Question

If one has twice the length of the other, what is the ratio of the diameter of the longer wire to the diameter of the shorter wire

Answer 1

Explanation:

The resistance of a wire is given by the formula as follows :

$R=\rho \dfrac{l}{A}$

l is length of wire

A is area of cross section, $A=\pi r^2$

Since, r=d/2, d = diameter

$A=\pi \dfrac{d^2}{4}$

It is given in the problem that the resistance of two aluminum wires is same.

$R_1=R_2\\\\\rho\dfrac{L_1}{\pi d_1^2/4}=\rho\dfrac{L_2}{\pi d_2^2/4}\\\\\dfrac{L_1}{L_2}=\dfrac{d_1^2}{d_2^2}$

We have, L₁=2L₂

So,

$\dfrac{2L_2}{L_2}=\dfrac{d_1^2}{d_2^2}\\\\\dfrac{d_1^2}{d_2^2}=2\\\\\dfrac{d_1}{d_2}=\sqrt{2}$

So, the ratio of the diameter of the longer wire to the diameter of the shorter wire is $\sqrt2:1$