Rods x and y of equal dimensions but of diffe | vedclass

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(A) Let the thermal resistance of rod $x$ be $R = \frac{\ell}{k_x A}$. Since the thermal conductivity of $x$ is $3$ times that of $y$ $(k_x = 3k_y)$,

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$89^{\circ}C$ and $73^{\circ}C$ respectively

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(A) Let the thermal resistance of rod $x$ be $R = \frac{\ell}{k_x A}$. Since the thermal conductivity of $x$ is $3$ times that of $y$ $(k_x = 3k_y)$, the thermal resistance of rod $y$ is $3R$.From the figure, the network between $B$ and $E$ consists of two parallel branches: one with rods $y$ and $y$ (total resistance $3R + 3R = 6R$) and one with rods $x$ and $x$ (total resistance $R + R = 2R$).The equivalent resistance $R_{BE}$ between $B$ and $E$ is $\frac{1}{R_{BE}} = \frac{1}{6R} + \frac{1}{2R} = \frac{1+3}{6R} = \frac{4}{6R} = \frac{2}{3R}$, so $R_{BE} = 1.5R$.The total resistance of the circuit is $R_{total} = R_{AB} + R_{BE} + R_{EF} = R + 1.5R + 3R = 5.5R = \frac{11R}{2}$.The total heat current $H = \frac{100 - 40}{5.5R} = \frac{60}{5.5R} = \frac{120}{11R}$.Now, $T_B = 100 - H \cdot R = 100 - \frac{120}{11R} \cdot R = 100 - 10.91 \approx 89.09^{\circ}C$.And $T_E = 40 + H \cdot (3R) = 40 + \frac{120}{11R} \cdot 3R = 40 + 32.73 \approx 72.73^{\circ}C$.Thus, $T_B \approx 89^{\circ}C$ and $T_E \approx 73^{\circ}C$.

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