A) Al, Cu, Ag
B) Ag, Cu, Al
C) Cu, Ag, Al
D) Al, Ag, Cu
Correct Answer: A
Solution :
\[\frac{{{K}_{1}}A({{\theta }_{1}}-\theta )}{{{l}_{1}}}=\frac{{{K}_{2}}A(\theta -{{\theta }_{2}})}{{{l}_{2}}}\] is better conductor than \[{{l}_{1}}={{l}_{2}}\] and \[\Rightarrow {{K}_{1}}A({{\theta }_{1}}-\theta )={{K}_{2}}A(\theta -{{\theta }_{2}})\] is better conductor than \[\Rightarrow \theta =\frac{{{K}_{1}}{{\theta }_{1}}+{{K}_{2}}{{\theta }_{2}}}{{{K}_{1}}+{{K}_{2}}}\]. Hence conductivity in increasing order is \[K=\frac{2{{K}_{1}}{{K}_{2}}}{{{K}_{1}}+{{K}_{2}}}=\frac{2.K.2K}{K+2K}=\frac{4}{3}K\].
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