Q. 103.5( 12 Votes )
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).
Initial temperature, T1 = 40°C = 40 + 273.15 K = 313.15K
Final temperature, T2 = 250°C = 250 + 273.15 = 523.15 K
Length of the brass rod at T1, L1 = 50 cm = 0.5 m
Diameter of the brass rod at T1, d1 = 3.0 mm = 3 × 10-3 m
Length of the steel rod at T2, L2 = 50 cm = 0.5 m
Diameter of the steel rod at T2, d2 = 3.0 mm = 3 × 10-3 m
Coefficient of linear expansion of brass, α1 = 2.0 × 10–5 K–1 Coefficient of linear expansion of steel, α2 = 1.2 × 10–5 K–1
Change in temperature, ΔT = T2 – T1 = 523.15 K-313.15K=210 K
For the expansion in the brass rod,
Change in length (∆L1)/Original length (L1) = α1ΔT
⇒ ΔL1 = α1L1ΔT
⇒ ΔL1 = (2.0 × 10–5 K–1)×(0.5 m)×(210 K)
⇒ ΔL1 = 0.0021 m = 0.21 cm
For the expansion in the steel rod,
Change in length (∆L2)/Original length (L2) = α2ΔT
⇒ ∆L2 = α2L2ΔT
⇒ ∆L2 = (1.2 × 10–5 K–1)×(0.5 m)×(210 K)
⇒ ∆L2 = 0.00126 m = 0.126 cm
Total change in length, ΔL = ∆L1 + ∆L2
⇒ ΔL = 0.21 cm + 0.126 cm
⇒ ∆L= 0.336 cm
No thermal stress develops at the junction since both the ends of the rod expand freely.
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