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Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityMiscellaneousMedium3 minQB

PhysicsMediumsingle choice

The wire shown in figure has a uniform cross-section $A$ .

wire with x=0 and x=L marked

Resistivity of the material of wire is given by $ρ = ρ_{0} (\frac{L}{L + x})$ . A potential difference $V$ is applied across the wire :-

Options:

Answer:

Solution:

$R = \int_{0}^{L} \frac{ρ _{0} L}{L + x} \frac{d x}{A} = \frac{ρ _{0} L ln 2}{A}$ - option (A).

Current density $J = I / A$ is uniform through the constant cross-section (B wrong); $E (x) = \frac{V}{ln 2 ( L + x )}$ gives $E (0) = \frac{V}{L ln 2}$ and $E (L) = \frac{V}{2 L ln 2}$ (C, D wrong).

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Miscellaneous

⏱ 3mℹ️ Source: QB

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