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(A) The potential gradient $x$ is given by the formula: $x = \frac{V_{wire}}{L} = \frac{E}{R + R_h} \cdot \frac{R}{L}$.Given: $E = 2\,V$,$R_h = 490\,\

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(A) The potential gradient $x$ is given by the formula: $x = \frac{V_{wire}}{L} = \frac{E}{R + R_h} \cdot \frac{R}{L}$.Given: $E = 2\,V$,$R_h = 490\,\Omega$,$L = 1\,m = 100\,cm$,and $x = 0.2\,mV/cm = 0.2 	imes 10^{-3}\,V/cm$.Substituting the values:$0.2 	imes 10^{-3} = \frac{2}{R + 490} \cdot \frac{R}{100}$.$0.2 	imes 10^{-3} 	imes 100 = \frac{2R}{R + 490}$.$0.02 = \frac{2R}{R + 490}$.$0.02(R + 490) = 2R$.$0.02R + 9.8 = 2R$.$9.8 = 1.98R$.$R = \frac{9.8}{1.98} \approx 4.949\,\Omega$.Rounding to the nearest provided option,the resistance is $4.9\,\Omega$.

$A$ potentiometer wire of length $1\,m$ is connected in series with a resistance of $490\,\Omega$ and a battery of $2\,V$. If the potential gradient is $0.2\,mV/cm$,then the resistance of the potentiometer wire is ............$\Omega$. (in $.9$)

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