Thermal Properties of Matter
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Question : 10
Total: 22
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 arefree to expand (Coefficient of linear expansion of brass = 2.0 × 10 – 5 K – 1 , = 1.2 × 10 – 5 K – 1 ) .
Solution:
For brass rod : α = 2.0 × 10 − 5 K − 1
l 1 = 50 c m ;
Δ T = 250 − 40 = 210 ° C
The length of the brass rod at 250°C is given byl 2 = l 1 ( 1 + α Δ T )
= 50 ( 1 + 2.0 × 10 – 5 × 210 )
= 50.21 c m
For steel rod:α ′ = 1.2 × 10 − 5 K − 1
l 1 ′ = 50 c m ;
Δ T ′ = 250 − 40 = 210 ° C
The length of the steel rod at 250°C is given by
l 2 ′ = l 1 ′ ( 1 + α ′ Δ T ′ )
= 50 ( 1 + 1.2 × 10 − 5 × 210 )
= 50.126 c m
Therefore, the length of the combined rod at 250°C
= l 2 + l 2 ′
= 50.21 + 50.126
= 100.336 c m
As the length of the combined rod at40 ° C = 50 + 50 = 100
The change in length of the combined rod at 250°C
= 100.336 – 100.0 = 0.336 c m
No thermal stress is developed at the junction since the rod freely expand.
The length of the brass rod at 250°C is given by
For steel rod:
The length of the steel rod at 250°C is given by
Therefore, the length of the combined rod at 250°C
As the length of the combined rod at
The change in length of the combined rod at 250°C
No thermal stress is developed at the junction since the rod freely expand.
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