1. Given situation

A battery of emf $E$ is connected across a conductor $PQ$ whose cross-sectional area changes along its length. The radii are $r_1$ and $r_2$ with $r_2 > r_1$ (so the wire becomes thicker as we move from $P$ to $Q$).

Thus, as we move from $P$ to $Q$:

• area $A$ increases,
• current $I$ in steady state remains the same through all cross-sections.

2. Drift velocity

For a conductor,

where:

• $n$ = number density of electrons,
• $e$ = charge of electron,
• $A$ = area of cross-section,
• $v_d$ = drift velocity.

Since $I$, $n$, and $e$ are constant,

So $v_d \propto \frac{1}{A}$.

As we move from $P$ to $Q$, area increases, hence drift velocity decreases.

So Option A is false.

3. Electric field

Using microscopic Ohm's law,

where current density

Therefore,

Since $A$ increases from $P$ to $Q$, $E$ decreases.

So Option B is true.

4. Electron current

In steady state, current is the same at every cross-section of a series conductor; otherwise charge would accumulate. Thus electron current does not decrease.

So Option C is false.

5. Check option D

Since only B is true, Option D (All of these) is false.

6. Final answer

The correct option is:

7. Comparison with stored answer

Stored correct answer: $A$

My derived answer is $B$.

Reason for disagreement:

• If the wire gets thicker from $P$ to $Q$ ($r_2 > r_1$), area increases.
• Since $v_d = I/(neA)$, drift velocity must decrease, not increase.
• Electric field varies as $E = \rho I/A$, so it decreases.

Hence the stored answer appears inconsistent with the standard relations for current in a non-uniform conductor.