The elastic potential energy stored in a steel wire of length $20 \mathrm{~m}$ stretched through $2 \mathrm{~cm}$ is $80 \mathrm{~J}$. The cross sectional area of the wire is __________ $\mathrm{mm}^{2}$.

$\left(\right.$ Given, $\left.y=2.0 \times 10^{11} \mathrm{Nm}^{-2}\right)$

Given, energy per unit volume = $\frac{1}{2} \times \text{stress} \times \text{strain}$ <br/><br/> The stress can be given as $\text{stress} = Y \times \text{strain}$, where Y is the Young's modulus. <br/><br/> The energy stored in the wire can be written as: <br/><br/> $$\text{Energy} = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}$$ <br/><br/> Substituting the stress formula, we get: <br/><br/> $\text{Energy} = \frac{1}{2} \times Y \times \text{strain}^2 \times A \times L$ <br/><br/> We are given that the energy stored is $80 \ \text{J}$, the original length of the wire is $20 \ \text{m}$, the elongation is $2 \ \text{cm}$, and the Young's modulus is $2.0 \times 10^{11} \ \text{Nm}^{-2}$. We need to find the cross-sectional area (A) of the wire. <br/><br/> $$80 = \frac{1}{2} \times 2 \times 10^{11} \times \left(\frac{2 \times 10^{-2}}{20}\right)^2 \times A \times 20$$ <br/><br/> Now we can solve for A: <br/><br/> $$A = \frac{80 \times 20^2}{(2.0 \times 10^{11}) \times (2 \times 10^{-2})^2} = 40 \times 10^{-6} \ \text{m}^2$$ <br/><br/> To convert the area to $\text{mm}^2$, we multiply by $10^6$: <br/><br/> $A = 40 \times 10^{-6} \times 10^6 = 40 \ \text{mm}^2$ <br/><br/> So, the cross-sectional area of the wire is $40 \ \text{mm}^2$.