Transmission Of Heat Question 77
Question: Two rods of same length and cross section are joined along the length. Thermal conductivities of first and second rod are $ K _{1},,\text{and},,K _{2} $ . The temperature of the free ends of the first and second rods are maintained at $ {\theta _{1}},,\text{and }{\theta _{2}} $ respectively. The temperature of the common junction is
[MP PET 2003]
Options:
A) $ \frac{{\theta _{1}}+{\theta _{2}}}{2} $
B) $ \frac{K _{2}K _{2}}{K _{1}+K _{2}}({\theta _{1}}+{\theta _{2}}) $
C) $ \frac{K _{1}{\theta _{1}}+K _{2}{\theta _{2}}}{K _{1}+K _{2}} $
D) $ \frac{K _{2}{\theta _{1}}+K _{1}{\theta _{2}}}{K _{1}+K _{2}} $
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Answer:
Correct Answer: C
Solution:
At steady state, rate of heat flow for both blocks will be same i.e., $ \frac{K _{1}A({\theta _{1}}-\theta )}{l _{1}}=\frac{K _{2}A(\theta -{\theta _{2}})}{l _{2}} $ (given $ l _{1}=l _{2} $ )
$ \Rightarrow K _{1}A({\theta _{1}}-\theta )=K _{2}A(\theta -{\theta _{2}}) $
$ \Rightarrow \theta =\frac{K _{1}{\theta _{1}}+K _{2}{\theta _{2}}}{K _{1}+K _{2}} $
BETA
