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

The correct option is A $45\Omega$

Since, volume is constant, i.e.,
$\pi r_1^2{l}_1=\pi r_2^2{l}_2$
$\Rightarrow {\left(\frac{r_1}{r_2}\right)}^2=\left(\frac{l_2}{l_1}\right)=\left(\frac{3l}{l}\right)=3$
Now, $\frac{R_1}{R_2}=\frac{\rho \frac{l_1}{A_1}}{\rho \frac{l_2}{A_2}}$
$=\frac{\frac{l_1}{r_l^2}}{\frac{l_2}{r_2^2}}=\frac{l_1}{l_2}\times \frac{r_2^2}{r_1^2}$
$\Rightarrow \frac{R_1}{R_2}=\frac{1}{3}\times \frac{1}{3}$
$\Rightarrow R_2=9R_1=9\times 5=45\Omega$