Question

A certain piece of copper is to be shaped into a conductor of minimum resistance. Its length and diameter should be respectively

• A. $l,d$
• B. $2l,d$
• C. $\frac{1}{2}l,2d$
• D. $2l,\frac{1}{2}d$

Answer 1

The correct option is C

$\frac{1}{2}l,2d$

Explanation for the correct option:

Step 1: Given

The length of the conductor is = $l$

The diameter of the conductor is = $d$

Step 2: Formula Used

The resistance of a conductor is given by $R=\frac{\rho l}{A}$, where $\rho$ is resistivity of the conductor, $l$ is length and $A$ is area of the conductor. This can be further reduced to the expression

$\begin{array}{rcl}R& =& \frac{\rho l}{\pi r^2}\\ & =& \frac{\rho l}{\pi {\left({\displaystyle \frac{d}{2}}\right)}^2}\\ & =& \frac{4\rho l}{\pi d^2}\end{array}$

Where $l$ is the length of the conductor and $d$ is diameter

Step 3: Calculation

Find the resistance when length is $l$ and diameter is $d$.

$\begin{array}{rcl}R& =& \frac{4\rho l}{\pi d^2}\end{array}$

Find the resistance when length is $2l$ and diameter is $d$.

$\begin{array}{rcl}R& =& \frac{4\rho \left(2l\right)}{\pi d^2}\\ & =& \frac{8\rho l}{\pi d^2}\end{array}$

Finding the resistance when length is $\frac{1}{2}l$ and diameter is $2d$.

$\begin{array}{rcl}R& =& \frac{4\rho \left({\displaystyle \frac{1}{2}}l\right)}{\pi {\left(2d\right)}^2}\\ & =& \frac{2\rho l}{4\pi d^2}\\ & =& \frac{\rho l}{2\pi d^2}\end{array}$

Finding the resistance when length is $2l$ and diameter is $\frac{1}{2}d$.

$\begin{array}{rcl}R& =& \frac{4\rho \left(2l\right)}{\pi {\left(\frac{1}{2}d\right)}^2}\\ & =& \frac{8\rho l}{{\displaystyle \frac{1}{4}}\pi d^2}\\ & =& \frac{32\rho l}{\pi d^2}\end{array}$

From the above values of resistance, it is clear that the least value is when length is $\frac{1}{2}l$ and diameter is $2d$.

Hence, the correct option is option (C).