The lengths of two wires made of the same material are in the ratio $2 : 1$ . If the masses of the two wires are in the ratio $1 : 8$ , find the ratio of their resistances.

1 : 16

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32 : 1

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16 : 1

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1 : 32

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Solution

The correct option is B $32 : 1$
mass $M = σ V$ where $σ = d e n s i t y$

$⟹ M = σ A L ⟹ A = \frac{M}{σ L}$

$R = ρ \frac{L}{A} = σ ρ \frac{L^{2}}{M}$

$⟹ \frac{R_{1}}{R_{2}} = \frac{M_{2}}{M_{1}} {(\frac{L_{1}}{L_{2}})}^{2}$

$⟹ \frac{R_{1}}{R_{2}} = 8 \times 4 = 32$

$⟹ R_{1} : R_{2} = 32 : 1$

Answer-(B)

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