A 20 cm long uniform wire of resistance 5 Ω is stretched to a uniform wire of 40 cm length. What will be the resistance of new wire?

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# A 20 cm long uniform wire of resistance 5 Ω is stretched to a uniform wire of 40 cm length. What will be the resistance of new wire?

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## Brainigniter

length = l = 20 cmArea of cross-section = AResistance = R =^{ρl}/_{A}= 5 ΩNew length = l’ = 40 cm = 2lNew area of cross-section = A’When the wire is stretched, its volume remains the same.So, A’ × l’ = A × l⇒ A’ × 40 = A × 20⇒ A’ =^{A}/_{2}New resistance= R’=^{ρl’}/_{A’}=^{ρ(2l)}/_{(A/}_{2})= 4^{ρl}/_{A}= 4R = 4 × 5 = 20 ΩTherefore, resistance of the new wire is 20 Ω.