Two wires of 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
(A). $\dfrac{{{\rho _1}{l_2} + {\rho _2}{l_1}}}{{{\rho _1}{\rho _2}}}$
(B). $\dfrac{{{\rho _1}{l_1} + {\rho _2}{l_2}}}{{{l_1} + {l_2}}}$
(C). $\dfrac{{{\rho _1}{l_1} - {\rho _2}{l_2}}}{{{l_1} - {l_2}}}$
(D). $\dfrac{{{\rho _1}{l_2} + {\rho _2}{l_1}}}{{{l_1} + {l_2}}}$

- Hint: Resistance of a wire depends on factors like its resistivity, area of cross section, length. Resistivity differs from material to material. The wires are connected in series, and they have same radius (area) therefore this is equivalent to a circuit with length${l_1} + {l_2}$, area $A$ with resistivity $\rho {}_{equivalent.}$ .We use the equation of $R{}_{effective}$when connected in series.

Formula used
$R = \dfrac{{\rho \times l}}{A}$, where $R$is the resistance of the wire, $\rho $ is the resistivity of the wire, $l$ is the length of the wire and $A$ is the area of cross section of the wire.
$R{}_{effective} = {R_1} + {R_2} + {R_3}.... + {R_n}$ .Here $R{}_{effective}$ is the effective resistance when n resistors are connected in series.$R{}_1$, ${R_2}$…. ${R_n}$ are their respective resistances.

Complete step-by-step solution -
Resistance is the measure of opposition of current in a circuit. The higher the resistance the lower will be the current flow and vice versa. Resistivity is the resistive power of the material to oppose or hinder the flow of current. Resistance is directionally proportional to the length i.e. the longer the wire, more resistance it experiences and hence lower current flow and vice versa. Similarly as the area of the cross section increases the resistance decreases. This implies that the area of cross section is inversely proportional to area of cross section. Thus we can say$R = \dfrac{{\rho \times l}}{A}$, where $R$is the resistance of the wire, $\rho $ is the resistivity of the wire, $l$ is the length of the wire and $A$ is the area of cross section of the wire. Let $A$ be the area of both the wires, their respective lengths be ${l_1}$ and ${l_2}$ , their resistivities be $\rho {}_1$ and $\rho {}_2$
Therefore ${R_1} = \dfrac{{{\rho _1} \times {l_1}}}{A}$and${R_2} = \dfrac{{{\rho _2} \times {l_2}}}{A}$ . Let the effective resistance be ${R_{effective}} = \dfrac{{{\rho _{equivalent}} \times ({l_1} + {l_2})}}{A}$
We know,$R{}_{effective} = {R_1} + {R_2} + {R_3}.... + {R_n}$ .Here $R{}_{effective}$ is the effective resistance when n resistors are connected in series.$R{}_1$, ${R_2}$…. ${R_n}$ are their respective resistances.
Here there are two resistances therefore $R{}_{effective} = {R_1} + {R_2}$
Substituting $\dfrac{{{\rho _{equivalent}} \times ({l_1} + {l_2})}}{A} = \dfrac{{{\rho _1} \times {l_1}}}{A} + \dfrac{{{\rho _2} \times {l_2}}}{A}$
$A$$A$ area of cross section , which is common in both sides cancels out and by rearranging we get
${\rho _{equivalent}} = \dfrac{{{\rho _1} \times {l_1} + {\rho _2} \times {l_2}}}{{({l_1} + {l_2})}}$
Thus the correct answer is B

Note: All materials resist flow of current but at different levels. Based on how freely the current is flowing in a wire it is classified into groups. Conductors offer very little resistance and electrons move freely in a conductor. Aluminum , gold are some of the examples. Insulators offer high resistance and resistance to the flow of electrons. Wood, plastic are some examples. A semiconductor material has a resistance value falling between that of a conductor and an insulator, such as glass.