Potentiometer null method for emf measurement Steps:

• Identify the driving cell (known emf E, negligible resistance) connected across the uniform wire of length L to create a potential gradient.
• Connect the unknown cell (emf E₀) in opposition to the wire via a galvanometer and jockey J.
• Slide J to find the null point where galvanometer current is zero, meaning no current draws from the test cell.
• Calculate E₀ = (l/L) × E, where l is the balancing length from the starting end. Why D is correct:
• It depicts the standard setup with driving cell across AB wire, test cell from A through galvanometer to J, enabling null deflection without current in test cell, per potentiometer principle. Why the others are wrong:
• A: Test cell powers the wire, reversing roles and drawing current from E₀, violating null condition.
• B: Galvanometer shunts the wire, bypassing potential comparison for E₀.
• C: Cells in series with wire, causing current flow through test cell and inaccurate emf measurement. Final answer: D