When a wire is compressed and its radius becomes 2R then its resistance will be:
Correct answer: C. 1/16 R
- A. 16R
- B. 4R
- C. 1/16 R
- D. 1/4R
Explanation
Assuming the volume of the wire is conserved during compression, if the radius doubles (r' = 2r), the area becomes 4 times larger (A' = 4A). To keep the volume constant, the length must become one-fourth (L' = L/4). The new resistance R' = ρ(L/4)/(4A) = (1/16)(ρL/A) = R/16.
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