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(D) Under steady-state conditions,the rate of heat flow $H$ through a rod is constant and is given by the formula:$H = \frac{KA(T_1 - T_2)}{L} = \frac

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(D) Under steady-state conditions,the rate of heat flow $H$ through a rod is constant and is given by the formula:$H = \frac{KA(T_1 - T_2)}{L} = \frac{KA(T_1 - T_x)}{x}$where $T_1$ is the temperature at the hot end $(x=0)$,$T_2$ is the temperature at the cold end $(x=L)$,and $T_x$ is the temperature at distance $x$.Rearranging the equation for $T_x$:$T_x = T_1 - \left( \frac{T_1 - T_2}{L} ight) x$This is a linear equation of the form $y = mx + c$,where the slope $m = -\frac{T_1 - T_2}{L}$ is negative and constant.Therefore,the temperature $	heta$ decreases linearly with distance $x$,which corresponds to a straight line with a negative slope.

Heat flows through a long metallic rod from one end to the other under steady-state conditions. The variation of temperature $\theta$ with distance $x$ from the hot end is shown in which of the following graphs?

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