Question Number 41443 by Necxx last updated on 07/Aug/18
$${Suppose}\:{that}\:{you}\:{wish}\:{to}\:{fabricate} \\ $$$${a}\:{uniform}\:{wire}\:{out}\:{of}\:\mathrm{1}\:{gram}\:{of} \\ $$$${copper}.{If}\:{the}\:{wire}\:{is}\:{to}\:{have}\:{a} \\ $$$${resistance}\:{of}\:{R}=\mathrm{0}.\mathrm{5}\Omega,{and}\:{all}\:{of} \\ $$$${the}\:{copper}\:{is}\:{to}\:{be}\:{used}.{What}\:{will} \\ $$$${be}\left({i}\right){the}\:{length}\:{and}\:\left({ii}\right){the}\:{diameter} \\ $$$${of}\:{the}\:{wire}. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 07/Aug/18
$${m}=\Pi{r}^{\mathrm{2}} {ld}\:\:\:\:{r}={radius}\:{of}\:{wire} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{l}={length}\:{of}\:{wire} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{d}={density}\:{of}\:{wire} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\rho={resistivity}\:{of}\:{wire} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{R}={resistance}\:{of}\:{wire} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}={mass}\:{of}\:{wire} \\ $$$${R}=\frac{\rho{l}}{\Pi{r}^{\mathrm{2}} } \\ $$$${m}.{R}={d}.\rho.{l}^{\mathrm{2}} \\ $$$${l}=\sqrt{\frac{{m}.{R}}{{d}.\rho}}\:\: \\ $$$${r}^{\mathrm{2}} =\frac{{m}}{\Pi{ld}} \\ $$$$\:\:{r}^{\mathrm{2}} =\frac{{m}}{\Pi{d}.}×\sqrt{\frac{{d}.\rho}{{m}.{R}}}\:=\frac{\mathrm{1}}{\Pi}.\sqrt{\frac{{m}.\rho}{{d}.{R}}}\: \\ $$$${r}=\frac{\mathrm{1}}{\:\sqrt{\Pi}}×\left(\frac{{m}.\rho}{{d}.{R}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$${dia}=\frac{\mathrm{2}}{\:\sqrt{\Pi}}×\left(\frac{{m}.\rho}{{d}.{R}}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$
Commented by Necxx last updated on 07/Aug/18
$${yeah}….{Thanks} \\ $$