# A material B has twice the specific resistance of 'A'. A circular wire made of 'B' has twice the diameter. of a wire made of A. Then for the two wires to have the same resistance, find the ratio lb/la.

Solution

The solution of your fiirst query is provided below

Let (ρ_{A}, l_{A}, r_{A}, A_{A}) and (ρ_{B}, l_{B}, r_{B}, A_{B}) be specific resistances, lengths, radii and areas of wires A and B, respectively.

$Resis\mathrm{tan}ceofA={R}_{A}=\frac{{\rho}_{A}{l}_{A}}{{{\mathrm{\pi r}}_{\mathrm{A}}}^{2}}\phantom{\rule{0ex}{0ex}}Resis\mathrm{tan}ceofB={R}_{B}=\frac{{\rho}_{B}{l}_{B}}{{{\mathrm{\pi r}}_{B}}^{2}}\phantom{\rule{0ex}{0ex}}Asgiven\phantom{\rule{0ex}{0ex}}{\rho}_{B}=2{\rho}_{A}\phantom{\rule{0ex}{0ex}}{r}_{B}=2{r}_{A}\phantom{\rule{0ex}{0ex}}{R}_{A}={R}_{B}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}Therefore\phantom{\rule{0ex}{0ex}}\frac{{\rho}_{A}{l}_{A}}{{{\mathrm{\pi r}}_{\mathrm{A}}}^{2}}=\frac{{\rho}_{B}{l}_{B}}{{{\mathrm{\pi r}}_{B}}^{2}}\phantom{\rule{0ex}{0ex}}\frac{{\rho}_{A}{l}_{A}}{{{\mathrm{\pi r}}_{\mathrm{A}}}^{2}}=\frac{2{\rho}_{A}{l}_{B}}{{\pi \left(2{r}_{A}\right)}^{2}}\phantom{\rule{0ex}{0ex}}\frac{{l}_{B}}{{l}_{A}}=\frac{2}{1}\phantom{\rule{0ex}{0ex}}$

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