Question

Two wires of equal cross-section one made up of aluminium and another of brass are joined end to end. When the combination of wires is kept under tension the elongation in wires is found to be equal. Find the ratio of lengths of two wires. $\left( {{Y}_{AL}}=7\times {{10}^{10}}N{{m}^{-2}}\text{ and }{{Y}_{brass}}=9.1\times {{10}^{10}}N{{m}^{-2}} \right)$

Answer 1

Hint: If we have an elastic wire of length, L and cross-sectional area, A and if we stretch it by a length $\Delta L$, then a restoring force gets generated inside the wire. This force tries to restore the wire back to its original length. We shall analyze this force and its relationship with the Young’s modulus further.

Complete answer:
Young’s modulus of elasticity, $Y=\dfrac{stress}{strain}$
Stress is defined as the restoring force per unit area. It is the quantity which tells us how quickly the material will snap back (or restore) to its original orientation.
Strain is the change in length per length or the relative change in length. This quantity tells us how much the material has been deformed.
Stress $=\dfrac{F}{A}$ and Strain $=\dfrac{\Delta L}{L}$
Where,
$F=$ force applied on the object
$A=$ area of cross-section
$\Delta L=$ change in length due to application of force
$L=$ natural length of object
$\Rightarrow Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$
$\Rightarrow Y=\dfrac{FL}{A\Delta L}$
It is given that on applying tension, elongation of wires (or change in length) is equal in wires of aluminium and brass which have equal cross-sectional area.
$\Rightarrow \Delta {{L}_{AL}}=\Delta {{L}_{brass}}$
Also, ${{A}_{AL}}={{A}_{brass}}$ and ${{F}_{AL}}={{F}_{brass}}$
Hence, taking ratio of Young’s modulus of aluminium and brass, we get
$\begin{align}
  & \Rightarrow \dfrac{{{Y}_{AL}}}{{{Y}_{brass}}}=~\dfrac{\left( \dfrac{{{F}_{AL}}{{L}_{AL}}}{{{A}_{AL}}\Delta {{L}_{AL}}} \right)}{\left( \dfrac{{{F}_{brass}}{{L}_{brass}}}{{{A}_{brass}}\Delta {{L}_{brass}}} \right)} \\
 & \Rightarrow \dfrac{{{Y}_{AL}}}{{{Y}_{brass}}}=\dfrac{{{F}_{AL}}{{L}_{AL}}}{{{A}_{AL}}\Delta {{L}_{AL}}}.\dfrac{{{A}_{brass}}\Delta {{L}_{brass}}}{{{F}_{brass}}{{L}_{brass}}} \\
\end{align}$
From above information, cancelling $\Delta {{L}_{AL}}$and $\Delta {{L}_{brass}}$, ${{F}_{AL}}$and ${{F}_{brass}}$ , ${{A}_{AL}}$ and ${{A}_{brass}}$respectively,
$\Rightarrow \dfrac{{{Y}_{AL}}}{{{Y}_{brass}}}=~\dfrac{{{L}_{AL}}}{{{L}_{brass}}}$
Given that ${{Y}_{AL}}=7\times {{10}^{10}}N{{m}^{-2}}\text{ and }{{Y}_{brass}}=9.1\times {{10}^{10}}N{{m}^{-2}}$,
$\Rightarrow \dfrac{{{L}_{AL}}}{{{L}_{brass}}}=\dfrac{7\times {{10}^{10}}N{{m}^{-2}}}{9.1\times {{10}^{10}}N{{m}^{-2}}}$
$\Rightarrow \dfrac{{{L}_{AL}}}{{{L}_{brass}}}=\dfrac{70}{91}$
Therefore, the ratio of lengths of aluminium and brass wire is $\dfrac{70}{91}$.

Note:
Technically, the more we deform the material, the more it will try to restore back to its original form, that is, the more strain we induce, the more stress will be experienced by the material. This implies that stress is linearly related to strain. This is valid for only small strain values because on applying a huge amount of strain, the material gets permanently deformed. The experiments conducted also support this fact.