WAEC Physics 2016 Paper

Concept:The resistance $R$ of a conductor is directly proportional to its length $L$ and inversely proportional to its cross-sectional area $A$.Explanation:The basic resistance formula is $R=\dfrac{\rho L}{A}$, where $\rho=3.0\times10^{-8}\,\Omega\text{m}$ is the resistivity of aluminium.The cable is circular, with diameter $d=4\times10^{-3}$ m.Therefore, its cross-sectional area is $A=\pi r^2=\dfrac{\pi d^2}{4}$.Substituting $A$ into the resistance formula gives $R=\dfrac{4\rho L}{\pi d^2}$.Rearrange to make $L$ the subject: $L=\dfrac{R\pi d^2}{4\rho}$.Substitute the given values: $R=21\,\Omega$, $\pi=3.14$, $d=4\times10^{-3}$ m, and $\rho=3.0\times10^{-8}\,\Omega\text{m}$.$$L=\frac{21\times3.14\times(4\times10^{-3})^2}{4\times(3.0\times10^{-8})}$$Simplify: $(4\times10^{-3})^2=16\times10^{-6}$.$$L=\frac{21\times3.14\times16\times10^{-6}}{12\times10^{-8}}$$$$L=8.792\times10^{3}\ \text{m}$$Rounding to two significant figures, $L\approx8.8\times10^{3}$ m.Answer:B. $8.8\times10^{3}$ m