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(C) Let $	heta$ be the temperature at the middle point $C$. Since the two parts of the rod are in series,the rate of heat flow $(H)$ through both par

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$90$

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(C) Let $	heta$ be the temperature at the middle point $C$. Since the two parts of the rod are in series,the rate of heat flow $(H)$ through both parts must be the same.The rate of heat flow is given by $H = \frac{KA \Delta T}{L}$.For the first part (length $1 	ext{ m}$,area $2A$,temperature difference $100 - 	heta$):$H_1 = \frac{K(2A)(100 - 	heta)}{1}$For the second part (length $1 	ext{ m}$,area $A$,temperature difference $	heta - 70$):$H_2 = \frac{KA(	heta - 70)}{1}$Equating $H_1 = H_2$:$K(2A)(100 - 	heta) = KA(	heta - 70)$$2(100 - 	heta) = 	heta - 70$$200 - 2	heta = 	heta - 70$$3	heta = 270$$	heta = 90^{\circ}	ext{C}$

$A$ metal rod of length $2 \text{ m}$ has cross-sectional areas $2A$ and $A$ as shown in the figure. The ends are maintained at temperatures $100^{\circ}\text{C}$ and $70^{\circ}\text{C}$. The temperature at the middle point $C$ is...... $^{\circ}\text{C}$

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