In the experiment of Ohm's law,… - JEE Main 2021 Physics PYQ

JEE Main 2021 — Physics Electromagnetism

2021mcqmedium

In the experiment of Ohm's law, a potential difference of $5.0V$ 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 $2.00A.$ The maximum permissible percentage error in the resistivity of the conductor is :-

Official previous-year question

Held on 18 Mar 2021 · Verified 6 Jul 2026.

Options

A
$3.9$
B
$8.4$
C
$7.5$
D
$3.0$

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Solution

$R=\frac{\rho \ell }{A}=\frac{V}{I}$

$\rho =\frac{AV}{I\ell }=\frac{\pi {d}^{2}V}{4I\ell }(A=\frac{\pi {d}^{2}}{4})$

$\therefore \frac{\Delta \rho }{\rho }=\frac{2\Delta d}{d}+\frac{\Delta V}{V}+\frac{\Delta I}{I}+\frac{\Delta \ell }{\ell }$

$\frac{\Delta \rho }{\rho }=2(\frac{0.01}{5.00})+\frac{0.1}{5.0}+\frac{0.01}{2.00}+\frac{0.1}{10.0}$

$\frac{\Delta \rho }{\rho }=0.004+0.02+0.005+0.01$

$\frac{\Delta \rho }{\rho }=0.039$

$%$ error $=\frac{\Delta \rho }{\rho }\times 100=0.039\times 100=3.90%.$

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Solution

$R=\frac{\rho \ell }{A}=\frac{V}{I}$

$\rho =\frac{AV}{I\ell }=\frac{\pi {d}^{2}V}{4I\ell }(A=\frac{\pi {d}^{2}}{4})$

$\therefore \frac{\Delta \rho }{\rho }=\frac{2\Delta d}{d}+\frac{\Delta V}{V}+\frac{\Delta I}{I}+\frac{\Delta \ell }{\ell }$

$\frac{\Delta \rho }{\rho }=2(\frac{0.01}{5.00})+\frac{0.1}{5.0}+\frac{0.01}{2.00}+\frac{0.1}{10.0}$

$\frac{\Delta \rho }{\rho }=0.004+0.02+0.005+0.01$

$\frac{\Delta \rho }{\rho }=0.039$

$%$ error $=\frac{\Delta \rho }{\rho }\times 100=0.039\times 100=3.90%.$

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