| 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 \[\frac{ACB}{ADB}=3\]. |
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A) \[\frac{7}{3}k\]
B) \[2k\]
C) \[\frac{5}{2}k\]
D) \[3k\]
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\[{{H}_{2}}=2H-{{H}_{1}}=\frac{7kA}{3l}({{T}_{1}}-{{T}_{2}})\]\[=\frac{k'A}{l}({{T}_{1}}-{{T}_{2}})\] \[\Rightarrow \] \[k'=\frac{7}{3}k\]
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