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1. Understand the Relationships:
- We're told that the resistance of a wire varies directly with its length and inversely with its cross-sectional area.
- Mathematically, we can express this relationship as [tex]\( R = k \frac{L}{A} \)[/tex], where [tex]\( R \)[/tex] is the resistance, [tex]\( L \)[/tex] is the length, [tex]\( A \)[/tex] is the cross-sectional area, and [tex]\( k \)[/tex] is a constant of proportionality.
2. Use Given Values to Determine the Constant [tex]\( k \)[/tex] for the Material:
- For the first wire:
- Length [tex]\( L_1 = 100 \)[/tex] meters
- Cross-sectional area [tex]\( A_1 = 1 \)[/tex] square millimeter
- Resistance [tex]\( R_1 = 2 \)[/tex] ohms
- Substitute these values into the relationship to find [tex]\( k \)[/tex]:
[tex]\[ 2 = k \frac{100}{1} \][/tex]
- Solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{2 \times 1}{100} = 0.02 \][/tex]
3. Find the Resistance of the Second Wire Using the Same Constant [tex]\( k \)[/tex]:
- For the second wire:
- Length [tex]\( L_2 = 250 \)[/tex] meters
- Cross-sectional area [tex]\( A_2 = 0.5 \)[/tex] square millimeters
- Using the relationship [tex]\( R = k \frac{L}{A} \)[/tex]:
[tex]\[ R_2 = 0.02 \times \frac{250}{0.5} \][/tex]
4. Calculate the Resistance:
[tex]\[ R_2 = 0.02 \times \frac{250}{0.5} = 0.02 \times 500 = 10 \text{ ohms} \][/tex]
Therefore, the resistance of a wire of the same material that is 250 meters long with a cross-sectional area of 0.5 square millimeters would be 10 ohms.
1. Understand the Relationships:
- We're told that the resistance of a wire varies directly with its length and inversely with its cross-sectional area.
- Mathematically, we can express this relationship as [tex]\( R = k \frac{L}{A} \)[/tex], where [tex]\( R \)[/tex] is the resistance, [tex]\( L \)[/tex] is the length, [tex]\( A \)[/tex] is the cross-sectional area, and [tex]\( k \)[/tex] is a constant of proportionality.
2. Use Given Values to Determine the Constant [tex]\( k \)[/tex] for the Material:
- For the first wire:
- Length [tex]\( L_1 = 100 \)[/tex] meters
- Cross-sectional area [tex]\( A_1 = 1 \)[/tex] square millimeter
- Resistance [tex]\( R_1 = 2 \)[/tex] ohms
- Substitute these values into the relationship to find [tex]\( k \)[/tex]:
[tex]\[ 2 = k \frac{100}{1} \][/tex]
- Solving for [tex]\( k \)[/tex]:
[tex]\[ k = \frac{2 \times 1}{100} = 0.02 \][/tex]
3. Find the Resistance of the Second Wire Using the Same Constant [tex]\( k \)[/tex]:
- For the second wire:
- Length [tex]\( L_2 = 250 \)[/tex] meters
- Cross-sectional area [tex]\( A_2 = 0.5 \)[/tex] square millimeters
- Using the relationship [tex]\( R = k \frac{L}{A} \)[/tex]:
[tex]\[ R_2 = 0.02 \times \frac{250}{0.5} \][/tex]
4. Calculate the Resistance:
[tex]\[ R_2 = 0.02 \times \frac{250}{0.5} = 0.02 \times 500 = 10 \text{ ohms} \][/tex]
Therefore, the resistance of a wire of the same material that is 250 meters long with a cross-sectional area of 0.5 square millimeters would be 10 ohms.
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