Question

Two unequal bars in length and cross-section with thermal conductivities $K_1$ and $K_2$ are connected in series so that one end is maintained at ${\theta }_1{}^{\circ }C$ and other end at ${\theta }_2{}^{\circ }C$. If the ratio of length of two bars is $n_1$ and ratio of cross-sectional areas of two bars $n_1$ is, if temperature of junction point is unknown and $\frac{n_1}{n_2}=\frac{2K_1}{K_2}$ , find the temperature of the junction point.

• A. $\frac{3({\theta }_1+{\theta }_2)}{2}$
• B. $\frac{2{\theta }_1+{\theta }_2}{3}$
• C. $\frac{2({\theta }_1+{\theta }_2)}{3}$
• D. $\frac{{\theta }_1+2{\theta }_2}{3}$

Answer 1

The correct option is D $\frac{{\theta }_1+2{\theta }_2}{3}$

Step-1: Find relation between thermal resistance between rods.

Given :
$\frac{l_1}{l_2}=n_1$,

$\frac{A_1}{A_2}=n_2$,

$\frac{n_1}{n_2}=\frac{2K_1}{K_2}$.

$\frac{n_1}{n_2}=\frac{2K_1}{K_2}$

$n_1=\frac{2K_1}{K_2}.n_2$

$\frac{l_1}{l_2}=\frac{2K_1}{K_2}=\frac{A_1}{A_2}$

$\frac{l_1}{K_1{A}_1}=\frac{2l_2}{K_2{A}_2}$

$R_1=2R_2$

Step-2: Find junction temperature ${\theta }_J$ of rods.

Formula used: $I=\Delta \theta /R$

If $R_2=R$ , then $R_1=2R$

So,

$I=\frac{{\theta }_1-{\theta }_J}{2R}=\frac{{\theta }_J-{\theta }_2}{R}$

${\theta }_1-{\theta }_J=2{\theta }_J-2{\theta }_2$

$\therefore {\theta }_J=\frac{{\theta }_1+2{\theta }_2}{3}$

Final answer: (b)