JEE PYQ: Thermodynamics Question 34
Question 34 - 2019 (09 Apr Shift 2)
Two materials having coefficients of thermal conductivity ‘3K’ and ‘K’ and thickness ‘$d$’ and ‘3$d$’, 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, $(\theta_2 > \theta_1)$. The temperature at the interface is:
(1) $\frac{\theta_1}{10} + \frac{9\theta_2}{10}$
(2) $\frac{\theta_2 + \theta_1}{2}$
(3) $\frac{\theta_1}{6} + \frac{5\theta_2}{6}$
(4) $\frac{\theta_1}{3} + \frac{2\theta_2}{3}$
Show Answer
Answer: (1)
Solution
$H_1 = H_2$: $\frac{3kA(\theta_2 - \theta)}{d} = \frac{kA(\theta - \theta_1)}{3d}$
$(3k)A\left(\frac{\theta_2 - \theta}{d}\right) = kA\left(\frac{\theta - \theta_1}{3d}\right)$
$9(\theta_2 - \theta) = (\theta - \theta_1)$
$\theta = \frac{\theta_1 + 9\theta_2}{10}$
BETA
