A) \[\frac{1}{3}{{K}_{A}}\]
B) \[3{{K}_{A}}\]
C) \[2{{K}_{A}}\]
D) \[\frac{2}{3}{{K}_{A}}\]
Correct Answer: A
Solution :
\[\] Given, \[{{K}_{B}}={{K}_{A}}/2,\] and \[{{K}_{B}}=3{{K}_{C}}\] \[\therefore \] \[{{K}_{C}}={{K}_{A}}/6\] Rods are in series form so \[\frac{L}{K}=\frac{{{l}_{1}}}{{{K}_{A}}}+\frac{{{l}_{2}}}{{{K}_{B}}}+\frac{{{l}_{3}}}{{{K}_{C}}}\] \[(\because l={{l}_{1}}={{l}_{2}}={{l}_{3}})\] \[\frac{3l}{K}=\frac{l}{{{K}_{A}}}+\frac{l}{{{K}_{A}}/2}+\frac{l}{{{K}_{A}}/6}\] or \[\frac{3l}{K}=\frac{9l}{{{K}_{A}}}\] or \[K=\frac{{{K}_{A}}}{3}\]
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