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### Thermal Properties of Matter NCERT Solutions Class 11 Physics - Solved Exercise Question 11.10

Question 11.10:
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).
Solution:
Initial temperature, T1 = 40°C
Final temperature, T2 = 250°C
Change in temperature, ΔT = T2T1 = 210°C
Length of the brass rod at T1, l1 = 50 cm
Diameter of the brass rod at T1, d1 = 3.0 mm
Length of the steel rod at T2, l2 = 50 cm
Diameter of the steel rod at T2, d2 = 3.0 mm
Coefficient of linear expansion of brass, α1 = 2.0 × 10–5K–1
Coefficient of linear expansion of steel, α2 = 1.2 × 10–5K–1
For the expansion in the brass rod, we have:
Change in length (∆l1) / Original length (l1)  =  α1ΔT
∆l1 = 50 X (2.1 X 10-5) X 210
= 0.2205 cm
For the expansion in the steel rod, we have:
Change in length (∆l2) / Original length (l2)  =  α2ΔT
∆l1 = 50 X (1.2 X 10-5) X 210
= 0.126 cm
Total change in the lengths of brass and steel,
Δl = Δl1 + Δl2
= 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.