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(B) Since both metal wires are of identical dimensions,their length and area of cross-section are the same. Let them be $l$ and $A$ respectively.The r

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$\frac{2{\sigma_1}{\sigma_2}}{{\sigma_1} + {\sigma_2}}$

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(B) Since both metal wires are of identical dimensions,their length and area of cross-section are the same. Let them be $l$ and $A$ respectively.The resistance of the first wire is $R_1 = \frac{l}{\sigma_1 A}$ ...$(i)$The resistance of the second wire is $R_2 = \frac{l}{\sigma_2 A}$ ...$(ii)$Since they are connected in series,their effective resistance is $R_s = R_1 + R_2$.$R_s = \frac{l}{\sigma_1 A} + \frac{l}{\sigma_2 A} = \frac{l}{A} \left( \frac{1}{\sigma_1} + \frac{1}{\sigma_2} \right)$ ...$(iii)$If $\sigma_{eff}$ is the effective conductivity of the combination,then the total length is $2l$ and the total resistance is $R_s = \frac{2l}{\sigma_{eff} A}$ ...$(iv)$Equating equations $(iii)$ and $(iv)$,we get:$\frac{2l}{\sigma_{eff} A} = \frac{l}{A} \left( \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2} \right)$$\frac{2}{\sigma_{eff}} = \frac{\sigma_1 + \sigma_2}{\sigma_1 \sigma_2}$$\sigma_{eff} = \frac{2 \sigma_1 \sigma_2}{\sigma_1 + \sigma_2}$

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

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