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Resistivity of the material of a conductor of uniform cross-section varies along its length as $P = P_{0} (1 + α x)$ . Find the resistance of the conductor, if its length is $l$ and area of cross-section A.

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$\frac{ρ_{o} ℓ}{A} [1 + \frac{α ℓ}{2}]$

$\frac{ρ_{o} ℓ}{A} [1 + α ℓ]$

$\frac{ρ_{o} ℓ}{A} [1 + 2 α ℓ]$

$\frac{ρ_{o} ℓ}{A} [1 + \frac{α ℓ^{2}}{2}]$

answer is A.

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Detailed Solution

$R = \int_{0}^{l} \frac{ρ d x}{A} = \int_{o}^{l} \frac{ρ_{o}}{A} (1 + α x) d x$

$= \frac{ρ_{o}}{A} (l + \frac{α l^{2}}{2})$

$R = \frac{ρ_{o} l}{A} (1 + \frac{α l^{}}{2})$

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