Question

Two rods of different materials have coefficient of linear expansion ${{\alpha }_{1}},{{\alpha }_{2}}$ and Young’s moduli ${{Y}_{1}},{{Y}_{2}}$ respectively are fixed between two rigid walls. The rods are heated such that they undergo the same increase in temperature. There is no bending of rods. If ${{\alpha }_{1}}:{{\alpha }_{2}}=2:3$, the thermal stress developed in the two rods are equal provided ${{Y}_{1}}:{{Y}_{2}}$ is equal to:
$\begin{align}
  & (A)2:3 \\
 & (B)1:1 \\
 & (C)3:2 \\
 & (D)4:9 \\
\end{align}$

Answer 1

Hint: It has been given to us that the temperature gradient of the two rods are equal and since they are fixed in between two walls, it means their initial lengths are also the same. We will use the formula which relates Young’s modulus with stress and strain for the rods and then proceed ahead in our solution.

Complete answer:
Let the length of both the rods be $L$. Then, the change in their lengths upon heating shall be:
For the first rod we have:
$\Rightarrow (\vartriangle {{L}_{1}})=L{{\alpha }_{1}}(\vartriangle T)$
For the second rod, we have:
$\Rightarrow (\vartriangle {{L}_{2}})=L{{\alpha }_{2}}(\vartriangle T)$
Now, strain is given by change in length upon initial length. Then,
Strain in the first rod can be calculated as follows:
$\begin{align}
  & \Rightarrow \dfrac{\vartriangle {{L}_{1}}}{L}=\dfrac{L{{\alpha }_{1}}(\vartriangle T)}{L} \\
 & \Rightarrow \dfrac{\vartriangle {{L}_{1}}}{L}={{\alpha }_{1}}(\vartriangle T) \\
\end{align}$
And strain in the second rod can be calculated as follows:
$\begin{align}
  & \Rightarrow \dfrac{\vartriangle {{L}_{2}}}{L}=\dfrac{L{{\alpha }_{2}}(\vartriangle T)}{L} \\
 & \Rightarrow \dfrac{\vartriangle {{L}_{2}}}{L}={{\alpha }_{2}}(\vartriangle T) \\
\end{align}$
Now, since stress is given as a product of Young’s modulus and strain and it has been given to be equal for both the rods, we can write as follows:
$\Rightarrow {{Y}_{1}}\left( \dfrac{\vartriangle {{L}_{1}}}{L} \right)={{Y}_{2}}\left( \dfrac{\vartriangle {{L}_{2}}}{L} \right)$
Putting the value of these terms, we get:
$\begin{align}
  & \Rightarrow {{Y}_{1}}{{\alpha }_{1}}(\vartriangle T)={{Y}_{2}}{{\alpha }_{2}}(\vartriangle T) \\
 & \Rightarrow {{Y}_{1}}{{\alpha }_{1}}={{Y}_{2}}{{\alpha }_{2}} \\
 & \Rightarrow \dfrac{{{Y}_{1}}}{{{Y}_{2}}}=\dfrac{{{\alpha }_{2}}}{{{\alpha }_{1}}} \\
\end{align}$
Given that,
$\begin{align}
  & \Rightarrow {{\alpha }_{1}}:{{\alpha }_{2}}=2:3 \\
 & \therefore {{\alpha }_{2}}:{{\alpha }_{1}}=3:2 \\
\end{align}$
Using the value of this ratio in the above expression, we get:
$\Rightarrow \dfrac{{{Y}_{1}}}{{{Y}_{2}}}=\dfrac{3}{2}$
Hence, the ratio of ${{Y}_{1}}:{{Y}_{2}}$ comes out to be $3:2$ .

Hence, option (C) is the correct option.

Note:
We started the problem with all the different terms having different values and it seemed like a hard question in the beginning. But, in the end it turned out to be a simple problem of ratio and proportion. Lengthy questions are not always too hard to solve if one’s concept is clear. Here also, we saw the use of just two formulas and concluded our answer shortly. So, one shouldn’t avoid them just by looking at them.