Potentiometer balance condition equates potential drops proportionally.

Steps:

• Driver cell provides 2V across uniform wire QR of length R.
• At balance point P, voltmeter reads zero, so e.m.f. of X (0.5V) equals potential drop across QP.
• Potential drop across QP is (length QP / R) × 2V = 0.5V.
• Solve: QP / R = 0.5 / 2 = 1/4, so QP = R/4.

Why A is correct:

• Ratio of e.m.f.s directly gives length fraction per potentiometer principle: V_X / V_driver = l / L.

Why the others are wrong:

• B: Assumes QP as 75% of wire, implying 1.5V drop, exceeding X's e.m.f.
• C: Assumes QP as 33% of wire, implying ~0.67V drop, not matching 0.5V.
• D: Assumes QP as 50% of wire, implying 1V drop, twice X's e.m.f.

Final answer: A