A rod of length l and of a uniform cross-sectional area A is made of a material of non-uniform thermal conductivity. Its thermal conductivity is temperature dependent and varies as K=BT, where B is a constant. If ends of the rod are maintained at constant temperatures T1 and T2 with T1>T2 , rate of flow of heat in steady state will be :

Q=K.A−ΔTΔx ∫0LQdx=−BA∫T1T2dTT

A rod of length l and of a uniform cross-sectional area A is made of a material of non-uniform thermal conductivity. Its thermal conductivity is temperature dependent and varies as K=BT, where B is a constant. If ends of the rod are maintained at constant temperatures T1 and T2 with T1>T2 , rate of flow of heat in steady state will be :

Q=K.A−ΔTΔx ∫0LQdx=−BA∫T1T2dTT

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A rod of length l and of a uniform cross-sectional area A is made of a material of non-uniform thermal conductivity. Its thermal conductivity is temperature dependent and varies as $K = \frac{B}{T}$ , where B is a constant. If ends of the rod are maintained at constant temperatures $T_{1}$ and $T_{2}$ with $T_{1} > T_{2}$ , rate of flow of heat in steady state will be :

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$\frac{BA}{l} \ln (\frac{T_{1}}{T_{2}})$

$\frac{Bl}{A} \ln (T_{1} T_{2})$

$\frac{BA}{l} (\frac{T_{1}}{T_{2}})$

$BAlln (\frac{T_{1}}{T_{2}})$

answer is C.

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Detailed Solution

$Q = K . A (- \frac{ΔT}{Δx})$ $\int_{0}^{L} Qdx = - BA \int_{T_{1}}^{T_{2}} \frac{dT}{T}$

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