Question

The length of a metal wire is $L_1$ when the tension is $T_1$ and $L_2$ when the tension is $T_2$. The unstretched length of the wire is

• A. $\frac{L_{2}+L_{2}}{2}$
• B. $\sqrt{L_{1}L_{2}}$
• C. $\frac{T_{2}L_{1}-T_{1}L_{2}}{T_{2}-T_{1}}$
• D. $\frac{T_{2}L_{1}-T_{1}L_{2}}{T_{2}+T_{1}}$

Answer 1

Answer: (C)

$\frac{T_{2}L_{1}-T_{1}L_{2}}{T_{2}-T_{1}}$

Answer 2

Let the initial length of the metal wire is $L$.
The strain at tension $T_{1}$ is $\Delta L_{1}=L_{1}-L$
The strain at tension $T_{2}$ is $\Delta L_{2}=L_{2}-L$
Suppose, the Young's modulus of the wire is $Y$, then
$\frac{\frac{T_{1}}{A}}{\frac{\Delta L_{1}}{L}}=\frac{\frac{T_{2}}{A}}{\frac{\Delta L_{2}}{L}}$
where, $A$ is an cross-section of the wire. assume to be same at all the situations. $\Rightarrow \frac{T_{1}}{A} \times \frac{L}{\Delta L_{1}}$
$=\frac{T_{2}}{A} \times \frac{L}{\Delta L_{2}}$
$\Rightarrow \frac{T_{1}}{\left(L_{1}-L\right)}=\frac{T_{2}}{\left(L_{2}-L\right)}$
$T_{1}\left(L_{2}-L\right)=T_{2}\left(L_{1}-L\right) ;$
$L=\frac{T_{2} L_{1}-T_{1} L_{2}}{T_{2}-T_{1}}$