Question

The length of a metal wire is $l_{1}$, when the tension in it is $T_{1}$ and $l_{2}$ when the tension is $T_{2}$. The natural length of the wire is?
\[\begin{align}
  & A.\dfrac{{{l}_{1}}+{{l}_{2}}}{2} \\
 & B.\sqrt{\dfrac{{{l}_{1}}}{{{l}_{2}}}} \\
 & C.\dfrac{{{l}_{1}}{{T}_{2}}-{{l}_{2}}{{T}_{1}}}{{{T}_{2}}-{{T}_{1}}} \\
 & D.\dfrac{{{l}_{1}}{{T}_{2}}+{{l}_{2}}{{T}_{1}}}{{{T}_{2}}+{{T}_{1}}} \\
\end{align}\]

Answer 1

Hint: We know that tension on a string is a scalar quantity and is similar to the force which is acting on the string. Since the tension on the string due to the respective length is given, we can find the young’s modulus of the string and use it to calculate the natural length of the wire.
Formula: $Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$

Complete answer:
We know that any bulk material experiences bulk modulus, which gives the elasticity of the material. It is the measure of how strong or weak is any given substance when subjected to some force or tension. We know that the elastic moduli or the Young’s modulus of the material is defined as the ratio of tensile or compressive stress to the longitudinal strain.
i.e. $Y=\dfrac{stress}{strain}$, where stress is given as the force per unit area i.e.$stress=\dfrac{force}{area}$ and strain is given as the ratio of change in size or shape to the original shape or size i.e. $strain=\dfrac{change\; in\; shape}{original\; in\;shape}$.
Then $Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$.Also, Young’s modulus is a constant for any given material.
Here, it is given that the length $l_{1}$ experiences a tension $T_{1}$, and the length $l_{2}$ experiences a tension $T_{2}$, respectively. Let the natural length or the actual length of the wire be $L$, and let $A$ be the cross-sectional area of the given wire.
Then we can say that the young’s modulus due to $T_{1}$ as $Y=\dfrac{T_{1}\times L}{A(l_{1}-L)}$, similarly, the young’s modulus due to $T_{2}$ as $Y=\dfrac{T_{2}\times L}{A(l_{2}-L)}$.
Since the Young’s modulus is a constant for any given material, we can equate the above two equations.
$\implies \dfrac{T_{1}\times L}{A(l_{1}-L)} =\dfrac{T_{2}\times L}{A(l_{2}-L)}$.
$\implies\dfrac{T_{1}}{A(l_{1}-L)} =\dfrac{T_{2}}{A(l_{2}-L)}$.
$\implies T_{2}\times A(l_{1}-L) =T_{1}\times A(l_{2}-L)$.
$\implies T_{2}l_{1}-T_{1}\times l_{2}=T_{2}L-T_{1}L$.
$\therefore L=\dfrac{T_{2}l_{1}-T_{1}\times l_{2}}{T_{2}-T_{1}}$

Hence the correct option is \[C.\dfrac{{{l}_{1}}{{T}_{2}}-{{l}_{2}}{{T}_{1}}}{{{T}_{2}}-{{T}_{1}}}\]

Note:
Any bulk material experiences bulk modulus, which is related to the elasticity of the material. This is a very easy sum, provided one knows the formula of young’s modulus. Here, the length and the respective tensions area given, using which the natural length of the wire is found.