An aluminium (resistivity rho = 2.2 times 10^ | vedclass

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(C) The resistance $R$ of a wire is given by the formula $R = \rho \frac{l}{A}$,where $\rho$ is the resistivity,$l$ is the length,and $A$ is the cross

Vedclass · Accepted Answer

$280$

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(C) The resistance $R$ of a wire is given by the formula $R = ho \frac{l}{A}$,where $ho$ is the resistivity,$l$ is the length,and $A$ is the cross-sectional area.Given: $R = 4 \Omega$,$ho = 2.2 	imes 10^{-8} \Omega \cdot m$,and diameter $d = 1.4 	ext{ mm} = 1.4 	imes 10^{-3} 	ext{ m}$.The radius $r = \frac{d}{2} = 0.7 	imes 10^{-3} 	ext{ m}$.The cross-sectional area $A = \pi r^2 = \pi (0.7 	imes 10^{-3})^2 	ext{ m}^2$.Rearranging the formula for length: $l = \frac{R A}{ho}$.Substituting the values: $l = \frac{4 	imes \pi 	imes (0.7 	imes 10^{-3})^2}{2.2 	imes 10^{-8}}$.$l = \frac{4 	imes 3.14159 	imes 0.49 	imes 10^{-6}}{2.2 	imes 10^{-8}} \approx \frac{6.1575 	imes 10^{-6}}{2.2 	imes 10^{-8}} \approx 2.7988 	imes 10^2 \approx 280 	ext{ m}$.

An aluminium (resistivity $\rho = 2.2 \times 10^{-8} \Omega \cdot m$) wire of a diameter $1.4 \text{ mm}$ is used to make a $4 \Omega$ resistor. The length of the wire is: (in $\text{ m}$)

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