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?

The wire of a potentiometer has a resistance of $4 \Omega$ and a length of $1 \ m$. It is connected to a cell of emf $2 \ V$ and internal resistance $1 \Omega$. The current flowing through the potentiometer wire is: (in $A$)

To determine the internal resistance of a cell with a potentiometer,when the cell is shunted by a resistance of $5 \Omega$,the balancing length is $250 \ cm$. When the cell is shunted by $20 \Omega$,the balancing length of the potentiometer wire is $400 \ cm$. The internal resistance of the cell is: (in $\Omega$)

$A$ cell in the secondary circuit gives a null deflection for $2.5 \,m$ length of a potentiometer wire having a total length of $10 \,m$. If the length of the potentiometer wire is increased by $1 \,m$ without changing the cell in the primary circuit, the new position of the null point is: (in $m$)

Two cells of e.m.f. $E_1$ and $E_2$ $(E_1 > E_2)$ are connected as shown in the figure. When a potentiometer is connected between points $A$ and $B$, the balancing length of the potentiometer wire is $412 \text{ cm}$. When the same potentiometer is connected between points $A$ and $C$, the balancing length is $103 \text{ cm}$. The ratio $E_1 : E_2$ is:

In a potentiometer experiment, the balancing length with a cell is $250 \, cm$. On shunting the cell with a resistance of $2 \, \Omega$, the balancing length becomes $125 \, cm$. The internal resistance of the cell is: