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Step-by-Step Solution
Step 1: Understand the Problem
We have two wires made of the same material (hence same Young’s modulus) and of the same volume. Wire 1 has cross-sectional area $A$ while Wire 2 has cross-sectional area $3A$. When a force $F$ is applied to Wire 1, its length increases by $\Delta x$. We need to find the force required to stretch Wire 2 by the same amount $\Delta x$.
Step 2: Recall the Formula Relating Force and Longitudinal Strain
The extension $\Delta L$ (or here $\Delta x$) of a wire under a tensile force $F$ is given by the relation:
$$
\Delta L = \frac{F L}{A E},
$$
where
$L$ is the original length of the wire,
$A$ is its cross-sectional area,
$E$ is the Young’s modulus of the material.
Since both wires are made of the same material, $E$ remains the same for both.
Step 3: Express the Lengths of the Wires
Both wires have the same volume $V$. Therefore, for Wire 1:
$$
V = A \times L_1 \quad \Longrightarrow \quad L_1 = \frac{V}{A}.
$$
For Wire 2:
$$
V = (3A) \times L_2 \quad \Longrightarrow \quad L_2 = \frac{V}{3A}.
$$
Step 4: Write the Extension Formulas for Each Wire
For Wire 1 (with area $A$):
$$
\Delta x = \frac{F \, L_1}{A \, E}.
$$
For Wire 2 (with area $3A$), let the force needed to stretch it by the same amount $\Delta x$ be $F_2$, then:
$$
\Delta x = \frac{F_2 \, L_2}{(3A) \, E}.
$$
Step 5: Equate the Extensions (Same $\Delta x$)
Since both wires are stretched by the same amount $\Delta x$, we have
$$
\frac{F \, L_1}{A \, E} = \frac{F_2 \, L_2}{(3A) \, E}.
$$
Canceling $E$ from both sides and substituting $L_1 = \frac{V}{A}$ and $L_2 = \frac{V}{3A}$:
$$
\frac{F \, \left(\frac{V}{A}\right)}{A} = \frac{F_2 \, \left(\frac{V}{3A}\right)}{3A}.
$$
This simplifies to
$$
\frac{F \, V}{A^2} = \frac{F_2 \, V}{3A \times 3A} = \frac{F_2 \, V}{9A^2}.
$$
Step 6: Solve for $F_2$
From the above equation, we get
$$
F_2 = 9F.
$$
Hence, the force required to stretch Wire 2 by the same amount $\Delta x$ is $9F$.
Step 7: Final Answer
The required force to stretch Wire 2 by the same extension is $9F$.
