question_answer

Two metal wires of identical dimensions are connected in series. If \[{{\sigma }_{1}}\] and \[{{\sigma }_{2}}\] are the conductivities of the metal wires respectively, the effective conductivity of the combination is    [NEET (Re) 2015]

A)  \[\frac{2{{\sigma }_{1}}{{\sigma }_{2}}}{{{\sigma }_{1}}+{{\sigma }_{2}}}\]

B)       \[\frac{{{\sigma }_{1}}+{{\sigma }_{2}}}{2{{\sigma }_{1}}{{\sigma }_{2}}}\]

C)   \[\frac{{{\sigma }_{1}}+{{\sigma }_{2}}}{{{\sigma }_{1}}{{\sigma }_{2}}}\] 

D)        \[\frac{{{\sigma }_{1}}{{\sigma }_{2}}}{{{\sigma }_{1}}+{{\sigma }_{2}}}\]

Correct Answer: A

Solution :

Net resistance of a metal wire having resistivity \[\rho ,\] we have

\[{{R}_{1}}={{\rho }_{1}}\frac{L}{A}\]

Similarly \[{{R}_{2}}={{\rho }_{2}}\frac{L}{A}\]

Then, net effective resistance of two metal wires,

\[{{R}_{eq}}={{R}_{1}}+{{R}_{2}}\]

\[\Rightarrow \]   \[\rho \frac{2L}{A}={{\rho }_{1}}\frac{L}{A}+{{\rho }_{2}}\frac{L}{A}\]

\[\Rightarrow \]   \[2\rho ={{\rho }_{1}}+{{\rho }_{2}}\]

As, conductivity \[\sigma =\frac{1}{\rho },\] we have

\[\frac{2}{\sigma }=\frac{1}{{{\sigma }_{1}}}+\frac{1}{{{\sigma }_{2}}}\]

\[\Rightarrow \]   \[\frac{2}{\sigma }=\frac{{{\sigma }_{1}}+{{\sigma }_{2}}}{{{\sigma }_{1}}.{{\sigma }_{2}}}\]

\[\Rightarrow \]   Net effective conductivity of combined wires,

\[\sigma =\frac{2{{\sigma }_{1}}.{{\sigma }_{2}}}{{{\sigma }_{1}}+{{\sigma }_{2}}}\]