A) \[\frac{{{k}_{1}}+{{k}_{2}}}{{{k}_{1}}}\]
B) \[\frac{{{k}_{1}}+{{k}_{2}}}{{{k}_{2}}}\]
C) \[\frac{{{k}_{2}}}{{{k}_{1}}+{{k}_{2}}}\]
D) \[\frac{{{k}_{1}}}{{{k}_{2}}}\]
Correct Answer: D
Solution :
\[\theta =\frac{{{\theta }_{1}}+{{\theta }_{2}}}{2}\] \[2\theta ={{\theta }_{1}}+{{\theta }_{2}}\] \[{{\theta }_{1}}-\theta =\theta -{{\theta }_{2}}\] ?.(1) Bars are in series Heat current\[=H=\] \[\frac{{{\theta }_{1}}-\theta }{{{R}_{1}}}=\frac{\theta -{{\theta }_{2}}}{{{R}_{2}}}\] ?(2) From eq. (1) and (2) \[{{R}_{1}}={{R}_{2}}\] \[\frac{{{l}_{1}}}{{{K}_{1}}{{A}_{1}}}=\frac{{{l}_{2}}}{{{K}_{2}}{{A}_{2}}}\] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{{{K}_{1}}}{{{K}_{2}}}\times \frac{{{A}_{1}}}{{{A}_{2}}}\] \[{{n}_{1}}=\frac{{{K}_{1}}}{{{K}_{2}}}.{{n}_{2}}\] \[\frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{K}_{1}}}{{{K}_{2}}}\]You need to login to perform this action.
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