Elasticity

Step 1: Formula for ElongationThe amount a wire stretches depends on Young's modulus formula:$Y=\;\frac{F l}{\Delta l A}=\;\frac{4 F l}{\Delta l \pi d^2}$Step 2: What stays the same?Both wires are made of the same material, have the same length, and the same stretching force. This means their Young's modulus $(Y)$, original length $(l)$, and force $(F)$ are the same.Step 3: Setting up the equation for both wiresSince Young's modulus is the same for both wires, we set the formulas equal for each wire:$Y_1=Y_2$$\;\frac{4 F l}{\Delta l_1 \pi d_1^2}=\;\frac{4 F l}{\Delta l_2 \pi d_2^2}$Step 4: Rearranging to compare elongationsWe can simplify to find how the elongation ( $\Delta l$ ) relates to diameter ( $d$ ):$\Delta l_1 \cdot d_1^2 = \Delta l_2 \cdot d_2^2$So,$\Delta l_2 = \Delta l_1 \cdot \left( \;\frac{d_1}{d_2} \right)^2$Step 5: Substitute valuesWe are told $\Delta l_1 = 1\,\text{mm}$ (the first wire stretches 1 mm ) and $d_2 = 4 d_1$ (the second wire's diameter is 4 times bigger).$\Delta l_2 = 1 \cdot \left( \;\frac{d_1}{4 d_1} \right)^2 = 1 \cdot \left( \;\frac{1}{4} \right)^2 = \;\frac{1}{16} \text{mm}$Final AnswerThe second, thicker wire stretches only $\;\frac{1}{16} \text{mm}$ when the same force is applied.