Answer to Question #188078 in Electric Circuits for Subham Acharya

Question

In the manufacture of copper wire a 5 m length of thick circular rod, which has a resistance of

 0.02 ohm is drawn out without change in volume until its new diameter is one-tenth of what it

 was. Calculate the length of the drawn conductor and its resistance. [Ans: 200 Q]

Answer 1

Explanations & Calculations

• If the diameter of the rod is $\small d$ and the new length of the drawn wire is $\small L$, then equalling the volume of both the products, the new length can be calculated.

$\qquad\qquad \begin{aligned} \small V&=\small \pi r^2l=\pi(\frac{d}{2})^2.l\\ &\propto d^2l\\\\ \small V_1&\propto\small d^2.5m\cdots(1)\\ \small V_2 &\propto\small (\frac{d}{10})^2.L\cdots(2) \\\\ \small 5d^2&=\small \frac{d^2.L}{10^2}\\ \small L&=\small 500\,m \end{aligned}$

• Resistance of the rod/wire is given by

$\qquad\qquad \begin{aligned} \small R&=\small \frac{\rho l}{A}=\frac{\rho l}{\pi r^2}=\frac{\rho l}{\pi\frac{d^2}{4}}\\ &\propto\small \frac{l}{d^2} \end{aligned}$

• Then the resistance of the rod is

$\qquad\qquad \begin{aligned} \small 0.02\Omega&\propto\frac{5}{d^2} \end{aligned}$

• That of the wire is,

$\qquad\qquad \begin{aligned} \small R&\propto \small \frac{500}{(\frac{d}{10})^2}\propto \frac{500\times100}{d^2} \end{aligned}$

• On division of the two equations

$\qquad\qquad \begin{aligned} \small \frac{R}{0.02}&=\small 100\times100\\ \small R&=\small \bold{200\Omega} \end{aligned}$