The resistance R of a wire is given by the relation `R= (rhol)/( pir^(2))`. Percentage error in the measurement of `rho, l and r` is 1 % , 2% and 3% respectively. Then the percentage error in the measurement of R is

Options

A. 6
B. 9
C. 8
D. 10

Explanation

Resistance of a wire `R=(pl)/(pir^(2))`
Take log to base e on both sides of equation (i), we get
`lnR=lnp+lnl-lnpi-2lnr`
Differentiate both sides of equation (ii), we get
`(DeltaR)/(R)=(DeltaR)/(p)+2(Deltal)/(l)-2(Deltar)/(r)`.....(iii)
Errors are always added, (for maximum error)
`therefore(DeltaR)/(R)xx100=(DeltaP)/(P)xx100+(Deltal)/(l)xx100+|(-2Deltar)/(r)xx100|`
`=1%+2%+2(3%)=9%`