Question

Two rods of different material having linear expansion coefficients ${\alpha _1}$ and ${\alpha _2}$ and young’s modulus ${Y_1}$ and ${Y_2}$ are fixed between two rigid walls. Both the rods are heated to the same temperature. If there is no bending of the rods, the thermal stress developed in them are equal provided that:
a) $\dfrac{{{Y_1}}}{{{Y_2}}} = \dfrac{{{\alpha _1}}}{{{\alpha _2}}}$
b) $\dfrac{{{Y_1}}}{{{Y_2}}} = \dfrac{{{\alpha _1}}}{{{\alpha _1} + {\alpha _2}}}$
c) $\dfrac{{{Y_1}}}{{{Y_2}}} = \dfrac{{{\alpha _2}}}{{\left( {{\alpha _1} + {\alpha _2}} \right)}}$
d) $\dfrac{{{Y_1}}}{{{Y_2}}} = \dfrac{{{\alpha _2}}}{{{\alpha _1}}}$

Answer 1

Hint: Use the concepts of thermal expansion, thermal stress and Hooke’s law in this question for the determination of correct answer from the given options. The thermal stresses generated in the rods are equal. So, equate the thermal stress of two rods so that we obtain the relation between young’s modulus and linear expansion coefficient of the rods.

Complete step by step answer:
It s given in the question that the linear expansion coefficients of the two rods are ${\alpha _1}$ and ${\alpha _2}$, the Young’s modulus of the two rods are ${Y_1}$ and ${Y_2}$. So we will use this information in the solution.
We know that when the heat is given to the rods, then they will start to expand due to thermal expansion, so from the expression of the change in length of the first rod due to heating, we get
$ \Delta {L_1} = {L_1}{\alpha _1}\Delta T\\$
$\implies \dfrac{{\Delta {L_1}}}{{{L_1}}} = {\alpha _1}\Delta T$ …… (1)
Here $\Delta {L_1}$ is the change in length, ${L_1}$ is the initial length of the rod and $\Delta T$ is the temperature difference.
Similarly, write the expression of the change in length of the second rod due to heating.
$\Delta {L_2} = {L_2}{\alpha _2}\Delta T\\$
$ \implies\dfrac{{\Delta {L_2}}}{{{L_2}}} = {\alpha _1}\Delta T$ …… (2)
Here $\Delta {L_2}$ is the change in length, ${L_2}$ is the initial length of the rod and $\Delta T$ is the temperature difference.
We know that the $\dfrac{{\Delta L}}{L}$ is known as strain, so we will use the equation (1) and (2) in the Hooke's law so that we can determine the develop thermal stress in both rods. So, apply the Hooke's law for the first rod; therefore, we get
$\begin{array}{l}
{\sigma _1} = {Y_1} \times \dfrac{{\Delta {L_1}}}{{{L_1}}}\\
{\sigma _1} = {Y_1} \times {\alpha _1}\Delta T\end{array}$ …… (3)
Here, ${\sigma _1}$ is the stress developed in the first rod.
Now apply the Hooke's law for the second rod, therefore, we get
$ {\sigma _2} = {Y_2} \times \dfrac{{\Delta {L_2}}}{{{L_2}}}\\$
$ \implies {\sigma _2} = {Y_2} \times {\alpha _2}\Delta T$ …… (4)
Here, ${\sigma _2}$ is the stress developed in the second rod.
It is given in the question that thermal stress developed in the rods are equal if there is no bending and the rods are heated at the same temperature, so we will use equations (3) and (4) to determine the required relation.
Therefore, we get
$ {\sigma _1} = {\sigma _2}\\$
$ \implies{Y_1} \times {\alpha _1}\Delta T = {Y_2} \times {\alpha _2}\Delta T$

The rods are heated at the same temperature, so the temperature differences in the above equation will become equal, so
$ {Y_1} \times {\alpha _1} = {Y_2} \times {\alpha _2}\\$
$ \implies \dfrac{{{Y_1}}}{{{Y_2}}} = \dfrac{{{\alpha _2}}}{{{\alpha _1}}}$
Therefore, the relation between Young’s modulus and linear expansion coefficients of the two rods is \[\dfrac{{{Y_1}}}{{{Y_2}}} = \dfrac{{{\alpha _2}}}{{{\alpha _1}}}\]

So, the correct answer is “Option D”.

Note:
Here, we use Hooke’s law because it relates the stress and strain of the rods with the help of an elastic constant known as Young’s modulus. Remember the Hooke’s law and expression of linear expansion for these types of questions. If values of coefficients are given in the question, then put them during the calculation, so that we obtain the relation between Young’s modulus of two rods.