Transmission Of Heat Question 323
Question: A ring consisting of two parts ADB and ACB of same conductivity k carries an amount of heat H. The ADB part is now replaced with another metal keeping the temperatures $ T _{1} $ and $ T _{2} $ constant. The heat carried increases to 2H. What should be the conductivity of the new ADB part? Given $ \frac{ACB}{ADB}=3 $ .
Options:
A) $ \frac{7}{3}k $
B) $ 2k $
C) $ \frac{5}{2}k $
D) $ 3k $
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Answer:
Correct Answer: A
Solution:
[a] $ H _{1}+H _{2}=\frac{kA(T _{1}-T _{2})}{3l}+\frac{kA(T _{1}-T _{2})}{l} $
$ =\frac{4}{3l}kA(T _{1}-T _{2}) $ . In later case$ H _{2}=2H-H _{1}=\frac{7kA}{3l}(T _{1}-T _{2}) $
$ =\frac{k’A}{l}(T _{1}-T _{2}) $
$ \Rightarrow $ $ k’=\frac{7}{3}k $
BETA
