Question

Two materials having coefficients of thermal conductivity $3K$ and $K$ and thickness $d$ and $3\text{d}$ respectively, are joined to form a slab as shown in the figure. The temperatures of the outer surfaces are ${\theta }_2$ and ${\theta }_1$ respectively, $\left({\left(\theta \right)}_2>{\left(\theta \right)}_1\right).$ The temperature at the interface is

• A. $\frac{{\theta }_2+{\theta }_1}{2}$
• B. $\frac{{\theta }_1}{6}+\frac{5{\theta }_2}{6}$
• C. $\frac{{\theta }_1}{3}+\frac{2{\theta }_2}{3}$
• D. $\frac{{\theta }_1}{10}+\frac{9{\theta }_2}{10}$

Answer 1

Answer: (D)

Let the temperature of the junction $T^{\circ }C$.
Rate of heat flow in Rod $1$ = rate of heat flow in Rod $2$
$\frac{3kA}{d}\left({\left(\theta \right)}_2-T\right)=\frac{kA}{3d}\left(T-{\left(\theta \right)}_1\right)$
$\Rightarrow 9\left({\left(\theta \right)}_2-T\right)=\left(T-{\left(\theta \right)}_1\right)$
$\Rightarrow 10T=9{\theta }_2+{\theta }_1$
$\Rightarrow T=\frac{9{\theta }_2+{\theta }_1}{10}=\frac{{\theta }_1}{10}+\frac{9{\theta }_2}{10}$