Question

The resistance, R, of a wire has been measured as (40.0 ± 0.02)Ω . Its length, and radius, have been measured as

l = (120.0 ± 0.1) cm and r = (2.00 ± 0.01) mm.

The relative percentage error in the resistivity, p \((p=\frac{Rl}{\pi r^2}),\) of the material of this wire, would be

(1) 6.08%

(2) 1.6%

(3) 1.08%

(4) 0.6%

Answer 1

(2) 1.6%

We have,

p = Rl/πr²

∴ \(\frac{Δp}{p}=\frac{ΔR}{R}+\frac{Δl}{l}+2\frac{Δr}{r}\)

\(=\frac{0.2}{40.0}+\frac{0.1}{120.0}+2\times \frac{0.01}{2.00}\)

= 0.005 + 0.008 + 0.01 

= 0.0158 

∴ Relative percentage error 

= 1.6%