Question

Two metal rods of the same lengths and area of cross-sections are fixed end to end between rigid supports. The materials of the rods have Young's moduli $Y_{1}$ and $Y_{2}$ and coefficients of linear expansion are $\alpha _{1}\text{ and }\alpha _{2}$ . The junction between the rods does not shift if the rods are cooled. Then

• A. $Y_{1}\alpha _{1}=Y_{2}\alpha _{2}$
• B. $Y_{1}\alpha _{2}=Y_{2}\alpha _{1}$
• C. $Y_{1}\alpha _{1}^{2}=Y_{2}\alpha _{2}^{2}$
• D. $Y_{1}^{2}\alpha _{1}=Y_{2}^{2}\alpha _{2}$

Answer 1

Answer: (A)

Thermal strain in a linear rod is given, $\Delta l=l\alpha t$
where $l$ is the length of the rod and $t$ is the temperature change.
Here $Y_{1}$ and $Y_{2}$ are Young's moduli of the material of the rods and $\alpha _{1} $ and $\alpha _{2}$ are the coefficients of thermal expansion.
Using Young's modulus, $Y=\frac{\text{stress}}{\text{strain}}=\frac{T/A}{\alpha t}$
$\Rightarrow T=YA\alpha t$ , where $T$ is the tension in rod.
Tension must be the same in both the rods for their junction to be in equilibrium.
$\therefore Y_{1}A\alpha _{1}t=Y_{2}A\alpha _{2}t\Rightarrow Y_{1}\alpha _{1}=Y_{2}\alpha _{2}$