Question

An aluminium and copper wire of same cross-sectional area, but having different lengths in the ratio $2:3$ are joined end to end. This composite wire is hung from a rigid support and a load is suspended from the free ends. If the increase in length of the composite wire is $2.1mm$, the increase in lengths of aluminium and copper wires is: $[Y_{al}=20\times 10^{11}N/m^{2}$ and $Y_{cu}=12 \times 10^{11} N/m^{2}]$

& A.0.7mm,1.4mm \\\ & B.0.9mm,1.2mm \\\ & C.1.0mm,1.1mm \\\ & D.\text{ }0.6mm,1.5mm \\\ \end{aligned}$$

Answer 1

From the formula of young’s modulus, we get, $Y\propto \dfrac{\Delta L}{L}$ and $Y\propto\dfrac{F}{A}$. To find the ratio of $\Delta L$of copper and aluminium, we can use this ratio i.e. $Y\propto \dfrac{\Delta L}{L}$ . Also given that, the increase in length of the composite wire is$2.1mm$, using which we can find the length $L$.

Formula used:
$Y\propto \dfrac{L}{\Delta L}$

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 ratio 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 $Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta L}{L}}$ Or$Y\propto\dfrac{F}{A}$ and $Y\propto \dfrac{ L}{\Delta L}$
Given that, the cross-sectional area of aluminium and copper is the same as $A$. Also the ratio of initial lengths of the wires is $2:3$. Then, let the length of aluminium and copper wires be $2x,3x$ respectively.
Also $Y_{al}=20\times 10^{11}N/m^{2}$ and $Y_{cu}=12 \times 10^{11} N/m^{2}$
Using $Y\propto \dfrac{ L}{\Delta L}$, we can say that,$\dfrac{Y_{A}}{Y_{C}}=\dfrac{\dfrac{ L_{A}}{\Delta L_{A}}}{\dfrac{ L_{C}}{\Delta L_{C}}}$ or $\dfrac{\Delta L_{C}}{\Delta L_{A}}=\dfrac{ Y_{A} \times L_{C}}{Y_{C}\times L_{A}}$
Then, $\dfrac{\Delta L_{C}}{\Delta L_{A}}=\dfrac{20\times 10^{11} \times 3}{12\times 10^{11} \times 2}=\dfrac{5}{2}$
Then, $2x+5x=2.1mm$ or $7x=2.1$ or $x=0.3$
Then the increase in lengths of aluminium and copper wires i.e. $\Delta L_{C}=5x=5\times 0.3=1.5mm$ and $\Delta L_{A}=2x=2\times 0.3=0.6mm$

Hence the answer is, $\text{D}\text{. }0.6mm,1.5mm$

Note:
A bulk like aluminium and copper experience 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 $Y\propto\dfrac{F}{A}$ The increase in length of the composite wire is $2.1mm$ implies $2x+5x=2.1mm$. This is needed to find $x$. Students need to first define the physical quantities being discussed about properly and then use the understanding and mathematical expression of these quantities such as stress, strain and Young’s modulus to solve the question. Additionally, students try to skip steps when solving these types of questions that can result in silly mistakes. So avoid skipping steps.