Transmission Of Heat Question 78
Question: Consider a compound slab consisting of two different materials having equal thickness and thermal conductivities K and 2K respectively. The equivalent thermal conductivity of the slab is
[CBSE PMT 2003]
Options:
A) $ \sqrt{2K} $
B) $ 3K $
C) $ \frac{4}{3}K $
D) $ \frac{2}{3}K $
Show Answer
Answer:
Correct Answer: C
Solution:
$ K=\frac{2K _{1}K _{2}}{K _{1}+K _{2}}=\frac{2.K.2K}{K+2K}=\frac{4}{3}K $
BETA
