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Difficulty: HardResistance and Resistivity

Consider the physical principles governing electrical resistance (R=ρLAR = \rho \frac{L}{A}) and material resistivity. Match each specific physical modification of a cylindrical wire conductor with its resulting net effect on the wire's total electrical resistance. Which pairings accurately reflect these proportional and thermal relationships?

  • Uniformly stretching a solid cylindrical wire to double its original length while preserving its total volumeIncreases total resistance by a factor of 4 (4R4R)
  • Doubling the cross-sectional diameter of a wire while keeping its length constantDecreases total resistance to one-fourth of its original value (0.25R0.25R)
  • Doubling both the length and the radius of a cylindrical wire simultaneouslyDecreases total resistance to one-half of its original value (0.5R0.5R)
  • Significantly increasing the operating temperature of a pure copper conductor without altering its dimensionsIncreases total resistance due to an increase in material resistivity (ρ\rho)

Answer

Stretching a wire to double length at constant volume pairs with increasing resistance by a factor of 4; doubling wire diameter pairs with decreasing resistance to one-fourth; doubling both length and radius pairs with decreasing resistance to one-half; and heating a copper conductor pairs with increasing resistance due to higher resistivity.
Each match strictly obeys the physical governing equation for resistance R=ρLAR = \rho \frac{L}{A}. For cylindrical geometry, A=πr2=πd24A = \pi r^2 = \frac{\pi d^2}{4}. Uniform stretching alters length and cross-sectional area inversely (2L×0.5A→4R2L \times 0.5A \to 4R). Diameter changes scale quadratically (2d→4A→0.25R2d \to 4A \to 0.25R). Simultaneous doubling of length and radius yields a ratio of 2/4=0.5R2/4 = 0.5R. Thermal elevation in conductors increases resistivity (ρ\rho) due to enhanced electron-phonon collisions.

Step-by-Step Solution

1
Analyze volume conservation under mechanical stretching
Stretching to double length (L′=2LL' = 2L) causes cross-sectional area to halve (A′=A/2A' = A/2), giving R′=ρ2LA/2=4RR' = \rho \frac{2L}{A/2} = 4R.
Since volume V=A⋅LV = A \cdot L remains constant, stretching alters both length and cross-sectional area simultaneously.
2
Calculate the geometric effect of doubling diameter
Doubling diameter (d′=2dd' = 2d) quadruples area (A′=4AA' = 4A), resulting in R′=ρL4A=0.25RR' = \rho \frac{L}{4A} = 0.25R.
Cross-sectional area varies quadratically with diameter (A=πd24A = \frac{\pi d^2}{4}).
3
Calculate simultaneous scaling of length and radius
Doubling length (2×2\times) and radius (4×4\times area) produces R′=24R=0.5RR' = \frac{2}{4}R = 0.5R.
The length factor of 2 in the numerator is divided by the area factor of 4 in the denominator.
4
Evaluate the thermal effect on metallic conductors
Increasing temperature increases copper's resistivity ρ\rho, raising resistance without altering geometry.
Metals possess a positive temperature coefficient of resistance where lattice vibrations hinder electron drift.

Key Concept

Conductor resistance is directly proportional to length and resistivity, and inversely proportional to the square of radius or diameter (R=ρLπr2R = \rho \frac{L}{\pi r^2}), with metallic resistivity increasing as temperature rises.
Estimated Time:1m 30s
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Topics

Resistance and Resistivity

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