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

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$\frac{ρ_{0} l}{A} [1 + \frac{α l}{2}]$

$\frac{ρ_{0} l}{A} [1 + α l]$

$\frac{ρ_{0} l}{A} [1 + \frac{α l^{2}}{2}]$

$\frac{ρ_{0} l}{A} [1 + 2 α l]$

answer is A.

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Detailed Solution

Resistance of an element of thickness dx at a distance x from on end is

$dR = \frac{ρ d x}{A} = \frac{ρ_{0} (1 + α x)}{A} d x$

Total resistance of the rod will be

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

$\frac{ρ_{0} l}{A} [1 + \frac{α l^{2}}{2}]$

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