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Step-by-Step Solution
Step 1: Identify the Physical Situation
A thin metal wire is subjected to two different tensions, $T_1$ and $T_2$. When the tension is $T_1$, the wire’s length is $l_1$, and when the tension is $T_2$, its length is $l_2$. We want to find the wire’s actual (unstretched) length, which we will call $l_0$.
Step 2: Express the Extension in Terms of Young’s Modulus
For a wire of original length $l_0$, cross-sectional area $A$, and Young’s modulus $Y$, the extension $\Delta l$ when a tension $T$ is applied can be written (using Hooke’s law and the definition of Young’s modulus) as:
$\displaystyle \Delta l \;=\; l - l_0 \;=\; \frac{T\,l_0}{A\,Y}$,
where $l$ is the new length under tension.
Step 3: Write the Extension Equations for Each Tension
When the tension is $T_1$, the length of the wire is $l_1$, so:
$\displaystyle l_1 - l_0 \;=\; \frac{T_1\,l_0}{A\,Y}.$
When the tension is $T_2$, the length is $l_2$, so:
$\displaystyle l_2 - l_0 \;=\; \frac{T_2\,l_0}{A\,Y}.$
Step 4: Form the Ratio of the Two Extensions
Divide one extension equation by the other to eliminate $A$ and $Y$:
$\displaystyle \frac{(l_1 - l_0)}{(l_2 - l_0)} \;=\; \frac{\frac{T_1\,l_0}{A\,Y}}{\frac{T_2\,l_0}{A\,Y}}
\;=\;\frac{T_1}{T_2}.$
Step 5: Rearrange to Solve for $l_0$
From the equation above, cross-multiplying and rearranging terms give:
$\displaystyle (l_1 - l_0)\,T_2 \;=\; (l_2 - l_0)\,T_1.$
Expanding both sides:
$\displaystyle l_1\,T_2 - l_0\,T_2 \;=\; l_2\,T_1 - l_0\,T_1.$
Rearrange to isolate $l_0$:
$\displaystyle l_0\,(T_1 - T_2) \;=\; T_1\,l_2 - T_2\,l_1.$
Thus,
$\displaystyle l_0 \;=\;\frac{T_1\,l_2 - T_2\,l_1}{T_1 - T_2}.$
Step 6: State the Final Answer
Therefore, the actual (unstretched) length of the metal wire is:
$\displaystyle \boxed{\frac{T_1\,l_2 \;-\; T_2\,l_1}{T_1 \;-\; T_2}}.$
