The specific resistance `rho` of a thin wire of radius r cm, resistance R ohm and length L is given by `rho = (pi r^2R)/(L). If L =78+-0.01cm` `r = 0.26 +- 0.02 and R =32 +- 1 Omega,` What is the percentage error in `rho?`

Answer

`18.51%`

Explanation

Max. percentage error in `rho = (pi r^2 R)/(L) is `
`(Deltarho)/(rho)xx100 =+-(2(Deltar)/(r ) +(DeltaR)/(R ) +(DeltaL)/(L))xx100`
`=+-((2xx0.02)/(0.26) +(1)/(32) +(0.01)/(78))xx100`
`=+-(15.38 +3.12 +0.012) = +-18.51%`