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

Updated On: Apr 27, 2024
  • $2 : 1$
  • $1 : 2$
  • $16 : 1$
  • $4 : 1$
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The Correct Option is C

Solution and Explanation

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 $
$\therefore$ $\left(\pi r^{2}_{1} \ell_{1}\right) \sigma =\left(\pi r^{2}_{2} \ell_{2}\right)\sigma$
$\Rightarrow$ $\frac{\ell_{1}}{\ell_{2}}=\left(\frac{r_{2}}{r_{1}}\right)^{2}$
$\therefore$ $\frac{R_{1}}{R_{2}}=\frac{\rho \frac{^{\ell_{1}}}{A_{1}}}{\rho \frac{\ell_{2}}{A_{2}}}$ $=\frac{\ell_{1}}{\ell_{2}} \times\frac{A_{2}}{A_{1}}$
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