WAEC Physics 2021 Paper

Concept:Use the resistance formula $R = \frac{\rho L}{A}$ and rearrange it to solve for the unknown length $L$.Explanation:The resistance $R$ of a wire depends on its resistivity $\rho$, length $L$, and cross-sectional area $A$.The relation is given by the equation $R = \frac{\rho L}{A}$.Rearrange this formula so that $L$ becomes the subject:$$L = \frac{R A}{\rho}.$$Substitute the values provided: $A = 2\pi \times 10^{-8}\ \mathrm{m^2}$, $\rho = 1.1 \times 10^{-8}\ \Omega\,\mathrm{m}$, and $R = 21\ \Omega$.Thus,$$L = \frac{21 \times (2\pi \times 10^{-8})}{1.1 \times 10^{-8}} = \frac{42\pi}{1.1}.$$The units cancel correctly, leaving the length in metres.Evaluating the expression: $\frac{42\pi}{1.1} \approx 119.9$, which rounds to $120\ \mathrm{m}$.Answer:C. 120 m