(i) Definition of drift velocity (ii) Derivation of expression of resistivity Factors affecting resistivity (iii) Reason of using constantan and manganin (i) Average velocity acquired by the electrons in the conductor in the presence of external electric field. [Alternatively: $v_d = \frac{-e E \tau}{m}$ where $\tau$ is the relaxation time.] (ii) $v_d = \frac{-e E \tau}{m}$ We have $E = -\frac{V}{\ell}$, where $V$ is potential across the length $l$ of the conductor $v_d = \frac{e V \tau}{m \ell}$ $\text{Current flowing}\; I = n e A v_d$ $I = n e A v_d \frac{e V r}{m \ell}$ $= \frac{n e^2 A V \tau}{m \ell}$ $\frac{I}{V} = \frac{n e^2 A \tau}{m \ell} = \frac{1}{R}$ ....(i) Also, $R = \rho \frac{\ell}{A}$ .....(ii) Comparing (i) and (ii), $\rho = \frac{m}{n e^2 \tau}$ Resistivity of the material of a conductor depends on the relaxation time, i.e., temperature and the number density of electrons. (iii) Because constantan and manganin show very weak dependence of resistivity on temperature. OR (i) Working principle of potentiometer (ii) Calculation of potential gradient and balance length (i) When constant current flows through a conductor of uniform area of cross section, the potential difference, across a length $l$ of the wire, is directly proportional to that length of the wire. $[V \propto l$ (Provided current and area are constant] (ii) Current flowing in the potentiometer wire $i = \frac{E}{R_{\text{total}}} = \frac{2.0}{15+10} = \frac{2}{25}\text{ A}$ $\therefore$ Potential difference across the two ends of the wire $V_{AB} = \frac{2}{25} \times 10\text{ V} = \frac{20}{25} = 0.8\text{ volt}$Hence potential gradient $K = \frac{V_{AB}}{l_{AB}} = \frac{0.8}{1.0} = 0.8\ \frac{\text{V}}{\text{m}}$Current flowing in the circuit containing experimental cell, $= \frac{1.5}{1.2+0.3} = 1\text{ A}$Hence, potential difference across length $AO$ of the wire $= 0.3 \times 1\text{ V} = 0.3\text{ V}$ $\Rightarrow 0.3 = K \times l_{AO}$ $= 0.8 \times I_{AO}$ $\Rightarrow l_{AO} = \frac{0.3}{0.8}\text{ m} = 0.375\text{ m}$ $=37.5\text{ cm}$