1 kg piece of copper is drawn into a wire of 1 mm thickness, and another 1 kg copper piece into a wire of 2 mm thickness. What is the ratio of the resistance of these two wires respectively?

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 D
16:1

The volume of wire = length x Cross-sectional Area.

Since volume is the same in both the cases as the density and mass of copper is same, the length of the second wire will be half of the length of the 1st wire.

$L (1 s t_{w} i r e) \times A (1 s t_{W} i r e)$ = $L (2 n d_{W} i r e) \times A (2 n d_{W} i r e)$

$L (1 s t_{w} i r e) \times A = L (2 n d W i r e) \times 4 A$

$\frac{L (1 s t_{w} i r e)}{4} = L (2 n d_{W} i r e)$

Using R = ρ $\frac{L}{A}$ ,

R(1st Wire) = ρ $\frac{L}{A}$

R(2nd Wire) = ρ $\frac{L}{4 \times 4 A}$

$\frac{R (1 s t W i r e)}{R (2 n d W i r e)}$ = 16:1

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