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(A) The potential gradient $x$ is defined as the potential drop per unit length of the wire,given by $x = \frac{V}{L}$.From Ohm's law,the potential dr

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$\frac{I\rho}{A}$

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(A) The potential gradient $x$ is defined as the potential drop per unit length of the wire,given by $x = \frac{V}{L}$.From Ohm's law,the potential drop $V$ across a wire of length $L$ is $V = IR$,where $R$ is the resistance of the wire.The resistance $R$ is given by the formula $R = \frac{\rho L}{A}$,where $\rho$ is the resistivity and $A$ is the area of cross-section.Substituting the expression for $R$ into the potential drop equation: $V = I \left( \frac{\rho L}{A} \right)$.Now,substituting $V$ into the potential gradient formula: $x = \frac{I \left( \frac{\rho L}{A} \right)}{L}$.Simplifying the expression,we get $x = \frac{I\rho}{A}$.

If the resistivity of a potentiometer wire is $\rho$ and the area of cross-section is $A$,then what will be the potential gradient along the wire,given that a current $I$ flows through it?

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