A) \[\frac{{{\rho }_{1}}{{l}_{1}}+{{\rho }_{2}}{{l}_{2}}}{{{l}_{1}}+{{l}_{2}}}\]
B) \[\frac{{{\rho }_{1}}{{l}_{2}}+{{\rho }_{2}}{{l}_{1}}}{{{l}_{1}}-{{l}_{2}}}\]
C) \[\frac{{{\rho }_{1}}{{l}_{2}}+{{\rho }_{2}}{{l}_{1}}}{{{l}_{1}}+{{l}_{2}}}\]
D) \[\frac{{{\rho }_{1}}{{l}_{1}}-{{\rho }_{2}}{{l}_{2}}}{{{l}_{1}}-{{l}_{2}}}\]
Correct Answer: A
Solution :
\[{{R}_{1}}=\frac{{{\rho }_{1}}{{l}_{1}}}{A}\,\text{and }{{R}_{2}}=\frac{{{\rho }_{2}}{{l}_{2}}}{A}\]. In series \[{{R}_{eq}}={{R}_{1}}+{{R}_{2}}\] \[\frac{{{\rho }_{eq.}}({{l}_{1}}+{{l}_{2}})}{A}=\frac{{{\rho }_{1}}{{l}_{1}}}{A}+\frac{{{\rho }_{2}}{{l}_{2}}}{A}\]\[\Rightarrow \,\,{{\rho }_{eq}}=\frac{{{\rho }_{1}}{{l}_{1}}+{{\rho }_{2}}{{l}_{2}}}{{{l}_{1}}+{{l}_{2}}}\].
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