A wire of resistance 160 is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be [2023]

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Solution

Q.
A wire of resistance 160 $Ω$ is melted and drawn in wire of one-fourth of its length. The new resistance of the wire will be [2023]

1 10 $Ω$

2 640 $Ω$

3 40 $Ω$

4 16 $Ω$

Ans.
(1)

$Volume = Constant$

$A_{1} L_{1} = A_{2} L_{2}$

$A_{1} L = A_{2} \frac{L}{4}$

$4 A_{1} = A_{2}$

$R_{1} = \frac{ρ L_{1}}{A_{1}}, R_{2} = \frac{ρ L_{2}}{A_{2}}$

$\frac{R_{2}}{R_{1}} = \frac{L_{2} A_{1}}{A_{2} L_{1}}$ $= \frac{1}{4} \frac{A_{1}}{4 A_{1} L}$

$R_{2} = \frac{1}{16} R_{1}$ $= 10 Ω$

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