Question

Three rods of material $\text{X}$ and three of material $\text{Y}$ are connected as shown in figure. All the rods are identical in length and cross-sectional area. If the end $A$ is maintained at ${60}^{\circ }\text{C}$ and the junction $E$ at ${10}^{\circ }\text{C}$, calculate the temperature at the junction $B$. The thermal conductivity of $\text{X}$ is $800{\text{Wm}}^{-1}$${}^{\circ }{\text{C}}^{-1}$ and that of $\text{Y}$ is $400{\text{Wm}}^{-1}$${}^{\circ }{\text{C}}^{-1}$.

• A. ${35}^{\circ }\text{C}$
• B. ${40}^{\circ }\text{C}$
• C. ${45}^{\circ }\text{C}$
• D. ${50}^{\circ }\text{C}$

Answer 1

The correct option is B ${40}^{\circ }\text{C}$

It is clear from the symmetry of the figure that the points $C$ and $D$ are equivalent in all respect and hence, they are at the same temperature. So, no heat will flow through the rod $CD$.

We can, therefore, neglect this rod in further analysis.

Let, $l$ and $A$ be the length and the area of cross-section of each rod.

The thermal resistances of $AB,BC$ and $BD$ are equal, each has a value,

$R_1=\frac{1}{K_{\text{X}}}\frac{l}{A}$

Similarly, thermal resistances of $CE$ and $DE$ are equal, each having a value.

$R_2=\frac{1}{K_{\text{Y}}}\frac{l}{A}$

As the rod $CD$ has no effect, we can say that the rods $BC$ and $CD$ are joined in series. Their equivalent thermal resistance is,

$R_3=R_{BC}+R_{CE}=R_1+R_2$

Also, the rods $BD$ and $DE$ together have an equivalent thermal resistance

$R_4=R_{BC}+R_{CE}=R_1+R_2$

The resistance $R_3$ and $R_4$ are joined in parallel and are equal, hence their equivalent thermal resistance is given by

$R_5=\frac{R_3}{2}=\frac{R_1+R_2}{2}$

The following Figure shows the successive steps in this reduction


Let $T_A,T_B,T_E$ be the temperature at $A,B,E$ respectively.

As the rod $AB$ and $BE$ are in series, the thermal current will be the same in both the rods.

$\Rightarrow \frac{T_A-T_B}{R_1}=\frac{T_B-T_E}{\left(\frac{R_1+R_2}{2}\right)}$

$\Rightarrow (T_A-T_B)K_{\text{X}}=\frac{2(T_B-T_E)}{(\frac{1}{K_{\text{X}}}+\frac{1}{K_{\text{Y}}})}$

Substituting the given values,

$\Rightarrow (60-T_B)\left(800\right)=\frac{2(T_B-10)\left(800\right)\left(400\right)}{1200}$

$\Rightarrow 180-3T_B=2T_B-20$

$\Rightarrow T_B=40{}^{\circ }\text{C}$

Hence, option $\left(B\right)$ is correct.