### Q.10:- A brass rod of length 50 cm and diameter 3.0 mm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250 °C, if the original lengths are at 40.0 °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 × 10^{–5 }K^{–1}, steel = 1.2 × 10^{–5 }K^{–1}).

**Answer:-**Initial temperature, *T*_{1} = 40°C

Final temperature, *T*_{2} = 250°C

Change in temperature, Δ*T* = *T*_{2} – *T*_{1 }= 210°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, α_{1} = 2.0 × 10^{–5}K^{–1}

Coefficient of linear expansion of steel, α_{2} = 1.2 × 10^{–5}K^{–1}

For the expansion in the brass rod, we have:

Change in length (∆*l*_{1}) / Original length (*l*_{1}) = α_{1}Δ*T*

∴ ∆*l*_{1} = 50 × (2.1 × 10^{-5}) × 210

= 0.2205 cm

For the expansion in the steel rod, we have:

Change in length (∆*l*_{2}) / Original length (*l*_{2}) = α_{2}Δ*T*

∴ ∆*l*_{1} = 50 × (1.2 × 10^{-5}) × 210

= 0.126 cm

Total change in the lengths of brass and steel,

Δ*l* = Δ*l*_{1} + Δ*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.