Two wires are made of the same material and have the same volume. The first wire has a cross-sectional area $A

Question

Two wires are made of the same material and have the same volume. The first wire has a cross-sectional area $A$ and the second wire has a cross-sectional area $3A$. If the length of the first wire is increased by $\Delta l$ on applying a force $F$, how much force is needed to stretch the second wire by the same amount?

Accepted Answer

1. **Young's Modulus Formula:** Young's modulus ($Y$) is a property of the material and is defined as stress divided by strain: $Y = \frac{F/A}{\Delta l/L} \implies F = \frac{YA\Delta l}{L}$ .
2. **Constraint (Same Volume):** The volume ($V$) of a wire is $V = A 	imes L$. Since both wires have the same volume, $A_1 L_1 = A_2 L_2$.
   Given $A_1 = A$ and $A_2 = 3A$, we have:
   $$ A \cdot L_1 = 3A \cdot L_2 \implies L_2 = \frac{L_1}{3} $$
3. **Comparison of Forces:** We need to find $F_2$ given that the extension $\Delta l$ and material ($Y$) are the same for both.
   $$ F_1 = \frac{Y A_1 \Delta l}{L_1} = F $$
   $$ F_2 = \frac{Y A_2 \Delta l}{L_2} $$
4. **Substitution:** Substitute $A_2 = 3A$ and $L_2 = L_1/3$ into the equation for $F_2$:
   $$ F_2 = \frac{Y (3A) \Delta l}{(L_1/3)} = 3 	imes 3 	imes \frac{Y A \Delta l}{L_1} = 9 \left( \frac{Y A \Delta l}{L_1} ight) $$
   $$ F_2 = 9 F $$

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