In the experiment of $Ohm's law,$ a potential difference of $5.0\, V$ is applied across the end of a conductor of length $10.0\, cm$ and diameter of $5.00\, mm .$ The measured current in the conductor is $2.00\, A$. The maximum permissible percentage error in the resistivity of the conductor is
JEE MAIN 2021, Diffcult
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$R =\frac{\rho \ell}{ A }=\frac{ V }{ I }$
$\rho=\frac{ AV }{ I \ell}=\frac{\pi d ^{2} V }{4 I \ell} \quad\left( A =\frac{\pi d ^{2}}{4}\right)$
$\therefore \frac{\Delta \rho}{\rho}=\frac{2 \Delta d }{ d }+\frac{\Delta V }{ V }+\frac{\Delta I }{ I }+\frac{\Delta \ell}{\ell}$
$\frac{\Delta \rho}{\rho}=2\left(\frac{0.01}{5.00}\right)+\frac{0.1}{5.0}+\frac{0.01}{2.00}+\frac{0.1}{10.0}$
$\frac{\Delta \rho}{\rho}=0.004+0.02+0.005+0.01$
$\frac{\Delta \rho}{\rho}=0.039$
$\%$ error $=\frac{\Delta \rho}{\rho} \times 100=0.039 \times 100=3.90 \%$
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