# 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).

Given,

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.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Thermal Properties of Matter | Matering important concepts48 mins
Laws of states of matter44 mins
Interactive quiz on liquefaction44 mins
Liquefaction of gases53 mins
Gaseous state- interactive quiz56 mins
Interactive quiz on Kinetic Theory and Real Gas equation52 mins
Interactive quiz on Measurement of Pressure & Dalton's Law.49 mins
Fluid Statics67 mins