Question

how resistance is inversely proportional to square of diameter

Answer 1

Dear Student,

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}, \\ \mathrm{R}=\frac{\mathrm{\rho l}}{\mathrm{A}} \\ \mathrm{where},\mathrm{\rho }=\mathrm{resistivity} \\ \mathrm{l}=\mathrm{length} \\ \mathrm{A}=\mathrm{area}\mathrm{of}\mathrm{cross}\mathrm{section} \\ \mathrm{Also},\mathrm{we}\mathrm{know}\mathrm{that}\mathrm{A}=\frac{{\mathrm{\pi d}}^2}{4} \\ \mathrm{where},\mathrm{d}=\mathrm{diameter} \\ \mathrm{So},\mathrm{R}=\frac{\mathrm{\rho l}}{\frac{{\mathrm{\pi d}}^2}{4}} \\ \mathrm{or},\mathrm{R}=\frac{4\mathrm{\rho l}}{{\mathrm{\pi d}}^2} \\ \mathrm{or},\mathrm{R}\mathrm{\alpha }\frac{1}{{\mathrm{d}}^2} \\ \mathrm{Hence},\mathrm{resistance}\mathrm{is}\mathrm{inversaly}\mathrm{proportional}\mathrm{to}\mathrm{the}\mathrm{square}\mathrm{of}\mathrm{diameter}. \end{gathered}$$