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^{2}_{1} \ell_{1} \right) \sigma $
Mass of Il nd wire $=\left(\pi r^{2}_{1} \ell_{1} \right) \sigma $
$\therefore $ $\left(\pi r^{2}_{1} \ell_{1}\right) \sigma =\left(\pi r^{2}_{2} \ell_{2}\right)\sigma$
$\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}$
$=\left(\frac{r_{2}}{r_{1}}\right)^{2} =\left(\frac{1}{1 /2}\right)^{4}$
$=16:1$