Question
A rod of length $l$ (laterally thermally insulated) of uniform cross-sectional area $A$ consists of a material whose thermal conductivity varies with temperature as $K=\frac{K_{o}}{a+b T}$, where $K_{0}, a$ and $b$ are constants. $T_{1}$ and $T_{2}\left(<T_{1}\right)$ are the temperature of two ends of rod. Then rate of flow of heat across the rod is(A) $\frac{A K_{0}}{b l}\left(\frac{a+b T_{1}}{a+b T_{2}}\right)$(B) $\frac{A K_{0}}{b l}\left(\frac{a+b T_{2}}{a+b T_{1}}\right)$(C) $\frac{A K_{0}}{b l} \ln \left[\frac{a+b T_{1}}{a+b T_{2}}\right]$(D) $\frac{A K_{0}}{a l} \ln \left[\frac{a+b T_{2}}{a+b T_{1}}\right]$
Step 1
Step 1: The rate of flow of heat across the rod is given by the formula $q=-KA\frac{dT}{dx}$, where $K$ is the thermal conductivity, $A$ is the cross-sectional area, and $\frac{dT}{dx}$ is the temperature gradient. Show more…
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