A copper rod and a steel rod of equal cross-s | vedclass

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(C) In steady state,the heat current $H$ flowing through both rods is the same.The heat current is given by $H = \frac{KA \Delta T}{L}$,which can be w

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(C) In steady state,the heat current $H$ flowing through both rods is the same.The heat current is given by $H = \frac{KA \Delta T}{L}$,which can be written as $\Delta T = H \cdot \frac{L}{KA}$.For a small segment of length $dx$,the temperature drop is $dT = H \cdot \frac{dx}{KA}$.Thus,the temperature gradient is $\frac{dT}{dx} = -\frac{H}{KA}$.Since $H$ and $A$ are constant,the magnitude of the temperature gradient is inversely proportional to the thermal conductivity $K$ (i.e.,$|\frac{dT}{dx}| \propto \frac{1}{K}$).Given $K_{Cu} > K_{Steel}$,the magnitude of the temperature gradient in the copper rod is smaller than that in the steel rod.This means the slope of the $T-x$ graph is less steep for the copper rod $(0 Graph $C$ correctly represents this behavior.

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