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(B) The resistance of a wire is given by $R = \frac{\rho l}{A}$,where $\rho$ is resistivity,$l$ is length,and $A$ is the cross-sectional area.When two

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$\frac{\rho_{1} l_{1}+\rho_{2} l_{2}}{l_{1}+l_{2}}$

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(B) The resistance of a wire is given by $R = \frac{\rho l}{A}$,where $\rho$ is resistivity,$l$ is length,and $A$ is the cross-sectional area.When two wires are connected in series,the total resistance $R_{eq}$ is the sum of individual resistances:$R_{eq} = R_{1} + R_{2}$Since the wires have the same radius,their cross-sectional areas are equal $(A_{1} = A_{2} = A)$.$R_{eq} = \frac{\rho_{1} l_{1}}{A} + \frac{\rho_{2} l_{2}}{A} = \frac{\rho_{1} l_{1} + \rho_{2} l_{2}}{A}$For the equivalent wire,the total length is $L = l_{1} + l_{2}$ and the area is $A$. The equivalent resistivity $\rho_{eq}$ is defined as:$R_{eq} = \frac{\rho_{eq} (l_{1} + l_{2})}{A}$Equating the two expressions for $R_{eq}$:$\frac{\rho_{eq} (l_{1} + l_{2})}{A} = \frac{\rho_{1} l_{1} + \rho_{2} l_{2}}{A}$$\rho_{eq} = \frac{\rho_{1} l_{1} + \rho_{2} l_{2}}{l_{1} + l_{2}}$

Two wires of the same radius having lengths $l_{1}$ and $l_{2}$ and resistivities $\rho_{1}$ and $\rho_{2}$ are connected in series. The equivalent resistivity will be

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