Question

One $kg$ of copper is drawn into a wire of $1$ mm diameter and a wire of $2$ mm diameter. The resistance of the two wires will be in the ratio

• A. $2:1$
• B. $1:2$
• C. $16:1$
• D. $4:1$

Answer 1

Answer: (C)

Mass of 1st wire $=\left(\pi r_1^2{\ell }_1\right)\sigma$
Mass of Il nd wire $=\left(\pi r_1^2{\ell }_1\right)\sigma$
$\therefore$ $\left(\pi r_1^2{\ell }_1\right)\sigma =\left(\pi r_2^2{\ell }_2\right)\sigma$
$\Rightarrow$ $\frac{{\ell }_1}{{\ell }_2}={\left(\frac{r_2}{r_1}\right)}^2$
$\therefore$ $\frac{R_1}{R_2}=\frac{\rho \frac{{}^{{\ell }_1}}{A_1}}{\rho \frac{{\ell }_2}{A_2}}$
$=\frac{{\ell }_1}{{\ell }_2}\times \frac{A_2}{A_1}$

$=\frac{{\ell }_1}{{\ell }_2}\times {\left(\frac{r_2}{r_1}\right)}^2$
$={\left(\frac{r_2}{r_1}\right)}^2={\left(\frac{1}{1/2}\right)}^4$

$=16:1$