Question

A composite wire consists of a steel wire of length $1.5m$ and a copper wire of length $2.0m$ with a uniform cross-sectional area of $2.5 \times {10^{ - 5}}{m^2}$ . It is loaded with a mass of $200kg$. Find the extension produced. Young’s modulus of copper is $1.0 \times {10^{11}}N{m^{ - 2}}$ and that of steel is $2.0 \times {10^{11}}N{m^{ - 2}}$ . Take $g = 9.8m{s^{ - 2}}$

Answer 1

Hint: To solve this question, we must observe that as the load and area of the cross-section are the same for both the copper wire and steel wire, the tensile stress on both the wires will be the same. Now put the calculated stress in the formula of Young’s modulus and solve the equations to get the result.

Formula Used:
$Y = \dfrac{{\left( {F/A} \right)}}{{\left( {\Delta l/l} \right)}}$
Where,
 $Y$ is Young’s modulus.
$F$is the force.
$A$ is the area of cross-section
$l$ is the original length.
$\Delta l$ is the change in length.

Complete step-by-step answer:
As the mass is hanging to the wire, we can infer that the force acting on it is only due to gravity. Therefore,
$F = mg......(1)$
Where,
$F$ is the force.
$m$ is the mass of the load.
$g$ is the acceleration due to gravity
In the question, it is given that the mass of the load is $200kg$and acceleration due to gravity is $9.8m{s^{ - 2}}$. We should put these values in equation (1) to get,
$F = 200 \times 9.8$
$ \Rightarrow F = 1960N......(2)$
Mathematically, tensile stress is given as:
Tensile stress,$\sigma = \dfrac{F}{A}......(3)$
Where,
$\sigma $ is tensile stress.
$F$is the force.
$A$ is the area of cross-section.
The area of cross-section is $2.5 \times {10^{ - 5}}{m^2}$ for both wires. Putting the values of force and area of the cross-section in equation (3) we get,
$\sigma = \dfrac{{1960}}{{2.5 \times {{10}^{ - 5}}}}$
$ \Rightarrow \sigma = 7.84 \times {10^7}N{m^{ - 2}}.......(4)$
Young’s Modulus is mathematically represented as:
$Y = \dfrac{{\left( {F/A} \right)}}{{\left( {\Delta l/l} \right)}}.......(5)$
Where,
 $Y$ is Young’s modulus.
$F$is the force.
$A$ is the area of cross-section
$l$ is the original length.
$\Delta l$ is the change in length.
Now, let the elongations in the copper wire be $\Delta {l_1}$ and the steel wire be $\Delta {l_2}$.

For copper wire:
$l = 2.0m$
$Y = 1.0 \times {10^{11}}N{m^{ - 2}}$
We will put all the given quantities along with the value of stress in equation (5) we get,
$1.0 \times {10^{11}} = \dfrac{{7.84 \times {{10}^7}}}{{\left( {\dfrac{{\Delta {l_1}}}{{2.0}}} \right)}}$
$ \Rightarrow \Delta {l_1} = \dfrac{{7.84 \times {{10}^7} \times 2.0}}{{1.0 \times {{10}^{11}}}}$
$ \Rightarrow \Delta {l_1} = 1.568mm$
For steel wire:
$l = 1.5m$
$Y = 2.0 \times {10^{11}}N{m^{ - 2}}$
We will put all the given quantities along with the value of stress in equation (5) we get,
$2.0 \times {10^{11}} = \dfrac{{7.84 \times {{10}^7}}}{{\left( {\dfrac{{\Delta {l_2}}}{{1.5}}} \right)}}$
$ \Rightarrow \Delta {l_2} = \dfrac{{7.84 \times {{10}^7} \times 1.5}}{{2.0 \times {{10}^{11}}}}$
$ \Rightarrow \Delta {l_2} = 0.588mm$
The total elongation of the composite wire,$\Delta l$ is given as,
$\Delta l = \Delta {l_1} + \Delta {l_2}$
$ \Rightarrow \Delta l = (1.568 + 0.588)mm$
$\Delta l = 2.156mm$

Note: Whenever an object is subjected to an external force, it experiences deformation i.e. change in its dimensions. When the force is exerted along the axis of force, it stretches the material. Such an external force acting per unit area which results in the elongation of the material is called tensile stress. Pascal is used as the SI unit of tensile stress. It is defined as the ratio of the force applied along the string to the area of cross-section of the string that is perpendicular to the applied force.
The ratio of the change in the dimension of an object due to tensile stress to its original dimension is called tensile strain. It is a dimensionless quantity.
Tensile strain, $\varepsilon = \dfrac{{\Delta l}}{l}$
Where,
$\varepsilon $ is the tensile strain.
$\Delta l$ is the change in length.
$l$ is the original length.
Young’s modulus is the ratio of tensile stress to tensile strain. It signifies the amount of compression or elongation a material can withstand with respect to its length. Its S.I. unit is Pascal.