The ends Q and R of two thin wires,PQ and RS, | vedclass

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(A) Let the temperature of the junction be $T$.Since the system is in steady state and thermally insulated,the rate of heat flow through $PQ$ equals t

Vedclass · Accepted Answer

$0.78$

Vedclass · Answer

(A) Let the temperature of the junction be $T$.Since the system is in steady state and thermally insulated,the rate of heat flow through $PQ$ equals the rate of heat flow through $RS$.$\frac{d Q}{d t} = \frac{K_{PQ} A (T - 10)}{L} = \frac{K_{RS} A (400 - T)}{L}$Given $K_{PQ} = 2 K_{RS}$,we have:$2(T - 10) = 400 - T$$2T - 20 = 400 - T$$3T = 420 \Rightarrow T = 140^{\circ} C$For wire $PQ$,the temperature gradient is $\frac{dT}{dx} = \frac{140 - 10}{1} = 130^{\circ} C/m$.The temperature at a distance $x$ from $P$ is $T(x) = 10 + 130x$.The change in length $dy$ of an element $dx$ is $dy = \alpha (T(x) - 10) dx = \alpha (130x) dx$.Integrating from $x = 0$ to $x = 1$:$\Delta L = \int_0^1 \alpha (130x) dx = 130 \alpha \left[ \frac{x^2}{2} ight]_0^1 = 65 \alpha$$\Delta L = 65 	imes 1.2 	imes 10^{-5} = 78 	imes 10^{-5} m = 0.78 	imes 10^{-3} m = 0.78 \,mm$.

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