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 is :

• A. $\frac{l_1+l_2}{2}$
• B. $\frac{T_1{l}_1-T_2{l}_2}{T_2-T_1}$
• C. $\frac{T_2{l}_1+T_1{l}_2}{T_1+T_2}$
  
• D. $\frac{T_2{l}_1-T_1{l}_2}{T_2-T_1}$
  

Answer 1

The correct option is D $\frac{T_2{l}_1-T_1{l}_2}{T_2-T_1}$


From Young's modulus relation,

$Y=\frac{\frac{F}{A}}{\frac{\Delta L}{L}}$

Let the original length be $l$.

For the case $\left(1\right)$,

$\frac{T_1}{A}=\frac{Y(l_1-l)}{l}$

For the case $\left(2\right)$,

$\frac{T_2}{A}=\frac{Y(l_2-l)}{l}$

So,
$\frac{T_1}{T_2}=\frac{l_1-l}{l_2-l}$

$\Rightarrow T_1{l}_2-T_{1}l=T_2{l}_1-T_{2}l$

$\Rightarrow l=\frac{T_2{l}_1-T_1{l}_2}{T_2-T_1}$