Wire Length w/ 0.068 Ohms & 2.8mm Diameter: ~11.56 ft

[unicorngirl]

The electrical resistance of a wire varies directly with the length of the wire and inversely with the square of the diameter of the wire. If a wire 50 feet long and 2 millimeters in diameter has a resistance of 0.265 ohms, find the length of a wire of the same material whose resistance is 0.068 ohms and whose diameter is 2.8 millimeters. The length of the wire is approximately ___ feet when the resistance is 0.068 ohms and the diameter is 2.8 millimeters. (Round to the nearest hundredth.)

 

[MarkFL]

Hello and welcome to MHB, unicorngirl! :D

I have moved your thread, since this problem type is generally encountered in an algebra course, and not a calculus-based statistics course.

We are told:

The electrical resistance of a wire varies directly with the length of the wire and inversely with the square of the diameter of the wire.

So, if we define:

$$R$$ = resistance
$$L$$ = length
$$D$$ = diameter

We may then take the above sentence, and express it mathematically as:

$$R=k\frac{L}{D^2}\tag{1}$$

where $k$ is the constant of proportionality.

To determine $k$, we may use the given information:

A wire 50 feet long and 2 millimeters in diameter has a resistance of 0.265 ohms.

Plug the values into (1)...what do you find for the magnitude and dimensions for $k$?

 

[unicorngirl]

Okay thank you! I wasn't sure where to post it.

I got .0212 for K. I hope I did that right.

 

[MarkFL]

Okay thank you! I wasn't sure where to post it.

I got .0212 for K. I hope I did that right.

Yes, the value you obtained is correct! (Yes)

We should at least be aware of the units for $k$:

$$k=\frac{D^2R}{L}=\frac{(2\text{ mm})^2(0.265\,\Omega)}{50\text{ ft}}=0.0212\frac{\text{mm}^2\Omega}{\text{ft}}$$

So now, using rational rather than decimal notation, we may state:

$$R=\frac{53D^2}{2500L}\tag{2}$$

Now, you have a question to answer:

Find the length of a wire of the same material whose resistance is 0.068 ohms and whose diameter is 2.8 millimeters.

So, solve (2) for $L$, and then plug in the given values for $R$ and $D$.

What do you find?