### Correct Answer: A A potentiometer measures an unknown electromotive force (e.m.f.) by finding a point on a resistance wire where the potential difference is equal to the unknown e.m.f. At this balance point, indicated by a zero reading on the galvanometer, no current flows in the secondary circuit. Therefore, the unknown e.m.f., *E*, is equal to the potential difference across the length of the wire XZ, *V*_XZ. To find *V*_XZ, we must first determine the current flowing through the primary circuit, which consists of the 1.5 V cell, its internal resistance, and the potentiometer wire XY. 1. **Calculate the total resistance of the primary circuit (*R*_total):** The internal resistance of the cell (0.50 Ω) and the resistance of the wire XY (0.50 Ω) are in series. *R*_total = *R*_internal + *R*_{wire XY} = 0.50 Ω + 0.50 Ω = 1.00 Ω 2. **Calculate the current in the primary circuit (*I*):** Using Ohm's law, *I* = *V* / *R*_total. The voltage is the e.m.f. of the driver cell. *I* = 1.5 V / 1.00 Ω = 1.5 A 3. **Calculate the resistance of the segment XZ (*R*_XZ):** The resistance wire is uniform, so its resistance is directly proportional to its length. The resistance of the segment XZ can be found using the ratio of its length to the total length. *R*_XZ = (*L*_XZ / *L*_XY) × *R*_XY *R*_XZ = (0.64 m / 0.96 m) × 0.50 Ω = (2/3) × 0.50 Ω ≈ 0.333 Ω 4. **Calculate the potential difference across XZ (*V*_XZ):** Using Ohm's law for the segment XZ, *V*_XZ = *I* × *R*_XZ. *V*_XZ = 1.5 A × ( (2/3) × 0.50 Ω ) = 1.5 A × (1/3) Ω = 0.50 V Since *E* = *V*_XZ at the balance point, the value of e.m.f. *E* is 0.50 V. ### Incorrect Answer: B This value, 0.75 V, is the potential difference across the entire length of the potentiometer wire XY. It is calculated as *V*_XY = *I* × *R*_XY = 1.5 A × 0.50 Ω = 0.75 V. The unknown e.m.f., *E*, is balanced against the potential difference of the segment XZ only, not the entire wire. This answer would be correct only if the balance length *L*_XZ was equal to the total length of the wire *L*_XY. ### Incorrect Answer: C This value, 1.0 V, is obtained if the internal resistance of the 1.5 V driver cell is incorrectly ignored. If the internal resistance were zero, the total resistance in the primary circuit would be only the wire's resistance, 0.50 Ω. The current would then be *I* = 1.5 V / 0.50 Ω = 3.0 A. The potential difference across XZ would be *V*_XZ = *I* × *R*_XZ = 3.0 A × ( (2/3) × 0.50 Ω ) = 1.0 V. However, the internal resistance is given and must be included in the total resistance of the primary circuit, as it affects the current and thus the potential difference along the wire. ### Incorrect Answer: D This value, 1.1 V, does not result from a correct application of the relevant physics principles or from a common conceptual mistake. A systematic calculation of the potential difference across the balanced length of the wire, taking into account all the given circuit parameters, yields 0.50 V. Therefore, 1.1 V is an incorrect value, likely arising from a miscalculation.

10: D.C. circuits