In the experiment of Ohm's law,a potential di | vedclass

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(A) The resistance $R$ is given by $R = \frac{\rho \ell}{A} = \frac{V}{I}$.Thus,the resistivity $\rho$ is given by $\rho = \frac{AV}{I\ell} = \frac{\p

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$3.9$

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(A) The resistance $R$ is given by $R = \frac{ho \ell}{A} = \frac{V}{I}$.Thus,the resistivity $ho$ is given by $ho = \frac{AV}{I\ell} = \frac{\pi d^2 V}{4I\ell}$,where $A = \frac{\pi d^2}{4}$.The relative error in resistivity is given by $\frac{\Delta ho}{ho} = 2\frac{\Delta d}{d} + \frac{\Delta V}{V} + \frac{\Delta I}{I} + \frac{\Delta \ell}{\ell}$.Given values: $V = 5.0\, V, \Delta V = 0.1\, V, \ell = 10.0\, cm, \Delta \ell = 0.1\, cm, d = 5.00\, mm, \Delta d = 0.01\, mm, I = 2.00\, A, \Delta I = 0.01\, A$.Substituting these values:$\frac{\Delta ho}{ho} = 2\left(\frac{0.01}{5.00}ight) + \frac{0.1}{5.0} + \frac{0.01}{2.00} + \frac{0.1}{10.0}$$\frac{\Delta ho}{ho} = 0.004 + 0.02 + 0.005 + 0.01 = 0.039$.The percentage error is $\frac{\Delta ho}{ho} 	imes 100 = 0.039 	imes 100 = 3.9\%$.

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