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Question: Answered & Verified by Expert
The two ends of a rod of length $L$ and a uniform crosssectional area $A$ are kept at two temperatures $T_{1}$ and $\mathrm{T}_{2}\left(\mathrm{~T}_{1}>\mathrm{T}_{2}\right)$. The rate of heat transfer, $\frac{\mathrm{dQ}}{\mathrm{dt}}$ through the rod in a steady state is given by:
PhysicsThermal Properties of MatterBITSATBITSAT 2021
Options:
  • A $\frac{k\left(T_{1}-T_{2}\right)}{L A}$
  • B $k L A\left(T_{1}-T_{2}\right)$
  • C $\frac{k A\left(T_{1}-T_{2}\right)}{L}$
  • D $\frac{k L\left(T_{1}-T_{2}\right)}{A}$
Solution:
2621 Upvotes Verified Answer
The correct answer is: $\frac{k A\left(T_{1}-T_{2}\right)}{L}$
$\frac{\mathrm{dQ}}{\mathrm{dt}}=\frac{\mathrm{kA}\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)}{\mathrm{L}}$ $\left[\left(\mathrm{T}_{1}-\mathrm{T}_{2}\right)\right.$ is the temperature difference $]$

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