The correct option is (A) 8 × 10–3 V cm–1.
Explanation:
i = [(EMF of battery) / (r + R)] = [2 / (1 + 4)] = (2/5)A = 0.4A
Potential difference across wire = 0.4 × 4 = 1.6V
Potential gradient = [(Potential difference across the wire) / (length of wire)]
= [(1.6V) / (200cm)] = 8 × 10–3 V/cm