One $k g$ of copper is drawn into a wire of $1 m m$ diameter and a wire of $2 m m$ diameter. The resistance of the two wires will be in the ratio

2 : 1

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1 : 2

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16 : 1

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4 : 1

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Solution

The correct option is C $16 : 1$
Mass of Ist wire $= (π r_{1}^{2} l_{1}) σ$
Mass of IInd wire $= (π r_{2}^{2} l_{2}) σ$
$∴ (π r_{1}^{2} l_{1}) σ = (π r_{2}^{2} l_{2}) σ$
$\Rightarrow \frac{l_{1}}{l_{2}} = {(\frac{r_{2}}{r_{1}})}^{2}$
$∴ \frac{R_{1}}{R_{2}} = \frac{ρ \frac{l_{1}}{A_{1}}}{ρ \frac{l_{2}}{A_{2}}}$
$= \frac{l_{1}}{l_{2}} \times \frac{A_{2}}{A_{1}} = \frac{l_{1}}{l_{2}} \times {(\frac{r_{2}}{r_{1}})}^{2}$
$= {(\frac{r_{2}}{r_{1}})}^{2} = {(\frac{1}{1 / 2})}^{4}$
$= 16 : 1$

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One kg of copper is drawn into a wire of 1mm diameter and a wire of 2mm diameter. The resistance of the two wires will be in the ratio

The correct option is C 16:1Mass of Ist wire =(πr21l1)σMass of IInd wire =(πr22l2)σ∴(πr21l1)σ=(πr22l2)σ⇒l1l2=(r2r1)2∴R1R2=ρl1A1ρl2A2 =l1l2×A2A1=l1l2×(r2r1)2 =(r2r1)2=(11/2)4 =16:1

One $k g$ of copper is drawn into a wire of $1 m m$ diameter and a wire of $2 m m$ diameter. The resistance of the two wires will be in the ratio