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The two ends of a rod of length L and a uniform cross-sectional area A are kept at two temperatures T₁ and T₂ (T₁ > T₂). The rate of heat transfer, dQ/dt, through the rod in a steady state is given by:

1. dQ/dt = k (T₁ - T₂)/LA
2. dQ/dt = kLA (T₁ - T₂)
3. dQ/dt = kA (T₁ - T₂)/L
4. dQ/dt = kL (T₁ - T₂)/A

Correct answer: dQ/dt = kA (T₁ - T₂)/L

Solution

The rate of heat transfer in a rod in steady state is given by Fourier's law of heat conduction: dQ/dt = kA(T₁ - T₂)/L, where k is the thermal conductivity, A is the cross-sectional area, and L is the length of the rod.

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