Question
The length of a metal wire is $l_1$ when the tension in it is $T_1$ and is $l_2$ when the tension is $T_2$. Find the original length of the wire.
✓
Answer
Let I and A be the original length and area of cross-section of the metal wire. Change in length in the first case = $(l_1- l)$ Change in length in the second case = $(l_2 - l)$
$\therefore\text{Y}=\frac{\text{T}_1}{\text{A}}\times\frac{\text{l}}{(\text{l}_1-\text{l)}}=\frac{\text{T}_2}{\text{A}}\times\frac{\text{l}}{(\text{l}_2-\text{l)}}$ or $\text{T}_1\text{l}_2-\text{T}_1\text{l}=\text{T}_2\text{l}_1-\text{T}_2\text{l}$
$\text{l}(\text{T}_2-\text{T}_1)=\text{T}_2\text{l}_1-\text{T}_1\text{l}_2$
$\Rightarrow\text{l}=\frac{\text{T}_2\text{l}_1-\text{T}_1\text{l}_2}{(\text{T}_2-\text{T}_1)}.$
$\therefore\text{Y}=\frac{\text{T}_1}{\text{A}}\times\frac{\text{l}}{(\text{l}_1-\text{l)}}=\frac{\text{T}_2}{\text{A}}\times\frac{\text{l}}{(\text{l}_2-\text{l)}}$ or $\text{T}_1\text{l}_2-\text{T}_1\text{l}=\text{T}_2\text{l}_1-\text{T}_2\text{l}$
$\text{l}(\text{T}_2-\text{T}_1)=\text{T}_2\text{l}_1-\text{T}_1\text{l}_2$
$\Rightarrow\text{l}=\frac{\text{T}_2\text{l}_1-\text{T}_1\text{l}_2}{(\text{T}_2-\text{T}_1)}.$
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