Question

The length of a metal wire is $l$ when the tension is $F$ and $xl$ when the tension is $yF$. Then the natural length of the wire is
A. $\dfrac{{\left( {x - y} \right)l}}{{x - 1}}$
B. $\dfrac{{\left( {y - x} \right)l}}{{y - 1}}$
C. $\dfrac{{\left( {x - y} \right)l}}{{x + 1}}$
D. $\dfrac{{\left( {y - x} \right)l}}{{y + 1}}$

Answer 1

The tension on a metal wire can change the length of wire and that happens due to the stress produced by the tension on the string. This stress acts against the property of elasticity of the material, which tries to resume the change to the normal condition. Here, you also need the Young’s modulus of elasticity which quantifies the amount of stress for the unit amount of strain that is the change in the length of the wire.

Formula Used:
If a force $F$ acting on a body with length $L$ and cross section area $A$, produces a change of length $\Delta L$, then the Young’s modulus $Y$ of elasticity I defined as
$Y = \dfrac{{Stress}}{{Strain}} \\\ \Rightarrow Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}} \\\$

Complete step by step answer:
Given:
When the tension on the metal wire is $F$ , the length is $l$. .
When the tension on the metal wire is $yF$ , the length is $xl$.
To get: The natural length of the wire.
Step 1:
Let the natural length of the wire is ${l_0}$ and the cross section area is $A$.
Let the Young’s modulus of the metal is $Y$.
From eq (1) express the amount of change in the length with other terms.
$Y = \dfrac{{\dfrac{F}{A}}}{{\dfrac{{\Delta L}}{L}}} \\\ \Rightarrow \dfrac{{Y\Delta L}}{L} = \dfrac{F}{A} \\\ \Rightarrow \Delta L = \dfrac{{FL}}{{YA}} \\\$
Step 2:
Calculate the change in the length of the wire for the given metal wire with tension $F$ from eq (2)
$\Delta L = \dfrac{{F{l_0}}}{{YA}} \\\ \\\$
By the problem,
$\Delta L = l - {l_0}$
Step 3:
Calculate the change in the length of the wire for the given metal wire with tension $yF$ from eq (2)
$\Delta L' = \dfrac{{yF{l_0}}}{{YA}} \\\ \\\$
By the problem,
$\Delta L' = xl - {l_0}$

Step 4:
Now divide the eq (3) by eq (5)
$\dfrac{{\Delta L}}{{\Delta L'}} = \dfrac{{\dfrac{{F{l_0}}}{{YA}}}}{{\dfrac{{yF{l_0}}}{{YA}}}}$
$\Rightarrow \dfrac{{\Delta L}}{{\Delta L'}} = \dfrac{1}{y}$
Step 5:
Now plug the values from eq (4) and eq (6) into eq (7)
$\dfrac{{\Delta L}}{{\Delta L'}} = \dfrac{1}{y} \\\ \Rightarrow \dfrac{{l - {l_0}}}{{xl - {l_0}}} = \dfrac{1}{y} \\\ \Rightarrow yl - y{l_0} = xl - {l_0} \\\ \Rightarrow yl - xl = y{l_0} - {l_0} \\\ \Rightarrow \left( {y - x} \right)l = \left( {y - 1} \right){l_0} \\\ \Rightarrow {l_0} = \dfrac{{\left( {y - x} \right)l}}{{y - 1}} \\\$
Final answer:
The natural length of the wire is (B) $\dfrac{{\left( {y - x} \right)l}}{{y - 1}}$.

Note: Here the strain of the wire for both the case with respectively $F$ and $yF$ tension are different. But the Young’s moduli are the same as the Young’ modulus of elasticity $Y$ is a property of matter/material hence, depends on only the metal at a certain temperature, pressure. Here, the volume constancy is not considered otherwise, there should be change in the cross section area as well which would unnecessarily complicate the problem.