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 ${T_2}$. What is the natural length of the wire?
A. $\dfrac{{{l_1} + {l_2}}}{2}$
B. $\sqrt {{l_1}{l_2}}$
C. $\dfrac{{{l_1}{T_2} - {l_2}{T_1}}}{{{T_2} - {T_1}}}$
D. $\dfrac{{{l_1}{T_2} + {l_2}{T_1}}}{{{T_2} + {T_1}}}$

Answer 1

Formula for Young Modulus is given as,
${\text{Young}}\;{\text{Modulus}}\; = \;\dfrac{{{\text{Normal}}\;{\text{Stress}}}}{{{\text{Longitudinal}}\;{\text{Strain}}}}$.
${\text{Normal}}\;{\text{Stress}} = \dfrac{T}{A}$
${\text{Longitudinal}}\;{\text{Strain}} = \dfrac{{\Delta l}}{l}$

Complete step by step solution:
Given,
The length of a metal wire is ${l_1}$ when the tension in it is ${T_1}$
And the length of the metal wire is ${l_2}$ when the tension in it is ${T_2}$
Let us assume that the original length of the metal wire be $l$.
And the area of the metal wire be $A$.

In first case, change in length in the metal wire,
$= {l_1} - l$
In second case, change in length in the metal wire,
$= {l_2} - l$

Young's modulus is a material property which measures a solid material's strength. It describes the relation between stress and strain in the linear elasticity regime of an uniaxial deformation in a material.

Formula for Young Modulus is given as,
${\text{Young}}\;{\text{Modulus}}\; = \;\dfrac{{{\text{Normal}}\;{\text{Stress}}}}{{{\text{Longitudinal}}\;{\text{Strain}}}}$
Normal stress is given by,
${\text{Normal}}\;{\text{Stress}} = \dfrac{T}{A}$
Longitudinal Strain is given by,
${\text{Longitudinal}}\;{\text{Strain}} = \dfrac{{\Delta l}}{l}$
Now, ${\text{Young}}\;{\text{Modulus}}\; = \;\dfrac{{\dfrac{T}{A}}}{{\dfrac{{\Delta l}}{l}}}$
Where, $T$ is tension,
$\Delta l$ is change in length

For first case,
${Y_1} = \dfrac{{{T_1}}}{A} \times \dfrac{l}{{{l_1} - l}}$ …… (i)
For second case,
${Y_2} = \dfrac{{{T_2}}}{A} \times \dfrac{l}{{{l_2} - l}}$ …… (ii)
For both the cases, Young Modulus remains same,
Therefore we equate equation (i) and (ii), as ${Y_1}$ is equal to ${Y_2}$

$$\dfrac{{{T_1}}}{A} \times \dfrac{l}{{{l_1} - l}} = \dfrac{{{T_2}}}{A} \times \dfrac{l}{{{l_2} - l}} \\\ \dfrac{{{T_1}}}{{{l_1} - l}} = \dfrac{{{T_2}}}{{{l_2} - l}} \\\$$

Further solving the above equation to determine the value of $l$

$$\dfrac{{{T_1}}}{{{l_1} - l}} = \dfrac{{{T_2}}}{{{l_2} - l}} \\\ {T_1}{l_2} - {T_1}l = {T_2}{l_1} - {T_2}l \\\ {T_2}l - {T_1}l = {T_2}{l_1} - {T_1}{l_2} \\\ l\left( {{T_2} - {T_1}} \right) = {T_2}{l_1} - {T_1}{l_2} \\\ l = \dfrac{{{T_2}{l_1} - {T_1}{l_2}}}{{{T_2} - {T_1}}} \\\$$

Therefore the length of the metal wire is $\dfrac{{{T_2}{l_1} - {T_1}{l_2}}}{{{T_2} - {T_1}}}$

Hence, the correct option is C.

Note: In both cases, we subtract the original length from the given two lengths to find the change in length. Hence, the change in lengths will be ${l_1} - l$ and ${l_2} - l$ respectively. The formula for Young Modulus is given by ${\text{Young}}\;{\text{Modulus}}\; = \;\dfrac{{\dfrac{T}{A}}}{{\dfrac{{\Delta l}}{l}}}$, after replacing the values of normal stress and longitudinal strain respectively. Here, $\Delta l$ represents the change in length in the metal wire.