A battery of emf E is connected across a conductor as shown. As one observes from A to B,
Column-I Column-II
(i) current (p) increases
(ii) drift velocity of electron (q) decreases
(iii) electric field (r) remains same
(iv) potential drop per unit length (s) cannot be determined
Now, match the given columns and select the correct option from the codes given below.

Moderate

Solution

Current through any cross-section should remain same as all the cross-sections are in series.
$I = {neAv}_{d},$ area increases drift velocity decreases. E is directly proportional to $v_{d},$ so E also decreases. lf E decreases, then potential difference across a segment of fixed length also decreases.
This can also be done like this:
$\begin{array}{l} dR = \frac{ρdx}{A}; A increases with x . \\ Potential across dx : dV = IdR = \frac{Iρdx}{A} \\ E = \frac{dV}{dx} = \frac{Iρ}{A} \propto \frac{1}{A} \end{array}$
So E and $\frac{d V}{d x}$ both decrease as A increases.

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	Column-I		Column-II
(i)	current	(p)	increases
(ii)	drift velocity of electron	(q)	decreases
(iii)	electric field	(r)	remains same
(iv)	potential drop per unit length	(s)	cannot be determined