In the experiment of Ohm's law, a potential difference of $5.0 \mathrm{~V}$ is applied across the end of a conductor of length $10.0 \mathrm{~cm}$ and diameter of $5.00 \mathrm{~mm}$. The measured current in the conductor is $2.00$ A. The maximum permissible percentage error in the resistivity of the conductor is :-
Correct Option: 1
$\mathrm{R}=\frac{\rho \ell}{\mathrm{A}}=\frac{\mathrm{V}}{\mathrm{I}}$
$\rho=\frac{\mathrm{AV}}{\mathrm{I} \ell}=\frac{\pi \mathrm{d}^{2} \mathrm{~V}}{4 \mathrm{I} \ell} \quad\left(\mathrm{A}=\frac{\pi \mathrm{d}^{2}}{4}\right)$
$\therefore \quad \frac{\Delta \rho}{\rho}=\frac{2 \Delta \mathrm{d}}{\mathrm{d}}+\frac{\Delta \mathrm{V}}{\mathrm{V}}+\frac{\Delta \mathrm{I}}{\mathrm{I}}+\frac{\Delta \ell}{\ell}$
$\frac{\Delta \rho}{\rho}=2\left(\frac{0.01}{5.00}\right)+\frac{0.1}{5.0}+\frac{0.01}{2.00}+\frac{0.1}{10.0}$
$\frac{\Delta \rho}{\rho}=0.004+0.02+0.005+0.01$
$\frac{\Delta \rho}{\rho}=0.039$
$\%$ error $=\frac{\Delta \rho}{\rho} \times 100=0.039 \times 100=3.90 \%$
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