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

Question

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

Dr. Abubacker · Accepted Answer

(d) $l = \frac{R\pi r^{2}}{ho} = \frac{4.2 	imes 3.14 	imes \left( 0.2 	imes 10^{- 3} ight)^{2}}{48 	imes 10^{- 8}} = 1.1m$

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

Explanation

Reviewed by Dr. Abubacker

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