A chunk of copper weighing $1 k g$ is drawn into a wire $1 m m^{2}$ thick. Another chunk of copper weighing $1 k g$ is drawn into a wire $2 m m^{2}$ thick. What is the ratio of their resistances?

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

Volume of the wire = length x cross-sectional area.

Let the volume of the first wire be $l_{1} \times a_{1}$

Let the volume of the second wire be $l_{2} \times a_{2}$

Given, $a_{1} = 1 m m^{2} a n d a_{2} = 2 m m^{2}$

Since volume is the same in both the cases, $l_{1} \times 1 = l_{2} \times 2$ $\Rightarrow l_{2} = \frac{l_{1}}{2}$

We know that the resistance of a wire $R = ρ \frac{l}{a}$ Where $ρ$ is the resistivity, L is the length of the wire and A is the cross-sectional area of the wire

Resistance of wire 1 = $R_{1} = ρ \frac{l_{1}}{a_{1}} = ρ \frac{l_{1}}{1}$

Resistance of wire 2 $R_{2} = ρ \frac{l_{2}}{a_{2}} = ρ \frac{\frac{l_{1}}{2}}{2} = ρ \frac{l_{1}}{4}$

i.e. $R_{2} = \frac{R_{1}}{4}$

Or, $R_{1} = 4 R_{2}$

Hence $R_{1} : R_{2} = 4 : 1$

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A chunk of copper weighing 1 kg is drawn into a wire 1 mm2 thick. Another chunk of copper weighing 1 kg is drawn into a wire 2 mm2 thick. What is the ratio of their resistances?

A chunk of copper weighing $1 k g$ is drawn into a wire $1 m m^{2}$ thick. Another chunk of copper weighing $1 k g$ is drawn into a wire $2 m m^{2}$ thick. What is the ratio of their resistances?