$A$ potentiometer having a potential gradient of $2\, mV/cm$ is used to measure the potential difference across a resistance of $10\, \Omega$. If a length of $50\, cm$ of the potentiometer wire is required to obtain the null point,the current passing through the $10\, \Omega$ resistor is (in $mA$):

In a potentiometer experiment,the balancing length for a cell is $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes half the initial balancing length. The internal resistance of the cell is: (in $\Omega$)

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?

For the measurement of potential difference,a potentiometer is preferred in comparison to a voltmeter because

The figure shows a $2.0 \; V$ potentiometer used for the determination of internal resistance of a $1.5 \; V$ cell. The balance point of the cell in open circuit is $76.3 \; cm$. When a resistor of $9.5 \; \Omega$ is used in the external circuit of the cell,the balance point shifts to $64.8 \; cm$ length of the potentiometer wire. Determine the internal resistance (in $\Omega$) of the cell.

When a cell of $E$.$M$.$F$. '$E_1$' is connected to a potentiometer wire, the balancing length is '$l_1$'. Another cell of $E$.$M$.$F$. '$E_2$' $(E_1 > E_2)$ is connected along with $E_1$ such that the two cells oppose each other, and the balancing length is '$l_2$'. The ratio $E_1 : E_2$ is