Two conducting wires made of the same material have lengths L and 1.2 L. If the volume of the longer wire is half that of the shorter wire, then what is the electric resistance of the longer wire in comparison with the shorter wire?
- A.288% larger
- B.200% larger
- C.188% larger
- D.108% larger
Correct answer
Explanation
The correct answer is C, 188% larger. The longer wire's resistance works out to 2.88 times that of the shorter one, which is 188% more.
Resistance is R = ρL/A, where ρ is the resistivity of the material. Since volume V = L × A, the area is A = V/L, so R = ρL²/V. For the shorter wire take length L and volume V. The longer wire has length 1.2L and volume V/2. Its resistance is ρ(1.2L)²/(V/2) = ρ × 1.44L² × 2/V = 2.88 ρL²/V. So R(longer) = 2.88 × R(shorter), and the increase is 2.88 − 1 = 1.88, or 188%. Both wires share ρ because they are made of the same material.
A is wrong because 288% is the new value as a percentage of the old one, not the increase over it.
B is wrong because 200% larger would mean three times the resistance.
D is wrong because 108% larger would mean only 2.08 times, which ignores the effect of length on area.
Exam tip: for a fixed volume, R grows as the square of length; "x% larger" means (ratio − 1) × 100.
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