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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 isA. 6B. 9C. 8D. 10 |
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Answer» 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%` |
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