Question

Calculate the percentage error in specific resistance , `rho = pi r^(2) R // l `, where r = radius of wire `= 0.26 +- 0.02 cm` , l = length of wire `= 156.0 +- 0.1 cm`, and R = resistance of wire `= 64 +- 2 Omega`.

Answer 1

Correct Answer - `~ 18.6%`
`P = ( pi r^(2) R )/( l) = (22)/(l) xx ((0.26)^(2) xx 64)/( 156)`
`(Delta P)/( P) = (2( 0.02))/(0.26) + (2)/(64) + (0.1) /(156)`
`(Delta P)/(P) xx 100 = 18.57 ~ 18.6%`