Question

Calculate the percentage error in specific resistance of the given material of wire whose

specific resistance can be found by the relation ρ = ( r

2 R)/l ,Where ‘r’ = radius of wire =

(0.26 ± 0.02)cm. ‘l’= length of wire= (15.6 ± 0.1) cm. R = resistance of wire = (64 ± 2) ohm​

Answer 1

Answer:

ρ = (πr²R)/l  

formula given

 applying dimension analysis,

since the constant pie is dimensionless then,

ρ = (r²R)/l

by introducing natural logarithm (In) both sides

Inρ = In r²+ In R - In l

Inρ=2 In r+In R-In l

then we differentiate,

dInρ=d2Inr+dInR-dInl

dρ = d 2r+dR-dl

Δp/p=Δ2r/r +ΔR/R-Δl/l

since with deal with measurements,then the errors are maximized

Δρ/ρ=2Δr/r±ΔR/R±Δl/l

before imputing our data we should first consider SI units

Δρ/ρ=(2 x 0.0002 /0.0026)m²+(2/64)ohms+(0.001/0.156)m

       =(2/13)m²+(1/32)ohms +(1/156)m

       =(224+91+18.67/2912)

       ∴=333.67/2912 ohms per metre

therefore

the percentage error = relative fractional error x 100%

where relative fraction error= absolute error/true error x 100%

(a ± b) where a is true error and  b absolute error

then relative fraction error= 333.67/2912 x 100%

                                           =0.11%

Explanation:

the percentage error in specific resistance of the given material of the wire is 0.11%