Question

If the measured value of resistance `R = 1.05 Omega`, wire diameter ` d = 0.60 mm` , and length `l = 75.3 cm`, then find the maximum permissible error in resistivity ,
` rho = ( R ( pi d^(2) // 4)) /( l)`.

Answer 1

`((d rho)/( rho))_(max) = ( Delta R )/( R ) + 2 ( Delta d) /(d) + ( Delta l)/(l)`
`R = 1.05 Omega rarr Delta R = 0.01 Omega `( least count)
`Delta d = 0.60 mm rarr Delta d = 0.01 mm `( least count)
`l = 75.3 rarr Delta l = 0.1 cm` ( least count )
`((d rho)/( rho))_( max) = ( 0.01 Omega )/( 1.05 Omega ) + 2 (( 0.01 mm)/( 0.60 mm)) + (0.1 cm)/(75.3 cm)`