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 $$1$$ m at $$100C$$. Now the end P is maintained at $$100C$$, while the end S is heated and maintained at $$4000C$$. 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

Question

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 $$1$$ m at $$100C$$. Now the end P is maintained at $$100C$$, while the end S is heated and maintained at $$4000C$$. 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

Infinity Learn · Accepted Answer

Since, both are connected in series, rate of heat flow is same.​⇒$$2kAT$$−$$10L=kA400$$−TL​⇒$$T=1400C$$​Temperature Gradient of PQ, △T△$$x=140$$−$$101=130So$$, temperature at distance x, ​$$T=10+130x$$​Consider an element of length dx at distance x from P.​Increase in its $$length=dL=dx$$αT−$$10$$⇒∫$$dL=$$α$$130$$∫$$01xdx$$​⇒△$$L=1.2$$×$$10$$−$$5$$×$$130$$×$$12=78$$×$$10$$−$$5m=0.78mm$$

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