Question

Three rods of material $X$ and three rods of material $Y$ are connected as shown in figure. All the rods are identical in length and cross-sectional area. If the end $A$ is maintained at ${60}^{\circ }C$ and the junction $E$ at ${10}^{\circ }C$, calculate the temperatures of the junctions $B,C,D$. The thermal conductivity of $X$ is $0.92$ $CGS$ units and that of $Y$ is $0.46CGS$ units.

Answer 1

$R_x\propto \frac{1}{K_x},R_y\propto \frac{1}{K_y}\frac{R_x}{R_y}=\frac{K_y}{K_x}=\frac{0.46}{0.92}=\frac{1}{2}$
Let $R_x=R\therefore R_y=2R$
Let us find out the total resistance.
$\sum R=R_y+$ effective resistance in the bridge
$\sum R=2R+\frac{2R\times 4R}{2R+4R}=2R+\frac{4}{3}R=\frac{10}{3}R$
and $△\theta =I\times R$
For $A$ and $B,{\theta }_A-{\theta }_B=60-{\theta }_B=2R\times I....\left(1\right)$
For $B$ and $C,{\theta }_B-{\theta }_C=2/3(I\times R)....\left(2\right)$
For $B$ and $C,{\theta }_C-{\theta }_E=2/3(I\times R).....\left(3\right)$
For $A$ and $E,{\theta }_A-{\theta }_E=60-10=50$
$\Rightarrow 10/3(R\times I)=50$
$\therefore R\times I=15$
$\therefore {\theta }_A-{\theta }_B=2\times 15=30\Rightarrow {\theta }_B={60}^{\circ }-{30}^{\circ }={30}^{\circ }C$,
Further, ${\theta }_B-{\theta }_C=\left(2/3\right)\times 15=10$
$\therefore {\theta }_C=30-10={20}^{\circ }C$
Obviously, ${\theta }_C={\theta }_D={20}^{\circ }C$.