What length of the wire of specific resistance 48×10−8Ωm is needed to make a resistance of 4.2Ω (diameter of wire = 0.4mm)

R=ρlA;l=RAρ=4.2×π0.42×10−3248×10−8=1.1m

What length of the wire of specific resistance 48×10−8Ωm is needed to make a resistance of 4.2Ω (diameter of wire = 0.4mm)

R=ρlA;l=RAρ=4.2×π0.42×10−3248×10−8=1.1m

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What length of the wire of specific resistance $48 \times 10^{- 8} Ω m$ is needed to make a resistance of $4.2 Ω$ (diameter of wire = 0.4mm)

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4.1m

2.1 m

3.1 m

1.1 m

answer is D.

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Detailed Solution

$R = \frac{ρl}{A}; l = \frac{RA}{ρ} = \frac{4.2 \times π {(\frac{0.4}{2} \times 10^{- 3})}^{2}}{48 \times 10^{- 8}} = 1.1 m$

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What length of the wire of specific resistance 48×10−8Ωm is needed to make a resistance of 4.2Ω (diameter of wire = 0.4mm)

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What length of the wire of specific resistance $48 \times 10^{- 8} Ω m$ is needed to make a resistance of $4.2 Ω$ (diameter of wire = 0.4mm)