Answer :
Sure, let’s go through the solution step-by-step.
We are given that the electric resistance [tex]\( R \)[/tex] of a wire varies directly with its length [tex]\( L \)[/tex] and inversely with the square of its diameter [tex]\( D \)[/tex]. This relationship can be represented mathematically as:
[tex]\[ R = k \frac{L}{D^2} \][/tex]
where [tex]\( k \)[/tex] is a constant of proportionality.
Step 1: Determine the constant of proportionality (k)
We are given the initial conditions:
- Length of the first wire [tex]\( L_1 = 80 \, \text{ft} \)[/tex]
- Diameter of the first wire [tex]\( D_1 = \frac{1}{8} \, \text{in} \)[/tex]
- Resistance of the first wire [tex]\( R_1 = \frac{1}{2} \, \text{ohm} \)[/tex]
Using the relationship:
[tex]\[ R_1 = k \frac{L_1}{D_1^2} \][/tex]
Substitute the given values into the equation:
[tex]\[ \frac{1}{2} = k \frac{80}{\left(\frac{1}{8}\right)^2} \][/tex]
Calculate [tex]\( \left(\frac{1}{8}\right)^2 \)[/tex]:
[tex]\[ \left(\frac{1}{8}\right)^2 = \frac{1}{64} \][/tex]
Now substitute this back into the equation:
[tex]\[ \frac{1}{2} = k \frac{80}{\frac{1}{64}} \][/tex]
Simplify the fraction:
[tex]\[ k = \frac{1}{2} \cdot \frac{1}{80} \cdot 64 \][/tex]
[tex]\[ k = 0.5 \cdot 64 \cdot \frac{1}{80} \][/tex]
[tex]\[ k = 32 \cdot \frac{1}{80} \][/tex]
[tex]\[ k = 32 \cdot \frac{1}{80} = 32 \cdot 0.0125 = 0.4 \][/tex]
Step 2: Determine the resistance of the second wire
We are given the second set of conditions:
- Length of the second wire [tex]\( L_2 = 120 \, \text{ft} \)[/tex]
- Diameter of the second wire [tex]\( D_2 = \frac{1}{4} \, \text{in} \)[/tex]
We need to find the resistance [tex]\( R_2 \)[/tex] for the second wire using the same relationship:
[tex]\[ R_2 = k \frac{L_2}{D_2^2} \][/tex]
Using the calculated [tex]\( k = 0.4 \)[/tex]:
[tex]\[ R_2 = 0.00009765625 \frac{120}{\left(\frac{1}{4}\right)^2} \][/tex]
Calculate [tex]\( \left(\frac{1}{4} \right)^2 \)[/tex]:
[tex]\[ \left(\frac{1}{4}\right)^2 = \frac{1}{16} \][/tex]
Substitute this back into the equation:
[tex]\[ R_2 = 0.00009765625 \frac{120}{\frac{1}{16}} \][/tex]
Simplify the fraction:
[tex]\[ R_2 = 0.00009765625 \cdot 120 \cdot 16 \][/tex]
[tex]\[ R_2 = 0.00009765625 \cdot 1920 \][/tex]
[tex]\[ R_2 = 0.1875 \][/tex]
So, the resistance in a piece of the same type of wire that is [tex]\(120 \, \text{ft}\)[/tex] long and has a diameter of [tex]\(\frac{1}{4} \, \text{in}\)[/tex] is [tex]\(0.1875 \, \text{ohms}\)[/tex].
We are given that the electric resistance [tex]\( R \)[/tex] of a wire varies directly with its length [tex]\( L \)[/tex] and inversely with the square of its diameter [tex]\( D \)[/tex]. This relationship can be represented mathematically as:
[tex]\[ R = k \frac{L}{D^2} \][/tex]
where [tex]\( k \)[/tex] is a constant of proportionality.
Step 1: Determine the constant of proportionality (k)
We are given the initial conditions:
- Length of the first wire [tex]\( L_1 = 80 \, \text{ft} \)[/tex]
- Diameter of the first wire [tex]\( D_1 = \frac{1}{8} \, \text{in} \)[/tex]
- Resistance of the first wire [tex]\( R_1 = \frac{1}{2} \, \text{ohm} \)[/tex]
Using the relationship:
[tex]\[ R_1 = k \frac{L_1}{D_1^2} \][/tex]
Substitute the given values into the equation:
[tex]\[ \frac{1}{2} = k \frac{80}{\left(\frac{1}{8}\right)^2} \][/tex]
Calculate [tex]\( \left(\frac{1}{8}\right)^2 \)[/tex]:
[tex]\[ \left(\frac{1}{8}\right)^2 = \frac{1}{64} \][/tex]
Now substitute this back into the equation:
[tex]\[ \frac{1}{2} = k \frac{80}{\frac{1}{64}} \][/tex]
Simplify the fraction:
[tex]\[ k = \frac{1}{2} \cdot \frac{1}{80} \cdot 64 \][/tex]
[tex]\[ k = 0.5 \cdot 64 \cdot \frac{1}{80} \][/tex]
[tex]\[ k = 32 \cdot \frac{1}{80} \][/tex]
[tex]\[ k = 32 \cdot \frac{1}{80} = 32 \cdot 0.0125 = 0.4 \][/tex]
Step 2: Determine the resistance of the second wire
We are given the second set of conditions:
- Length of the second wire [tex]\( L_2 = 120 \, \text{ft} \)[/tex]
- Diameter of the second wire [tex]\( D_2 = \frac{1}{4} \, \text{in} \)[/tex]
We need to find the resistance [tex]\( R_2 \)[/tex] for the second wire using the same relationship:
[tex]\[ R_2 = k \frac{L_2}{D_2^2} \][/tex]
Using the calculated [tex]\( k = 0.4 \)[/tex]:
[tex]\[ R_2 = 0.00009765625 \frac{120}{\left(\frac{1}{4}\right)^2} \][/tex]
Calculate [tex]\( \left(\frac{1}{4} \right)^2 \)[/tex]:
[tex]\[ \left(\frac{1}{4}\right)^2 = \frac{1}{16} \][/tex]
Substitute this back into the equation:
[tex]\[ R_2 = 0.00009765625 \frac{120}{\frac{1}{16}} \][/tex]
Simplify the fraction:
[tex]\[ R_2 = 0.00009765625 \cdot 120 \cdot 16 \][/tex]
[tex]\[ R_2 = 0.00009765625 \cdot 1920 \][/tex]
[tex]\[ R_2 = 0.1875 \][/tex]
So, the resistance in a piece of the same type of wire that is [tex]\(120 \, \text{ft}\)[/tex] long and has a diameter of [tex]\(\frac{1}{4} \, \text{in}\)[/tex] is [tex]\(0.1875 \, \text{ohms}\)[/tex].
