Question

A wire of length $L$ and density $\rho$ and Young's modulus $Y$ is hanging from a support. Find the elongation in the length of wire at which wire will break:
\[\begin{align}
  & A.\dfrac{{{L}^{2}}\rho g}{Y} \\
 & B.\dfrac{{{L}^{2}}\rho g}{2Y} \\
 & C.\dfrac{2{{L}^{2}}\rho g}{Y} \\
 & D.\dfrac{{{L}^{2}}\rho g}{4Y} \\
\end{align}\]

Answer 1

Hint: The elastic moduli or the Young’s modulus of the material is the ratio of tensile or compressive stress to the longitudinal strain. Using the given data and the formula of Young's modulus, we can find the maximum elongation of the wire, when the force is applied on the wire.

Formula used: $Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$

Complete step by step answer:
We know that a bulk like aluminium and copper experience bulk modulus, which is related to the elasticity of the material. We know that the elastic moduli or the Young’s modulus of the material is the ration of tensile or compressive stress to the longitudinal strain.
i.e. $Y=\dfrac{stress}{strain}$, where stress is the force per unit area i.e.$stress=\dfrac{force}{area}$ and strain is 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 $Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$
Here, it is given that the length of the wire is $L$ and the density of the wire is $\rho$ and young’s modulus is $Y$.
Let us assume that the force $F=mg$ acts on the wire, also let the area of the wire be $A=L^{2}$ , and $\Delta L$ is the elongation of the wire, now substituting the values, we get
$Y=\dfrac{\dfrac{mg}{L^{2}}}{\dfrac{\Delta L}{L}}$
We also know that the density $\rho$ is the ratio between the mass and the volume of the wire, if $L^{3}$ is the volume of the wire, then,
$\rho=\dfrac{m}{L^{3}}$
Rearranging, we can say $m=\rho\times L^{3}$
Substituting for the mass in the $Y$,we get $Y=\dfrac{\dfrac{\rho gL^{3}}{L^{2}}}{\dfrac{\Delta L}{L}}$
Rearranging for $\Delta L$, we get$\Delta L=\dfrac{\dfrac{\rho gL^{3}}{L^{2}}}{\dfrac{Y}{L}}$
$\implies\Delta L=\dfrac{\rho gL^{2}}{Y}$
Thus the maximum elongation possible is given as $\dfrac{\rho gL^{2}}{Y}$

So, the correct answer is “Option A”.

Note: Any bulk material experiences bulk modulus, which is related to the elasticity of the material. That strain has no units, thus, the units of young’s modulus is the same as the strain i.e.$N/m^{2}$. Also, $Y\propto \dfrac{ L}{\Delta L}$ and $Y\propto\dfrac{F}{A}$