Question

A wire of resistance $5\,\Omega$ is drawn out so that its new length is $3$ times its original length.What is the resistance of the new wire?

• A. $45\,\Omega$
• B. $15\,\Omega$
• C. $5/3\, \Omega$
• D. $5\, \Omega$

Answer 1

Answer: (A)

$45\,\Omega$

Answer 2

Since, volume is constant, i.e.,
$\pi r_{1}^{2} l_{1}=\pi r_{2}^{2} l_{2}$
$\Rightarrow \left(\frac{r_{1}}{r_{2}}\right)^{2}=\left(\frac{l_{2}}{l_{1}}\right)=\left(\frac{3 l}{l}\right)=3$
Now, $\frac{R_{1}}{R_{2}}=\frac{\rho \frac{l_{1}}{A_{1}}}{\rho \frac{l_{2}}{A_{2}}}$
$=\frac{l_{1} / r_{1}^{2}}{l_{2} / r_{2}^{2}}=\frac{l_{1}}{l_{2}} \times \frac{r_{2}^{2}}{r_{1}^{2}}$
$\Rightarrow \frac{R_{1}}{R_{2}}=\frac{1}{3} \times \frac{1}{3}$
$\Rightarrow R_{2}=9 R_{1}=9 \times 5=45 \,\Omega$