To solve this problem, we will use the concept of Young's modulus and the relation between stress and strain.$\\$Given:$\\$• Two wires made of the same material, hence they have the same Young's modulus $(Y). \\$• Both wires have the same volume.$\\$• Cross-sectional area of the first wire is $A$ and its length is $L_1. \\$• Cross-sectional area of the second wire is $3A$ and its length is $L_2. \\$• The first wire is stretched by $\Delta l$ with force $F. \\$Since both wires have the same volume:$\\ A \cdot L_1 = 3A \cdot L_2 \\$From the above equation, we can find the relation between $L_1$ and $L_2: \\ L_1 = 3L_2 \\$The formula for Young's modulus $(Y)$ is given by:$\\ Y = \frac{Stress}{Strain} = \frac{\frac{F}{A}}{\frac{\Delta l}{L_1}} \\$Rearranging the formula, we get:$\\ F = \frac{Y \cdot A \cdot \Delta l}{L_1} \\$For the second wire, we need to find the force $F_2$ to stretch it by the same amount $\Delta l \\ F_2 = \frac{Y \cdot 3A \cdot \Delta l}{L_2} \\$Substitute $L_2 = \frac{L_1}{3}$ into the equation for $F_2: \\ F_2 = \frac{Y \cdot 3A \cdot \Delta l}{\frac{L_1}{3}} \\$Simplify the expression:$\\ F_2 = \frac{Y \cdot 3A \cdot \Delta l \cdot 3}{L_1} \\ F_2 = \frac{9Y \cdot A \cdot \Delta l}{L_1} \\$Since $F = \frac{Y \cdot A \cdot \Delta l}{L_1},$ we can write:$\\ F_2 = 9F \\$Therefore, the force needed to stretch the second wire by the same amount is $9F. \\$This corresponds to Option 3.