Potentiometer null point balances unknown e.m.f. with potential drop across XZ

Steps:

• Total resistance of driver circuit: 0.50 Ω (internal) + 0.50 Ω (wire) = 1.0 Ω.
• Driver current: 1.5 V / 1.0 Ω = 1.5 A.
• Potential difference across wire XY: 1.5 A × 0.50 Ω = 0.75 V.
• Potential gradient: 0.75 V / 0.6 m = 1.25 V/m.
• Unknown e.m.f.: 1.25 V/m × 0.4 m = 0.50 V (balancing length 0.4 m yields exact match).

Why A is correct:

• At null deflection, unknown e.m.f. equals I × (resistance per unit length × XZ), per potentiometer principle, giving 0.50 V.

Why the others are wrong:

• B: 0.75 V matches full wire drop, requiring XZ = 0.6 m.
• C: 1 V requires XZ = 0.8 m, exceeding wire length.
• D: 1.5 V requires XZ = 1.2 m, impossible on 0.6 m wire.

Final answer: A