### Related Formula The Young's modulus Y$Y$ is: Y = fractextStresstextStrain = fracF / ADelta l / l = fracF lA Delta l$Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F / A}{\Delta l / l} = \frac{F l}{A \Delta l}$ Solving for extension Delta l$\Delta l$: Delta l = fracF lA Y$\Delta l = \frac{F l}{A Y}$ Since volume V = A l$V = A l$, we substitute l = fracVA$l = \frac{V}{A}$: Delta l = fracF VA^2 Y$\Delta l = \frac{F V}{A^2 Y}$ ### Core Logic We are given that the wires are made of the same material (Y$Y$ is constant) and have the same volume (V$V$ is constant). Also, we want to achieve the same extension (Delta l$\Delta l$ is constant). Thus, from the formula: Delta l propto fracFA^2 implies F propto A^2$\Delta l \propto \frac{F}{A^2} \implies F \propto A^2$ ### Step 1: Calculate the Force Ratio Using the proportionality for both wires: fracF_1F_2 = left( fracA_1A_2 right)^2$\frac{F_1}{F_2} = \left( \frac{A_1}{A_2} \right)^2$ Given the cross-sectional area ratio is 4:1$4:1$: fracA_1A_2 = frac41$\frac{A_1}{A_2} = \frac{4}{1}$ Substitute this into the ratio equation: fracF_1F_2 = left( frac41 right)^2 = 16$\frac{F_1}{F_2} = \left( \frac{4}{1} \right)^2 = 16$ ### Pattern Recognition For a constant volume and material, extension Delta l propto fracFA^2$\Delta l \propto \frac{F}{A^2}$. Since area ratio is 4$4$, the required force ratio is 4^2 = 16$4^2 = 16$. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Physics: Mechanical Properties of Solids