WAEC Physics 2017 Paper

Concept:Potentiometer wire has uniform resistance per unit length, so resistance of a segment is proportional to its length.Explanation:The resistance per metre of wire is $10\,\Omega\,\text{m}^{-1}$.Length of segment AJ is $l = 30\,\text{cm} = 0.30\,\text{m}$.Therefore, resistance of AJ is $R_{AJ} = 10\,\Omega\,\text{m}^{-1} \times 0.30\,\text{m} = 3.0\,\Omega$.Since the full wire AB has resistance $10\,\Omega$ and the e.m.f of the battery is $3.0\,\text{V}$, the current in the wire is $I = \dfrac{3.0\,\text{V}}{10\,\Omega} = 0.30\,\text{A}$.Hence, voltage across AJ is $V_{AJ} = I \times R_{AJ} = 0.30\,\text{A} \times 3.0\,\Omega = 0.9\,\text{V}$.Answer:Resistance of AJ is $3.0\,\Omega$ and voltage across AJ is $0.9\,\text{V}$.Correct option: A.