In the experiment of Ohm's law, a potential difference of 5.0 V is applied across the end of a condu

Question

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\

Professor PC Thomas Classes · Answer

(c) : $V=5.0 \mathrm{~V}, \Delta V=0.1 \mathrm{~V}$\ $l=10 \mathrm{~cm}=10 	imes 10^{-2} \mathrm{~m}, \Delta l=0.1 \mathrm{~cm}$\ diameter, $d=5.00 \mathrm{~mm}=5 	imes 10^{-3} \mathrm{~m}$\ $\Delta d=0.01 \mathrm{~mm}$\ $I=2.00 \mathrm{~A}, \Delta I=0.01 \mathrm{~A}$\ Resistance is given by, $R=ho \frac{l}{A}$.\ From Ohm's law, $R=\frac{V}{I}$\ So, $V=I ho \frac{l}{A}$ or, $ho=\frac{V A}{I l}, ho=\frac{V \pi d^{2}}{I l}$\ For maximum permissible error,\ $\frac{\Delta ho}{ho}=\frac{\Delta V}{V}+\frac{2 \Delta d}{d}+\frac{\Delta I}{I}+\frac{\Delta l}{l}$\ $=\frac{0.1}{5}+2 	imes \frac{0.01}{5}+\frac{0.01}{2}+\frac{0.1}{10}$\ $=0.02+0.004+0.005+0.01=0.039$\ Percentage error, $\frac{\Delta l}{l} 	imes 100=0.039 	imes 100=3.9$\ Maximum permissible error in the resistivity of the conductor is $3.9 \%$.\ Hence, the correct answer is option (c).\