$1 k g$ piece of copper is drawn into a wire $1 m m^{2}$ thick, and another $1 k g$ copper piece into a wire $2 m m^{2}$ thick. Ratio of resistances respectively is

$2 : 1$

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$4 : 1$

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$8 : 1$

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$16 : 1$

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Solution

The correct option is B
$4 : 1$

The volume of wire = Length $\times$ Cross-sectional Area.

Since volume is same in both the cases, the length of the second wire will be half of the length of 1st wire.

$L_{1} = 2 L_{2} a n d A_{2} = 2 A_{1}$

Using $R = ρ \frac{L}{A}$ , $R_{1} = ρ \frac{L_{1}}{A_{1}}$ and $R_{2} = ρ \frac{L_{2}}{A_{2}}$

$R_{1} : R_{2}$ = $4 : 1$ i.e $R_{1}$ is $4$ times greater than $R_{2}$

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1 kg piece of copper is drawn into a wire 1 mm2 thick, and another 1 kg copper piece into a wire 2 mm2 thick. Ratio of resistances respectively is

$1 k g$ piece of copper is drawn into a wire $1 m m^{2}$ thick, and another $1 k g$ copper piece into a wire $2 m m^{2}$ thick. Ratio of resistances respectively is