STANDARD Physics Resistance and Resistivity MCQ Question
A wire with 15 Ω resistance is stretched by one tenth of its original length and volume of wire is kept constant. Then its resistance will be
15.18 Ω
81.15 Ω
51.18 Ω
18.15 Ω
Correct Answer
Detailed Explanation
If the wire is stretched by (1/10)th of its original length then the new length of wire becomes l₂ = l + l/10 = 11l/10. As the volume of wire remains constant, πr₁²l₁ = πr₂²l₂. Solving for r₂ gives r₂² = 10r₁²/11. The resistance of stretched wire is calculated using R = ρl/A, resulting in 18.15 Ω.
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