A wire of resistance R is stretched to double its length. Assuming volume remains constant, the new resistance is: MCQ with Answer and Explanation

A wire of resistance R is stretched to double its length. Assuming volume remains constant, the new resistance is:
A. R/2
B. 4R
C. R
D. 2R
Answer: Option B
Solution (By JKExamLibrary)
Resistance R = ρL/A. Volume V = A·L constant. If L' = 2L, then A' = V/L' = V/(2L) = A/2. Thus R' = ρ(2L)/(A/2) = 4(ρL/A) = 4R. Resistance increases by factor of 4. Memory tip: 'Stretching wire: R ∝ L² when volume constant'. This proportional reasoning problem tests resistance concepts, frequently appearing in competitive exams. Always verify volume conservation assumption; if area changed independently, result would differ. This problem assesses understanding of geometric effects on electrical properties.

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