathrm{m}$ is $A=\pi r^2=3.14\times 10^{-6}{\mathrm{m}}^2$ The maximum force that can be applied without exceeding the elastic limit is therefore $$\begin{array}{rl}& F={\left(\frac{F}{A}\right)}_{\max }\times (A)\\ & =\left(2.5\times 10^8\mathrm{N}{\mathrm{m}}^{-2}\right)\left(3.14\times 10^{-6}{\mathrm{m}}^2\right)=785\mathrm{N}\end{array}$$ When this force is applied, the wire will stretch by $$\begin{array}{rl}& \Delta L=\frac{L}{Y}\frac{F}{A}=\frac{(3\mathrm{m})(785\mathrm{N})}{\left(2\times 10^{11}\mathrm{N}/{\mathrm{m}}^2\right)\left(3.14\times 10^{-6}{\mathrm{m}}^2\right)}\\ & =3.75\times 10^{-3}\mathrm{m}=3.75\mathrm{mm}\end{array}$$