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Step-by-Step Solution
Step 1: Understand the Given Information
• Both wires are made of the same material and have the same volume.
• Wire 1 has cross-sectional area $A$ and wire 2 has cross-sectional area $3A$.
• The length of wire 1 is increased by $\Delta l$ when a force $F$ is applied.
• We want the same increase $\Delta l$ in wire 2 and need to find the force $F'$ required for that.
Step 2: Relate Volume to Dimensions of Each Wire
Since both wires have the same volume and we denote the original length of wire 1 by $l$, its volume is $V = A \cdot l$.
For wire 2, its cross-sectional area is $3A$, so if we call its length $l_2$, then:
$V = 3A \cdot l_2$.
But $V$ is the same for both, so:
$A \cdot l = 3A \cdot l_2 \quad \Rightarrow \quad l_2 = \frac{l}{3}.$
Step 3: Recall the Formula for Elongation
The extension $\Delta l$ of a wire under a tensile force, using Young’s modulus $Y$, is given by:
$\Delta l = \frac{F \, l}{A \, Y},$
where $l$ is the length of the wire, $A$ is its cross-sectional area, and $Y$ is the Young's modulus of the material.
Step 4: Apply the Formula to Each Wire
For wire 1 (length $l$, area $A$, force $F$):
$\Delta l = \frac{F \, l}{A \, Y}. \quad \text{(Equation 1)}$
For wire 2 (length $\tfrac{l}{3}$, area $3A$, unknown force $F'$ to get the same elongation $\Delta l$):
$\Delta l = \frac{F' \left(\frac{l}{3}\right)}{(3A) \, Y}
\;=\;
\frac{F' \, l}{9 A \, Y}.\quad \text{(Equation 2)}$
Step 5: Equate the Two Elongations
Since both wires must experience the same increase $\Delta l$, we set Equation (1) equal to Equation (2):
$\frac{F \, l}{A \, Y} = \frac{F' \, l}{9A \, Y}.$
Canceling the common terms $\left(l, A, Y\right)$ on both sides, we get:
$F = \frac{F'}{9}
\quad \Rightarrow \quad
F' = 9F.$
Final Answer
To stretch the second wire by the same amount, the required force is $9F$.
