What Is the Ratio of Maximum to Minimum Resistance When a Wire Is Stretched to Double of Its Length?
To find the ratio of maximum to minimum resistance when a wire is stretched to double its length, we consider the relation of resistance with length and cross-sectional area.
Concept
Resistance of a wire is given by R = ρ L / A, where ρ is resistivity, L is length, and A is cross-sectional area.
When a wire is stretched to double its length, its volume remains constant, so A decreases.
Calculation
Let initial length = L, initial area = A, initial resistance = R₀.
After stretching, new length = 2L, volume = constant → A’ = A/2.
New resistance R’ = ρ * (2L) / (A/2) = ρ * 2L * 2 / A = 4 * (ρ L / A) = 4 R₀
Conclusion
The ratio of maximum to minimum resistance = R’ / R₀ = 4 : 1.
Stretching a wire increases its resistance because the length increases while cross-sectional area decreases, keeping the volume constant.
What Is the Ratio of Maximum to Minimum Resistance When a Wire Is Stretched to Double of Its Length?
Keywords
ratio of maximum to minimum resistance, wire resistance, resistance stretched wire, Class 10 physics, NEB physics, SEE physics, CTEVT physics, current electricity, resistivity, physics numericals, exam oriented physics, physics notes, R max R min, stretched wire, aklearningnepal
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wire resistance, stretched wire, maximum resistance, minimum resistance, ratio resistance, Class 10, NEB, SEE, CTEVT, physics notes, exam question, resistivity, current electricity, physics numericals, aklearningnepal
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What Is the Ratio of Maximum to Minimum Resistance When a Wire Is Stretched to Double of Its Length?