Q121 / 160 · WBJEE 2009 · Physics

One $\mathrm{Kg}$ of copper is drawn into a wire of $1 \mathrm{~mm}$ diameter and a wire of $2 \mathrm{~mm}$ diameter. The resistance of the two wires will be in the ratio

  • A
    $2: 1$
  • B
    $1: 2$
  • Correct answer
    C
    $16: 1$
  • D
    $4: 1$

Solution

Mass $=\left(\pi \mathrm{r}_1^2 \ell_1\right) \sigma$ (Ist wire) $\begin{aligned} & \text { Mass }=\left(\pi \mathrm{r}_1^2 \ell_2\right) \sigma(2 \text { nd wire) } \\ & \left(\pi \mathrm{r}_1^2 \ell_1\right) \sigma=\left(\pi \mathrm{r}_2^2 \ell_2\right) \sigma\end{aligned}$ $\begin{aligned} & \left(\pi \mathrm{r}_1^2 \ell_1\right) \sigma=\left(\pi \mathrm{r}_2^2 \ell_2\right) \sigma \\ & \frac{\ell_1}{\ell_2}=\left(\frac{r_2}{r_1}\right)^2\end{aligned}$ $\frac{R_1}{R_2}=\frac{\rho \frac{\ell_1}{A_1}}{\rho \frac{\ell_2}{A_2}}=\frac{\ell_1}{\ell_2} \times \frac{A_2}{A_1}=\frac{\ell_1}{\ell_2} \times\left(\frac{r_2}{r_1}\right)^2$ $\begin{aligned} & =\left(\frac{r_2}{r_1}\right)^4 \\ & \Rightarrow 16: 1\end{aligned}$

Hence, the correct answer is option: 'C'.

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