The length of metallic wire

Question:

The length of metallic wire is $l_{1}$ when tension in it is $T_{1}$. It is $l_{2}$ when the tension is $T_{2}$. The original length of the wire will be :

1. $\frac{l_{1}+l_{2}}{2}$

2. $\frac{\mathrm{T}_{1} l_{1}-\mathrm{T}_{2} l_{2}}{\mathrm{~T}_{2}-\mathrm{T}_{1}}$

3. $\frac{\mathrm{T}_{2} l_{1}+\mathrm{T}_{1} l_{2}}{\mathrm{~T}_{1}+\mathrm{T}_{2}}$

4. $\frac{T_{2} l_{1}-T_{1} l_{2}}{T_{2}-T_{1}}$

Correct Option: , 4

Solution:

(4)

From young's modulus relation $\left(\mathrm{y}=\frac{\frac{\mathrm{F}}{\mathrm{A}}}{\left(\frac{\Delta \mathrm{I}}{\mathrm{I}}\right)}\right)$

we can write for $1^{\text {st }}$ case

$\frac{T_{1}}{A}=\frac{y\left(\ell_{1}-\ell\right)}{\ell}$

we can write for $2^{\text {nd }}$ case

$\frac{T_{2}}{A}=\frac{y\left(\ell_{2}-\ell\right)}{\ell}$

$\frac{T_{1}}{T_{2}}=\frac{\ell_{1}-\ell}{\ell_{2}-\ell}$

$\mathrm{T}_{1} \ell_{2}-\mathrm{T}_{1} \ell=\mathrm{T}_{2} \ell_{1}-\mathrm{T}_{2} \ell$

$\frac{T_{2} l_{1}-T_{1} l_{2}}{T_{2}-T_{1}}=\ell$