Current Electricity

JEE Main level, single correct, medium

24 of 380

Question 24 of 380

In the experiment of Ohm's law, a potential difference of 5.0V5.0V is applied across the end of a conductor of length 10.0cm10.0cm and diameter of 5.00mm.5.00mm. The measured current in the conductor is 2.00A.2.00A. The maximum permissible percentage error in the resistivity of the conductor is :-

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Answer: (A) 3.93.9

R=ρℓA=VIR=\dfrac{\rho \ell }{A}=\dfrac{V}{I}
ρ=AVIℓ=πd2V4Iℓ(A=πd24)\rho =\dfrac{AV}{I\ell }=\dfrac{\pi {d}^{2}V}{4I\ell }\left(A=\dfrac{\pi {d}^{2}}{4}\right)
∴Δρρ=2Δdd+ΔVV+ΔII+Δℓℓ∴\dfrac{\Delta \rho }{\rho }=\dfrac{2\Delta d}{d}+\dfrac{\Delta V}{V}+\dfrac{\Delta I}{I}+\dfrac{\Delta \ell }{\ell }
Δρρ=2(0.015.00)+0.15.0+0.012.00+0.110.0\dfrac{\Delta \rho }{\rho }=2\left(\dfrac{0.01}{5.00}\right)+\dfrac{0.1}{5.0}+\dfrac{0.01}{2.00}+\dfrac{0.1}{10.0}
Δρρ=0.004+0.02+0.005+0.01\dfrac{\Delta \rho }{\rho }=0.004+0.02+0.005+0.01
Δρρ=0.039\dfrac{\Delta \rho }{\rho }=0.039
% error =Δρρ×100=0.039×100=3.90=\dfrac{\Delta \rho }{\rho }\times 100=0.039\times 100=3.90%.

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