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 $2$
$\frac{3 kA}{d}\left(\left(\theta \right)_{2} - T\right)=\frac{kA}{3 d}\left(T - \left(\theta \right)_{1}\right)$
$\Rightarrow 9\left(\left(\theta \right)_{2} - T\right)=\left(T - \left(\theta \right)_{1}\right)$
$\Rightarrow 10T=9\theta _{2}+\theta _{1}$
$\Rightarrow T=\frac{9 \theta _{2} + \theta _{1}}{10}=\frac{\theta _{1}}{10}+\frac{9 \theta _{2}}{10}$