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

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$\frac{1}{4} \frac{ρ l}{V A}$

$\frac{1}{4} \frac{V A}{ρ l}$

$\frac{3}{4} \frac{V A}{ρ l}$

$4 \frac{V A}{ρ l}$

answer is B.

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Detailed Solution

Resistance $R = \frac{ρ l}{A}$ (original)
A/c to question resistance = $\frac{ρ (2 l)}{A / 2} = \frac{4 ρ l}{A}$

$\Rightarrow I = \frac{V}{R} = \frac{V A}{4 ρ l}$

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