If a copper wire is stretched to make its cross sectional radius 0.1%thinner then what is the percentage increase in its resistance
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Answered by
297
When a wire is stretched it’s volume remains constant.
R = ρL/A
= (ρL/A) × A/A
= ρV/A²
= ρV / (π²r⁴)
From above
R ∝ 1/r⁴
ΔR/R = 4 × Δr/r
= 4 × 0.1%
= 0.4%
Resistance of wire is increased by 0.4%
R = ρL/A
= (ρL/A) × A/A
= ρV/A²
= ρV / (π²r⁴)
From above
R ∝ 1/r⁴
ΔR/R = 4 × Δr/r
= 4 × 0.1%
= 0.4%
Resistance of wire is increased by 0.4%
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