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

\frac{1}{4} \frac{ρ l}{V A}

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\frac{3}{4} \frac{V A}{ρ l}

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4 \frac{V A}{ρ l}

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\frac{1}{4} \frac{V A}{ρ l}

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Solution

The correct option is D $\frac{1}{4} \frac{V A}{ρ l}$
We know that
$R = ρ \frac{l}{A}$
Now, new length: $l^{'} = 2 l$
& new area of cross section, $A^{'} = A / 2$

$∴$ New resistance, $R^{'} = ρ . \frac{2 l}{\frac{A}{2}}$
$\Rightarrow R^{'} = 4 \frac{ρ l}{A}$
$\Rightarrow R^{'} = 4 R$

$∴$ Resultant current: $I = \frac{V}{4 R}$
$I = \frac{1}{4} \frac{V A}{ρ l}$
Hence, option (d) is correct.

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The correct option is D 14VAρlWe know that R=ρlA Now, new length: l′=2l & new area of cross section, A′=A/2 ∴ New resistance, R′=ρ.2lA2 ⇒R′=4ρlA ⇒ R′=4R ∴ Resultant current: I=V4R I=14VAρl Hence, option (d) is correct.