Question
Physics Question on work done thermodynamics
The ends Q and R of two thin wires, PQ and RS, are soldered (joined) together. Initially each of the wires has a length of 1m at 10∘C. Now the end P is maintained at 10∘C, while the end S is heated and maintained at 400∘C. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQ is 1.2×10−5K−1, the change in length of the wire PQ is
A
0.78 mm
B
0.90 mm
C
1.56 mm
D
2.34 mm
Answer
0.78 mm
Explanation
Solution
KPQ=2KRS
αPQ=1.2×10−5K−1
dtdH=3R390=KAdxdT
dxdT=3RKA390
=3×KA1×KA390
dxdT=3390
T=3390x+10
Change in length of element =dxαdT
Where dT=3390x+10−10
=3390x
dℓ=3α×390xdx
Δℓ=α3390[2x2]01
=31.2×10−5×390×21
Δℓ=61.2×390×10−5×103mm
=0.78mm