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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityMicroscopicMedium2 minPYQ_2021

PhysicsMediumsingle choice

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 :-

Options:

Answer:

Solution:

$R = \frac{ρ ℓ}{A} = \frac{V}{I}$
$ρ = \frac{A V}{I ℓ} = \frac{π d^{2} V}{4 I ℓ} (A = \frac{π d^{2}}{4})$
$∴ \frac{Δ ρ}{ρ} = \frac{2 Δ d}{d} + \frac{Δ V}{V} + \frac{Δ I}{I} + \frac{Δ ℓ}{ℓ}$
$\frac{Δ ρ}{ρ} = 2 (\frac{0.01}{5.00}) + \frac{0.1}{5.0} + \frac{0.01}{2.00} + \frac{0.1}{10.0}$
$\frac{Δ ρ}{ρ} = 0.004 + 0.02 + 0.005 + 0.01$
$\frac{Δ ρ}{ρ} = 0.039$
$%$ error $= \frac{Δ ρ}{ρ} \times 100 = 0.039 \times 100 = 3.90 % .$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Microscopic

⏱ 2mℹ️ Source: PYQ_2021

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