JAMB Physics 2010 Paper

Concept:A balanced metre bridge works on the Wheatstone bridge principle: the ratio of resistances in the two gaps equals the ratio of the corresponding wire lengths.Explanation:The two $8\,\Omega$ resistors on the left side are in parallel.Their equivalent resistance is:$$\frac{1}{R} = \frac{1}{8} + \frac{1}{8} = \frac{2}{8}$$$$R = 4\,\Omega$$Thus, the parallel combination gives $R = 4\,\Omega$, while the right gap has $R_1 = 8\,\Omega$.For a balanced metre bridge:$$\frac{R}{R_1} = \frac{x}{100 - x}$$Substitute the resistances:$$\frac{4}{8} = \frac{x}{100 - x}$$$$4(100 - x) = 8x$$$$400 - 4x = 8x$$$$400 = 12x$$$$x = \frac{400}{12} = 33.3\,\text{cm}$$Answer:The value of $x$ is $33.3\,\text{cm}$.Correct option: C. 33.3 cm