Question

The resistance of a wire varies directly as its length and inversely as the square of its diameter. A 32-ft wire with a 0.8-in diameter has a resistance of 0.0125 .
Find the resistance of a 62-ft wire with a diameter of 0.4 in.

Answer 1

Let the resistance \( R \) be modeled by the equation

\[
R = k \frac{L}{d^2}
\]

where \( L \) is the length of the wire, \( d \) is the diameter of the wire, and \( k \) is a constant of proportionality.

Given:
- For the first wire, \( L_1 = 32 \) ft, \( d_1 = 0.8 \) in, and \( R_1 = 0.0125 \) ohms.

We can solve for \( k \) using this information:

\[
0.0125 = k \frac{32}{(0.8)^2}
\]

Calculating \( (0.8)^2 \):

\[
(0.8)^2 = 0.64
\]

Substituting that into the equation gives:

\[
0.0125 = k \frac{32}{0.64}
\]

Calculating \( \frac{32}{0.64} \):

\[
\frac{32}{0.64} = 50
\]

Now we can rewrite the equation:

\[
0.0125 = k \cdot 50
\]

Solving for \( k \):

\[
k = \frac{0.0125}{50} = 0.00025
\]

Now, we want to find the resistance \( R_2 \) of a second wire with \( L_2 = 62 \) ft and \( d_2 = 0.4 \) in:

\[
R_2 = k \frac{L_2}{d_2^2}
\]

Calculating \( d_2^2 \):

\[
(0.4)^2 = 0.16
\]

Now substituting \( L_2 \), \( d_2 \), and \( k \):

\[
R_2 = 0.00025 \frac{62}{0.16}
\]

Calculating \( \frac{62}{0.16} \):

\[
\frac{62}{0.16} = 387.5
\]

Thus,

\[
R_2 = 0.00025 \cdot 387.5 = 0.096875
\]

Therefore, the resistance of the 62-ft wire with a diameter of 0.4 in is:

\[
\boxed{0.096875} \text{ ohms}
\]