4 wire is stretched so that its length is 20% more than its initial length the percentage increase in the resistance of the wire is
Answers
Percentage increase in resistance of wire is 44%.
Explanation:
=> Suppose the resistance of wire is R₁
Radius of the wire = r₁
Length of the wire = l
Resistivity of wire = ρ
=> Thus, R₁ = ρ * l / πr₁²
Volume remains constant.
V₀ = Vnew
πr₁²l = π r²new * l.new
=> Now, According to the question, wire is stretched so that its length is 20% more than its initial length.
Rnew = ρ * l.new/ πr²new
∴ πr₁²l = π * r²new * l.new
= π * r²new * (1.2 l )
∴ r₁² = r²new (1.2)
r²new = r₁² /1.2
r.new = r₁ / √1.2
=> New resistance of the wire:
Rnew = ρ . (1.2* l) / π (r₁ / √1.2)²
= ρ * l * 1.2 * 1.2 / πr²
= 1.44 ρ * l / πr²
= 1.44 R₁
=> Thus, percentage change in resistance:
% change in resistance = (Rnew - R₁ / R₁)
= 1.44 R₁ - R₁ / R₁
= 0.44 R₁ / R₁
= 0.44
= 44%
Thus, percentage increase in resistance of wire is 44%
Learn more:
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