Question

Two wires have lengths, resistances, resistivities – each in the ratio 1 : 2. What is the ratio of their diameters?

(a) 1 : 2 (b) 1 : 4 (c) 4 : 1 (d) $1:\sqrt{2}$

Answer 1

Dear Student

$$\begin{gathered} letR=\rho \frac{l}{A} \\ Ristheresis\tan ce \\ \rho istheresitivity \\ listhelength \\ Aisthecrosssectionalareaofthewire \\ A=\pi r^2 \\ ristheradiusofthewire \\ R=\rho \frac{l}{\pi r^2} \\ letforthefirstwire \\ {R}_1=R \\ {\rho }_1=\rho \\ {l}_1=l \\ {r}_1=r \\ forthesecondwire \\ {R}_2=2R \\ {\rho }_2=2\rho \\ {l}_2=2l \\ {r}_2=2r \\ taketheratio \\ \frac{R_1}{R_2}=\frac{{\rho }_1\frac{l_1}{A_1}}{{\rho }_2\frac{l_2}{A_2}} \\ \frac{R}{2R}=\frac{\rho \frac{l}{\pi {r_1}^2}}{2\rho \frac{2l}{4\pi {r_2}^2}}=\frac{1}{2}\times \frac{l}{\pi {r_1}^2}\times \frac{4\pi {r_2}^2}{2l}=\frac{{r_2}^2}{{r_1}^2} \\ \frac{{r_2}^2}{{r_1}^2}=\frac{1}{2} \\ \frac{r_2}{r_1}=\frac{1}{\sqrt{2}} \\ diameterd=2\times r \\ \frac{d_2}{d_1}=\frac{2\times r_2}{2\times r_1}=\frac{1}{\sqrt{2}} \\ \frac{d_2}{d_1}=\frac{1}{\sqrt{2}} \end{gathered}$$
d₁:d₂ =$\sqrt{2}$ :1

Regards