A wire 50, cm long and 1, mm^2 in cross-secti | vedclass

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(A) Given:Length $l = 50\, cm = 0.5\, m = 50 	imes 10^{-2}\, m$Area of cross-section $A = 1\, mm^2 = 1 	imes 10^{-6}\, m^2$Current $I = 4\, A$Voltag

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$1 	imes 10^{-6}\, \Omega\cdot m$

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(A) Given:Length $l = 50\, cm = 0.5\, m = 50 	imes 10^{-2}\, m$Area of cross-section $A = 1\, mm^2 = 1 	imes 10^{-6}\, m^2$Current $I = 4\, A$Voltage $V = 2\, V$Using Ohm's Law,the resistance $R$ is:$R = \frac{V}{I} = \frac{2}{4} = 0.5\, \Omega$The formula for resistance in terms of resistivity $ho$ is:$R = ho \frac{l}{A}$Rearranging for $ho$:$ho = \frac{R \cdot A}{l} = \frac{0.5 	imes 10^{-6}}{0.5} = 1 	imes 10^{-6}\, \Omega\cdot m$Thus,the resistivity of the wire is $1 	imes 10^{-6}\, \Omega\cdot m$.

$A$ wire $50\, cm$ long and $1\, mm^2$ in cross-section carries a current of $4\, A$ when connected to a $2\, V$ battery. The resistivity of the wire is:

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