The resistance of a conductor of diameter d and length l is R Ω. If the diameter of the conductor is halved and its length is doubled, the resistance will be
Correct answer: D. 8R ΩCivil Engineering
- A. R Ω
- B. 2R Ω
- C. 4R Ω
- D. 8R ΩCivil Engineering
Explanation
Using R = ρl/A, halving the diameter makes the cross-sectional area one-fourth, while doubling the length doubles the numerator. Therefore the new resistance is 2R ÷ 1/4 = 8R.
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