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Question
A wire is made by attaching two segments together end to end. One segment is made of aluminium and the other is steel. The effective coefficient of linear expansion of the two segment wire is 19×10−6/(∘C). The fractional length of aluminium is
(linear coefficients of thermal expansion of aluminium and steel are 23×10−6/(∘C) and 12×10−6/(∘C) respectively)
(linear coefficients of thermal expansion of aluminium and steel are 23×10−6/(∘C) and 12×10−6/(∘C) respectively)
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Solution
The change in length of wire =lAlαAlΔθ+lStαStΔθ associated with temperature change of Δθ,
where,
αAl and αSt are the coefficient of linear expansion of aluminium and steel, respectively
αAl=23×10−6/∘C
αSt=12×10−6/∘C
The effective coefficient of linear expansion of the two segments of wire =19×10−6/∘C
l1αAlΔθ+l2αStΔθ=(l1+l2)αΔθ
l1(l1+l2)=α−l2(l1+l2)αStαAl
Let [l1l1+l2=x](l2l1+l2)=1−x
x=α−(1−x)αStαAl
x=α−αStαAl−αSt=19×10−6−12×10−623×10−6−12×10−6=711
where,
αAl and αSt are the coefficient of linear expansion of aluminium and steel, respectively
αAl=23×10−6/∘C
αSt=12×10−6/∘C
The effective coefficient of linear expansion of the two segments of wire =19×10−6/∘C
l1αAlΔθ+l2αStΔθ=(l1+l2)αΔθ
l1(l1+l2)=α−l2(l1+l2)αStαAl
Let [l1l1+l2=x](l2l1+l2)=1−x
x=α−(1−x)αStαAl
x=α−αStαAl−αSt=19×10−6−12×10−623×10−6−12×10−6=711
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