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Question

Physics Question on work done thermodynamics

The ends QQ and RR of two thin wires, PQPQ and RSRS, are soldered (joined) together. Initially each of the wires has a length of 1m1\, m at 10C10^{\circ} C. Now the end PP is maintained at 10C10^{\circ} C, while the end SS is heated and maintained at 400C400^{\circ} C. The system is thermally insulated from its surroundings. If the thermal conductivity of wire PQPQ is twice that of the wire RS and the coefficient of linear thermal expansion of PQPQ is 1.2×105K11.2 \times 10^{-5} K ^{-1}, the change in length of the wire PQPQ is

A

0.78 mm

B

0.90 mm

C

1.56 mm

D

2.34 mm

Answer

0.78 mm

Explanation

Solution

KPQ=2KRSK _{ PQ }=2 K _{ RS }
αPQ=1.2×105K1\alpha_{ PQ }=1.2 \times 10^{-5} K ^{-1}
dHdt=3903R=KAdTdx\frac{ dH }{ dt }=\frac{390}{3 R }= KA \frac{ dT }{ dx }
dTdx=3903RKA\frac{ dT }{ dx }=\frac{390}{3 RKA }
=3903×1KA×KA=\frac{390}{3 \times \frac{1}{ KA } \times KA }
dTdx=3903\frac{ dT }{ dx }=\frac{390}{3}
T=390x3+10T =\frac{390 x }{3}+10

Change in length of element =dxαdT= dx \alpha dT
Where dT=390x3+1010dT =\frac{390 x }{3}+10-10
=390x3=\frac{390 x }{3}
d=α×390x3dxd \ell=\frac{\alpha \times 390 x }{3} dx
Δ=α3903[x22]01\Delta \ell=\alpha \frac{390}{3}\left[\frac{ x ^{2}}{2}\right]_{0}^{1}
=1.2×105×3903×12=\frac{1.2 \times 10^{-5} \times 390}{3} \times \frac{1}{2}
Δ=1.2×3906×105×103mm\Delta \ell=\frac{1.2 \times 390}{6} \times 10^{-5} \times 10^{3} mm
=0.78mm=0.78\, mm