Question

The resistivity of a coper rod 50 ft long and 0.25 in. in diameter

Answer 1

To solve this you need to know 3 things.

First, you need to know the resistivity of the copper. Since you did not provide the resistivity we are going to use the actual resistivity, which is
`1.7 * 10^(-8)`

Second, you need to know the length, which is 50 ft

Third, you need to know the area. We are not given the area but given the diameter so we need to find the area.

`A=\pi r^2`

`A=\pi .25^2`

Now that we have our resistance, length and area, it is time to use the following formula to calculate the resistivity. To calculate the resistivity, use the following formula.

`resistivity = (resistance * length)/(area)`

`resistivity = (1.7 * 10^(-8) * 50)/(\pi .25^2)`

`resistivity = 4.3290... * 10^(-6)`

Answer 2

we are given with a copper rod , which has a length of 50ft and diameter of 0.25 in.

As we know that the resistivity is given by the formula

`R=\rho(l)/(A)`

We also know that the resistivity of the copper is
`\rho=1.72*10^(-8)\Omega m`

Here we are given that
`l=50ft,r=(0.25)/(2)=0.124inches=0.0031496m`

Substitute these values we get

`R=1.72*10^(-8) (50)/(\pi * (0.0031496)^2)`

Simplify we get

R=0.0275955 ohm

`R=2.7*10^(-3)\Omega`