Question

Two materials having coefficients of thermal conductivity '3K' and 'K' and thickness 'd' and '3d', 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, $(\theta_2 > \theta_1)$. The temperature at the interface is :-

• A. $\frac{\theta_{2} + \theta_{1}}{2} $
• B. $ \frac{\theta_{1}}{10} + \frac{9\theta_{2}}{10}$
• C. $ \frac{\theta_{1}}{3} + \frac{2\theta_{2}}{3}$
• D. $ \frac{\theta_{1}}{6} + \frac{5\theta_{2}}{6}$

Answer 1

Answer: (B)

Let the temperature of interface be " $\theta$ "
$i_1 = i_2$ {Steady state conduction}
$\frac{3KA\left(\theta_{2}-\theta\right)}{d} = \frac{KA\left(\theta - \theta_{1}\right)}{3d} $
$\theta = \frac{9 \theta_{2}}{10} + \frac{\theta_{1}}{10} $