To solve this problem, we need to understand the relationship between the extension $\Delta l$ and the length $l$ of the wire.$\\$Given:$\\$• Volume of copper $V$ is constant.$\\$• Length of the wire $l. \\$• Force applied $F. \\$• Extension produced $\Delta l. \\$The volume of the wire is given by:$\\ V = A \cdot l \\$where $A$ is the cross-sectional area of the wire. Since $V$ is constant, we have:$\\ A = \frac{V}{l} \\$The extension $\Delta l$ is given by Hooke's Law:$\\ \Delta l = \frac{F \cdot l}{A \cdot Y} \\$where $Y$ is the Young's modulus of the material. Substitute $A = \frac{V}{l}: \\ \Delta l = \frac{F \cdot l}{\left(\frac{V}{l}\right) \cdot Y} \\$Simplify the expression:$\\ \Delta l = \frac{F \cdot l^2}{V \cdot Y} \\$This shows that $\Delta l$ is proportional to $l^2. \\$Therefore, the graph of $\Delta l$ versus $l^2$ will be a straight line.$\\$Thus, the correct option is Option 2: $\Delta l$ versus $l^2.$