The length of a rod under longitudinal tension $T_1$ is $L_1$ and that under longitudinal tension $T_2$ is…
The length of a rod under longitudinal tension $T_1$ is $L_1$ and that under longitudinal tension $T_2$ is $L_2$. What is the actual length of the rod, in the absence of tensions?
$\frac{L_1 T_1-L_2 T_2}{T_2-T_1}$
$\frac{L_1 T_2-L_2 T_1}{T_2+T_1}$
$\frac{L_1 T_1-L_2 T_2}{T_2+T_1}$
$\frac{L_1 T_2-L_2 T_1}{T_2-T_1}$
Solution
Let, the natural length of wire be $L_0$.
Using Hooke's law,
$
Y=\frac{T L_0}{A \Delta L}...(i)
$
where, $Y$ is Young's modulus of elasticity, $T$ is temperature, $A$ is area and $\Delta L$ is change in length, i.e. $\Delta L=L-L_0$
From Eq. (i), we get
$
L-L_0=\frac{T L_0}{A Y}
$
Case 1 When tension is $T_1$ and length of wire $L=L_1$,
$L_1-L_0=\frac{T_1 L_0}{A Y}$...(ii)
Case 2 When tension is $T_2$ and length of wire $L=L_2$
$L_2-L_0=\frac{T_2 L_0}{A Y}$...(iii)
Dividing Eq. (ii) by Eq. (iii), we get
$
\begin{aligned}
& \frac{L_1-L_0}{L_2-L_0}=\frac{T_1}{T_2} \\
\Rightarrow \quad & L_0=\frac{L_1 T_2-L_2 T_1}{T_2-T_1}
\end{aligned}
$