Question

A composite rod consists of a steel rod of length \[25\,{\text{cm}}\] and area \[2A\] and a copper rod of length \[50\,{\text{cm}}\] and area \[A\] . The composite rod is subjected to an axial load \[F\] . If the Young’s moduli of steel and copper are in the ratio \[2:1\] , then:
This question has multiple correct options.
A. The extension produced in copper rod will be more
B. The extension in copper and steel parts will be in the ratio \[1:2\]
C. The stress applied to copper rod will be more
D. No extension will be produced in the steel rod

Answer 1

Hint: We know that stress is defined as force experienced per unit area. The extension produced on a rod is directly proportional to the force but inversely proportional to the area and Young’s modulus. We will find out the respective extensions for the two rods and then find the ratio.

Formula used:
The relation which gives the value of stress is given below:
\[\sigma = \dfrac{F}{A}\] …… (1)
Where,
\[\sigma \] indicates stress.
\[F\] indicates force of the load.
\[A\] indicates the area.
Again, the formula which gives the extension is given below:
\[\Delta l = \dfrac{{Fl}}{{AY}}\] …… (2)
Where,
\[\Delta l\] indicates the extension.
\[F\] indicates the applied force.
\[l\] indicates original length.
\[A\] indicates the area.
\[Y\] indicates Young’s modulus.

Complete step by step answer:
In the given question, we are supplied with the following data:
There is a composite rod which consists of a steel rod and a copper rod.
The length of the steel rod is \[25\,{\text{cm}}\] and area \[2A\] .
The length of the copper rod is \[50\,{\text{cm}}\] and area \[A\] .
The composite rod is subjected to an axial load \[F\] .
The ratio of the Young’s moduli of steel to that of copper is \[2:1\] .
We are asked to find the statements which go well with the case.

To begin with,
First, we will go for the comparison of stress:
Since, for the two rods, force is exactly the same.
From the equation (1), we can write:
\[\sigma \propto \dfrac{1}{A}\]
It implies that, greater is the area lower is the stress experienced.
Since, the area of the steel rod is more than that of copper.
So, the stress applied to the copper rod will be more than the steel rod.

Again, for their respective extension:
\[{\left( {\Delta l} \right)_{{\text{steel}}}} = \dfrac{{F \times 25}}{{2A \times 2Y}}\]
\[{\left( {\Delta l} \right)_{{\text{copper}}}} = \dfrac{{F \times 50}}{{A \times Y}}\]
Now, we will find the ratio:
$\dfrac{{{{\left( {\Delta l} \right)}_{{\text{steel}}}}}}{{{{\left( {\Delta l} \right)}_{{\text{copper}}}}}} = \dfrac{{\dfrac{{F \times 25}}{{2A \times 2Y}}}}{{\dfrac{{F \times 50}}{{A \times Y}}}} \\
\Rightarrow \dfrac{{{{\left( {\Delta l} \right)}_{{\text{steel}}}}}}{{{{\left( {\Delta l} \right)}_{{\text{copper}}}}}} = \dfrac{{F \times 25}}{{2A \times 2Y}} \times \dfrac{{A \times Y}}{{F \times 50}} \\
\Rightarrow \dfrac{{{{\left( {\Delta l} \right)}_{{\text{steel}}}}}}{{{{\left( {\Delta l} \right)}_{{\text{copper}}}}}} = \dfrac{1}{{2 \times 2 \times 2}} \\
\Rightarrow \dfrac{{{{\left( {\Delta l} \right)}_{{\text{steel}}}}}}{{{{\left( {\Delta l} \right)}_{{\text{copper}}}}}} = \dfrac{1}{8} \\$
\[\therefore {\left( {\Delta l} \right)_{{\text{steel}}}}:{\left( {\Delta l} \right)_{{\text{copper}}}} = 1:8\]
Hence, from the above ratio it is clear that the extension produced in copper rod will be more.

The correct options are A and C.

Note:While solving the given problem, it is important to remember that stress is nothing but pressure experienced by a rod when subjected to some load. Young’s modulus defines the stiffness of a material. Higher the value of the Young’s modulus is, lower is the extension produced. The extension is inversely proportional to the area of the material. A material with a small area produces greater extension.