Heat And Thermodynamics Ques 46
- Two materials having coefficients of thermal conductivity " $3 K$ ’ 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, $\left(\theta _2>\theta _1\right)$. The temperature at the interface is
(a) $\frac{\theta _2+\theta _1}{2}$
(b) $\frac{\theta _1}{3}+\frac{2 \theta _2}{3}$
(c) $\frac{\theta _1}{6}+\frac{5 \theta _2}{6}$
(d) $\frac{\theta _1}{10}+\frac{9 \theta _2}{10}$
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Answer:
Correct Answer: 46.(d)
Solution:
Formula:
Series And Parallel Combination Of Rod :
- Let interface temperature in steady state conduction is $\theta$, then assuming no heat loss through sides;
$$ \begin{aligned} & \left(\begin{array}{l} \text { Rate of heat } \\ \text { flow through } \\ \text { first slab } \end{array}\right)=\left(\begin{array}{l} \text { Rate of heat } \\ \text { flow through } \\ \text { second slab } \end{array}\right) \\ & \Rightarrow \quad \frac{(3 K) A\left(\theta _2-\theta\right)}{d}=\frac{K A\left(\theta-\theta _1\right)}{3 d} \\ & \Rightarrow \quad 9\left(\theta _2-\theta\right)=\theta-\theta _1 \\ & \Rightarrow \quad 9 \theta _2+\theta _1=10 \theta \\ & \Rightarrow \quad \theta=\frac{9}{10} \theta _2+\frac{1}{10} \theta _1 \end{aligned} $$
BETA
