Step 1: Idea behind a steady drift:
Think of an electron moving through a lattice of ions. The electric field pushes it forward. Collisions with ions take away its extra forward momentum. After a short time the two effects balance, like a ball falling at its terminal speed in a thick liquid.
Step 2: Balance of momentum (part i):
Let $\tau$ be the mean time between collisions. In this time the field gives each electron a forward momentum $eE\tau$, and the collision removes it.
Rate of momentum gain = $eE$ (force from the field).
Rate of momentum loss = $m v_d / \tau$ (collisions).
In the steady state these are equal, so:
\[ eE = \frac{m v_d}{\tau} \quad \Rightarrow \quad v_d = \frac{eE\tau}{m} \]
The right side has no time in it, so the average velocity is the same at every instant. The electrons drift opposite to $\vec{E}$.
Step 3: Current from current density:
Charge density of the moving electrons is $n e$, where n is the number of free electrons per unit volume. Current density is charge density times velocity:
\[ J = n e v_d \]
For a uniform cylinder of cross-section A, the current is $I = J A$. So:
\[ I = n e A v_d \]
Step 4: Instant current in spite of slow drift (part ii):
A drift of a few mm per second means an electron needs hours to go along a few metres of wire. Yet a lamp glows the moment the switch is pressed.
The reason is that the signal does not travel with the electrons. The electric field travels along the wire as an electromagnetic disturbance at nearly the speed of light.
Every part of the wire already holds free electrons, just like a pipe already full of water. When the field reaches them, they all begin to move at once.
Step 5: Series wires of different radius (part iii):
Let the radii be $r_1 = 3k$ and $r_2 = 2k$. Their areas are $A_1 = 9\pi k^2$ and $A_2 = 4\pi k^2$.
The current is the same in a series circuit, and n is the same for copper. So $v_d A$ is the same for both wires:
\[ v_{d1} \times 9\pi k^2 = v_{d2} \times 4\pi k^2 \]
\[ \frac{v_{d1}}{v_{d2}} = \frac{4}{9} \]
The thicker wire has the smaller drift velocity, which fits the idea that the same charge passes through a wider space.
Final Answer:
Steady drift velocity is $v_d = eE\tau/m$, the current is $I = neAv_d$, and the drift velocities of the two wires are in the ratio 4 : 9.
\[ \boxed{v_{d1} : v_{d2} = 4 : 9} \]