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

Hint: 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.