JEE MAIN - Physics (2019 - 9th April Evening Slot - No. 5)
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 :-
_9th_April_Evening_Slot_en_5_1.png)
$${{{\theta _1}} \over {10}} + {{9{\theta _2}} \over {10}}$$
$${{{\theta _2} + {\theta _1}} \over 2}$$
$${{{\theta _1}} \over {6}} + {{5{\theta _2}} \over {6}}$$
$${{{\theta _1}} \over {3}} + {{2{\theta _2}} \over {3}}$$
Explanation
Let interface temperature in steady state conduction is $\theta$, then assuming no heat loss through sides;
$$ \left(\begin{array}{l} \text { Rate of heat } \\ \text { flow through } \\ \text { first slab } \end{array}\right)=\left(\begin{array}{l} \text { Rate of heat } \\ \text { flow through } \\ \text { second slab } \end{array}\right) $$
$$ \begin{aligned} & \Rightarrow \frac{(3 K) A\left(\theta_2-\theta\right)}{d}=\frac{K A\left(\theta-\theta_1\right)}{3 d} \\\\ & \Rightarrow \quad 9\left(\theta_2-\theta\right)=\theta-\theta_1 \\\\ & \Rightarrow \quad 9 \theta_2+\theta_1=10 \theta \\\\ & \Rightarrow \quad \theta=\frac{9}{10} \theta_2+\frac{1}{10} \theta_1 \end{aligned} $$
$$ \left(\begin{array}{l} \text { Rate of heat } \\ \text { flow through } \\ \text { first slab } \end{array}\right)=\left(\begin{array}{l} \text { Rate of heat } \\ \text { flow through } \\ \text { second slab } \end{array}\right) $$
$$ \begin{aligned} & \Rightarrow \frac{(3 K) A\left(\theta_2-\theta\right)}{d}=\frac{K A\left(\theta-\theta_1\right)}{3 d} \\\\ & \Rightarrow \quad 9\left(\theta_2-\theta\right)=\theta-\theta_1 \\\\ & \Rightarrow \quad 9 \theta_2+\theta_1=10 \theta \\\\ & \Rightarrow \quad \theta=\frac{9}{10} \theta_2+\frac{1}{10} \theta_1 \end{aligned} $$
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