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(B) The current $I$ flowing through the wire of resistance $R = 5 \, \Omega$ is given by the formula $I = \frac{E}{r + R}$, where $E = 3 \, V$ is the

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$2.5 \, V/m$

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(B) The current $I$ flowing through the wire of resistance $R = 5 \, \Omega$ is given by the formula $I = \frac{E}{r + R}$, where $E = 3 \, V$ is the $emf$ of the battery and $r = 1 \, \Omega$ is its internal resistance.Substituting the values: $I = \frac{3}{1 + 5} = \frac{3}{6} = 0.5 \, A$.The potential difference $V_{wire}$ across the entire length of the wire is $V_{wire} = I 	imes R = 0.5 	imes 5 = 2.5 \, V$.The length of the wire is $L = 1 \, m$.The potential gradient $k$ is defined as the potential drop per unit length: $k = \frac{V_{wire}}{L} = \frac{2.5 \, V}{1 \, m} = 2.5 \, V/m$.

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