A wire is stretched such that its length -increases by 0.2 %. Its resistance

Question

A wire is stretched such that its length -increases by 0.2 %. Its resistance (A) increases by 0.2 % (B) decreases by 0.4% (C) increases by 0.4% (D) increases by 0.1%

Tardigrade Admin · Accepted Answer

R = ρ (l/A) After stretching, area of cross-section changes as the volume remains constant, i e., V = Al or A = (V/l) ∴ R = ρ (l/V / l) = ρ (l2/V) or (Δ R/R) =2 (Δ l/l) = 2 × 0.2 % (∵ (Δ l/l) = 0.2 %) = 0.4 %

Q. A wire is stretched such that its length -increases by $0.2%$ . Its resistance

increases by 0.2 %

decreases by 0.4%

increases by 0.4%

increases by 0.1%

$R = ρ \frac{l}{A}$
After stretching, area of cross-section changes as the volume remains constant,
$i e ., V = A l$ or $A = \frac{V}{l}$
$∴ R = ρ \frac{l}{V / l} = ρ \frac{l ^{2}}{V}$
or $\frac{Δ R}{R} = 2 \frac{Δ l}{l} = 2 \times 0.2% (∵ \frac{Δ l}{l} = 0.2%)$
$= 0.4%$

Q. A wire is stretched such that its length -increases by 0.2%. Its resistance