A brass rod of length `50 cm` and diameter `3.0 cm` 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 length are at `40.0^(@)C`?
(Coefficient of linear expansion of brass `=2.0 xx 10^(-5)//^(@)C, steel = 1.2 xx 10^(-5)//^(@)C`

Explanation

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`
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"(Deltal_(1)))/("Original length "(l_(1)))=a_(1)DeltaT`
`:. Deltal_(1)=50xx(2.1xx10^(-5))xx210`
`=0.2205` cm
For the expansion in the steel rod, we have:
`("Change in length"(Deltal_(2)))/("Original length "(l_(2)))=a_(2)DeltaT`
`:. Deltal_(2)=50xx(2.1xx10^(-5))xx210`
`=0.126 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.