Q.(a) Define current density and relaxation time. [1]

Current density and relaxation time are defined first; combining the microscopic drift-velocity picture with these definitions gives resistivity in terms of carrier density and relaxation time; and combining two cells (with the same terminal current direction) in parallel via Kirchhoff's laws gives a single equivalent cell with combined emf and reduced internal resistance.

1. Definitions: Current density $\vec{J}$ at a point in a conductor is the current flowing per unit cross-sectional area, taken perpendicular to the direction of current flow: $J = I/A$ (a vector, pointing along the direction of conventional current flow). Its SI unit is A/m². Relaxation time $\tau$ is the average time interval between two successive collisions of a free (conduction) electron with the fixed ions/atoms of the conductor's lattice, as the electron drifts under an applied electric field.
2. Resistivity in terms of $n$ and $\tau$: Under an applied field $E$, each free electron (charge $-e$, mass $m$) experiences a force $-eE$, giving acceleration $a = -eE/m$. Between collisions (average time $\tau$), an electron starting from rest (on average, after a randomising collision) gains an average drift velocity $v_d = a\tau = \dfrac{eE\tau}{m}$ (magnitude) If $n$ is the number density of free electrons, the current density is $J = nev_d = ne \times \dfrac{eE\tau}{m} = \dfrac{ne^2\tau}{m}E$ Comparing with the microscopic form of Ohm's law, $J = \sigma E = E/\rho$: $\dfrac{1}{\rho} = \dfrac{ne^2\tau}{m} \quad\Rightarrow\quad \rho = \dfrac{m}{ne^2\tau}$ This shows resistivity is inversely proportional to both the free-electron density and the relaxation time — a material with fewer free electrons, or with electrons colliding more often (smaller $\tau$), has higher resistivity.
3. Two cells in parallel: Let two cells of emf $\varepsilon_1$, $\varepsilon_2$ and internal resistances $r_1$, $r_2$ be connected in parallel between two common terminals A and B, both driving current in the same direction into the external circuit, and let $I_1$, $I_2$ be the currents supplied by each cell, with $I = I_1+I_2$ the total current delivered, and $V$ the common terminal potential difference across A, B. For cell 1: $V = \varepsilon_1 - I_1 r_1 \Rightarrow I_1 = \dfrac{\varepsilon_1 - V}{r_1}$ …