What length of the wire of specific resistance 48 -10-8- m is needed to make a resistance of 4-2-diameter of wire -0-4mm

The correct option is D 1.1m R=plA;l=RAp=4.2×π ( 0.42×10^ 3 )^24.8×10^ 8=1.1m ...

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Standard XII

Physics

Conductivity and Resistivity

What length o...

Question

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 $= 0.4$ mm)

4.1m

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3.1m

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2.1m

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1.1m

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Solution

The correct option is D 1.1m
$R = \frac{p l}{A}; l = \frac{R A}{p} = \frac{4.2 \times π {(\frac{0.4}{2} \times 10^{- 3})}^{2}}{4.8 \times 10^{- 8}} = 1.1 m$

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

The correct option is D 1.1mR=plA;l=RAp=4.2×π(0.42×10−3)24.8×10−8=1.1m

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 $= 0.4$ mm)

The correct option is D 1.1m
$R = \frac{p l}{A}; l = \frac{R A}{p} = \frac{4.2 \times π {(\frac{0.4}{2} \times 10^{- 3})}^{2}}{4.8 \times 10^{- 8}} = 1.1 m$