Question

A square aluminium rod is $1m$ long and $5mm$ on edge. What must be the radius of another aluminium rod whose length is $1m$ and which has the same resistance as the previous rod?
A. ${\text{5}}{\text{.0mm}}$
B. ${\text{4}}{\text{.2mm}}$
C. ${\text{2}}{\text{.8mm}}$
D. ${\text{1}}{\text{.4mm}}$

Answer 1

Hint: Resistance is measured in resistivity. The electrical resistance of the conductor is directly proportional to the length of the conductor and inversely proportional to the area of the cross-section. It can be expressed as \[R = \dfrac{{\rho L}}{A}\] where $\rho $ is the resistivity of the material.

Complete step by step answer:
Let us suppose that the radius of the aluminium rod is $ = r$
The radius of the another rod is $ = r'$
Given that both the rods have the same length.
Therefore, $l = l' = 1m$
Since, both the rods are made up of the same material. Therefore they have the same resistivity.
$\rho = \rho '$
Hence, the resistance of both the rods are also equal.
$R = R'$
$ \Rightarrow \dfrac{{\rho l}}{A} = \dfrac{{\rho 'l'}}{{A'}}$
Take common factors from both the sides of the equations and remove them.
$ \Rightarrow A = A'$
$\pi r{'^2} = {\left( 5 \right)^2}$
Simplify the above equation and make the unknown radius r’ the subject –
$
  r{'^2} = \dfrac{{25}}{{\dfrac{{22}}{7}}} \\
 \implies r{'^2} = \dfrac{{25 \times 7}}{{22}} \\
 \implies r{'^2} = \dfrac{{175}}{{22}} \\
 $
Take square-root on both the sides of the equation –
$\sqrt {r{'^2}} = \sqrt {\dfrac{{175}}{{22}}} $
Square and square-root cancel each other on the left hand side of the equation –
\[
  r' = \sqrt {7.95} \\
  r' = 2.81mm \\
 \]

So, the correct answer is “Option C”.

Note:
Always check the given units for all the terms. Note down all the given terms and unknown terms and find the correlation amongst them for the required solution. Remember the basic formulas and the properties of physical terms to solve these types of word problems. Know the basic difference between the resistance and resistivity.