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A wire of resistance $160 Ω$ is melted and drawn into a wire of one-fourth of its length. The new resistance of the wire will be

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$640 Ω$

$16 Ω$

$10 Ω$

$40 Ω$

answer is C.

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Detailed Solution

$R = ρ \frac{L}{A}$

$LA = constant$

$R \propto L^{2}$

$\frac{160}{R} = \frac{L^{2}}{(\frac{L}{4})}$

$R = 10 Ω$

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