Initial temperature, T${}_1$ = 40°C

Final temperature, T${}_2$ = 250°C

Change in temperature, ΔT = T${}_2$ – T${}_1$ = 210°C

Length of the brass rod at T${}_1$, l${}_1$ = 50 cm

The 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

The diameter of the steel rod at T${}_2$, d${}_2$ = 3.0 mm

Coefficient of linear expansion of brass, ${\alpha }_1$ = 2.0 × 10${}^{-5}{K}^{-1}$

Coefficient of linear expansion of steel, ${\alpha }_2$ = 1.2 × 10${}^{-5}{K}^{-1}$

For the expansion in the brass rod,

$$\begin{gathered} \frac{\mathrm{Change} \mathrm{in} \mathrm{length} (∆{\mathrm{l}}_1)}{\mathrm{Orignal} \mathrm{length} \left({\mathrm{l}}_1\right)}=\mathrm{\alpha }∆\mathrm{T} \\ \therefore ∆{\mathrm{l}}_1=50\times (2.1\times {10}^{-5})\times 210=0.2205 \end{gathered}$$

For the expansion in the steel rod,

$\therefore ∆{\mathrm{l}}_1=50\times (2.1\times {10}^{-5})\times 210=0.126 \mathrm{cm}$

Total change in the lengths of brass and steel,

Δl = Δl${}_1$ + Δl${}_2$

= 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.