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A conducting wire of length ‘ $l$ ’, area of cross-section A and electric resistivity $ρ$ is connected between the terminals of a battery. A potential difference V is developed across its ends, causing an electric current. If the length of the wire of the same material is doubled and the area of cross-section is halved, the resultant current would be:

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$4 \frac{V A}{ρ l}$

$\frac{1}{4} \frac{V A}{ρ l}$

$\frac{3}{4} \frac{V A}{ρ l}$

$\frac{1}{4} \frac{ρ l}{V A}$

answer is B.

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Detailed Solution

$\begin{array}{l} i = \frac{V}{R} \\ New resistance = R^{'} = \frac{ρ (21)}{A / 2} = 4 R \\ New current = i = \frac{V}{4 R} = \frac{V A}{4 ρ l} \end{array}$

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