The length of a given cylindrical wire is increased by 100\% . Due to the consequent decrease in diameter, the percentage change in the resistance of - Physics MCQ AIEEE 2003 2026 | ExamTest.live

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The length of a given cylindrical wire is increased by 100% . Due to the consequent decrease in diameter, the percentage change in the resistance of the wire will be:

Solution & Step-by-step Explanation

Resistance R = ρ \frac{l}{A} .When a wire is stretched, its volume V = A l remains constant. So A = V / l . R = ρ \frac{l ^{2}}{V} ⟹ R \propto l^{2} Initial length l_{1}, new length l_{2} = l_{1} + 100% of l_{1} = 2 l_{1} . \frac{R _{2}}{R _{1}} = (\frac{l _{2}}{l _{1}})^{2} = (\frac{2 l _{1}}{l _{1}})^{2} = 4 R_{2} = 4 R_{1} .Percentage change = \frac{R _{2} - R _{1}}{R _{1}} \times 100 = \frac{4 R _{1} - R _{1}}{R _{1}} \times 100 = 300% .

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The length of a given cylindrical wire is increased by 100% . Due to the consequent decrease in diameter, the percentage change in the resistance of the wire will be:

200%

100%

50%

300%

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