JEE PYQ: Thermodynamics Question 11
Question 11 - 2021 (26 Feb Shift 1)
The temperature $\theta$ at the junction of two insulating sheets, having thermal resistances $R_1$ and $R_2$ as well as top and bottom temperatures $\theta_1$ and $\theta_2$ (as shown in figure) is given by :
(1) $\frac{\theta_1 R_2 + \theta_2 R_1}{R_1 + R_2}$
(2) $\frac{\theta_1 R_2 - \theta_2 R_1}{R_2 - R_1}$
(3) $\frac{\theta_2 R_2 - \theta_1 R_1}{R_2 - R_1}$
(4) $\frac{\theta_1 R_1 + \theta_2 R_2}{R_1 + R_2}$
Show Answer
Answer: (1)
Solution
Temperature at the junction is $\theta$.
Using the formula: $\frac{T_2 - T}{R_1} = \frac{T - T_1}{R_2}$
$\frac{\theta_2 - \theta}{R_2} = \frac{\theta - \theta_1}{R_1}$
$R_1(\theta_2 - \theta) = R_2(\theta - \theta_1)$
$R_1\theta_2 - R_1\theta = R_2\theta - R_2\theta_1$
$R_1\theta + R_2\theta = R_1\theta_2 + R_2\theta_1$
$\theta = \frac{R_1\theta_2 + R_2\theta_1}{R_1 + R_2}$
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