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

Step-by-step explanation:

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

`R=\rho (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 (d^2)/(4)`

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

`R_1=R_2\\\\\rho(L_1)/(\pi d_1^2/4)=\rho(L_2)/(\pi d_2^2/4)\\\\(L_1)/(L_2)=(d_1^2)/(d_2^2)`

We have, L₁=2L₂

So,

`(2L_2)/(L_2)=(d_1^2)/(d_2^2)\\\\(d_1^2)/(d_2^2)=2\\\\(d_1)/(d_2)=√(2)`

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