In the experiment of Ohm's law, a potential difference of 5.0 V is… | JEE Main 2021 (18 Mar, Shift 1)

Q85JEE Main 2021 · 18 MAR · SHIFT 1Difficult

Errors in Measurement

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:

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Answer: (a)

$\rho = \dfrac{RA}{l} = \dfrac{V\pi d^2}{4Il}$ , so $\dfrac{\Delta\rho}{\rho} = \dfrac{\Delta V}{V} + \dfrac{\Delta I}{I} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta l}{l}$ .
Errors from last digit: $\Delta V = 0.1$ , $\Delta I = 0.01$ , $\Delta d = 0.01$ mm, $\Delta l = 0.1$ cm.
$= \dfrac{0.1}{5.0} + \dfrac{0.01}{2.00} + 2\times\dfrac{0.01}{5.00} + \dfrac{0.1}{10.0} = 0.02 + 0.005 + 0.004 + 0.01 = 0.039$
So the error is $3.9\%$ , answer (a).

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