A metal wire has a resistance of $35 \,\Omega$. If its length is increased to double by drawing it, then its new resistance will be (in $\Omega$)

Question

AIIMS 2018, Diffcult

Download our app for free and get started

Solution

Given, $R_{1}=35 \Omega, l_{2}=2 l_{1}$

On increasing the length,

$m_{1}=m_{2}$

$\therefore \rho A_{1} l_{1}=\rho A_{2} l_{2}$

$\pi r_{1}^{2} l_{1}=\pi r_{2}^{2} l_{2}$

$\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{l_{2}}{l_{1}}$

$\frac{r_{1}^{2}}{r_{2}^{2}}=\frac{2 l_{1}}{l_{1}}$

$\frac{r_{1}^{2}}{r_{2}^{2}}=2 \ldots(i)$

$\frac{R_{1}}{R_{2}}=\frac{\rho \cdot \frac{l_{1}}{\pi r_{1}^{2}}}{\rho \cdot \frac{l_{2}}{\pi r_{2}^{2}}}$

$=\frac{l_{1}}{l_{2}} \cdot \frac{r_{2}^{2}}{r_{1}^{2}}=\frac{l_{1}}{2 l_{1}} \cdot \frac{1}{2}$

$\frac{R_{1}}{R_{2}}=\frac{1}{4}$

$R_{2}=4 R_{1}=4 \times 35=140 \Omega$

Download our app
and get started for free