JAMB Physics 2018 Paper

Concept:A metre bridge balances like a Wheatstone bridge, so resistance is found from the ratio of the two wire lengths separated by the balance point.Explanation:Let the unknown resistance be $R$ and the known standard resistance be $R_s = 2.0\,\Omega$.The standard resistor is connected at the $0.0\,\text{cm}$ end, so the balance length $L = 55.0\,\text{cm}$ corresponds to it.The remaining wire length is $100 - 55 = 45.0\,\text{cm}$, which corresponds to the unknown resistor.At balance, the metre bridge formula is:$$\frac{R}{R_s} = \frac{100 - L}{L} = \frac{45}{55}$$Substitute $R_s = 2.0\,\Omega$:$$R = 2.0 \times \frac{45}{55}$$$$R = \frac{90}{55} = 1.636...$$$$R \approx 1.64\,\Omega$$This is the correct nearest value.Answer:B. $1.64\,\Omega$