A brass rod of length $50\; cm$ and diameter $3.0 \;mm$ is jotned to a steel rod of the same length and diameter. What is the change in length of the combined rod at $250\,^{\circ} C ,$ if the original lengths are at $40.0\,^{\circ} C ?$ Is there a 'thermal stress' developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass $=2.0 \times 10^{-5} \;K ^{-1},$ steel $=1.2 \times 10^{-5}\; K ^{-1} J$

Initial temperature, $T_{1}=40^{\circ} C$

Final temperature, $T_{2}=250^{\circ} C$

Change in temperature, $\Delta T=T_{2}-T_{1}=210^{\circ} C$

Length of the brass rod at $T_{1}, l_{1}=50 cm$

Diameter of the brass rod at $T_{1}, d_{1}=3.0 mm$

Length of the steel rod at $T_{2}, l_{2}=50 cm$

Diameter of the steel rod at $T_{2}, d_{2}=3.0 mm$

Coefficient of linear expansion of brass, $\alpha_{1}=2.0 \times 10^{-5} K ^{-1}$

Coefficient of linear expansion of steel, $\alpha_{2}=1.2 \times 10^{-5} K ^{-1}$

For the expansion in the brass rod, we have:

$\frac{\text { Change in length }\left(\Delta I_{1}\right)}{\text { Original length }\left(I_{1}\right)}=\alpha_{1} \Delta T$

$\therefore \Delta l_{1}=50 \times\left(2.1 \times 10^{-5}\right) \times 210$

$=0.2205 cm$

For the expansion in the steel rod, we have:

$\frac{\text { Change in length }\left(\Delta l_{2}\right)}{\text { Original length }\left(l_{2}\right)}=\alpha_{2} \Delta T$

$\therefore \Delta l_{2}=50 \times\left(1.2 \times 10^{-5}\right) \times 210$

$=0.126 cm$

Total change in the lengths of brass and steel,

$\Delta l=\Delta l_{1}+\Delta l_{2}$

$=0.2205+0.126$

$=0.346 cm$

Total change in the length of the combined rod $=0.346\, cm$

Since the rod expands freely from both ends, no thermal stress is developed at the junction