In a potentiometer experiment,cells of e.m.f. $E_{1}$ and $E_{2}$ are connected in series $(E_{1} > E_{2})$,and the balancing length is $64 \ cm$. If the polarity of $E_{2}$ is reversed,the balancing length becomes $32 \ cm$. The ratio $\frac{E_{1}}{E_{2}}$ is:

In a potentiometer experiment,when three cells $A, B$ and $C$ are connected in series,the balancing length is found to be $420 \ cm$. If cells $A$ and $B$ are connected in series,the balancing length is $220 \ cm$ and for cells $B$ and $C$ connected in series,the balancing length is $320 \ cm$. The emf of cells $A, B$ and $C$ are respectively in the ratio of:

The resistivity of a potentiometer wire is $40 \times 10^{-8} \, \Omega \cdot m$ and its area of cross-section is $8 \times 10^{-6} \, m^2$. If $0.2 \, A$ current is flowing through the wire,the potential gradient will be:

Explain the comparison of the electromotive force (emf) of two cells using a potentiometer with a necessary diagram.

In a potentiometer experiment,the balancing point with a cell is at a length $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is: (in $\Omega$)

In a potentiometer experiment, a null point is obtained at a particular point for a cell on a potentiometer wire of length $L$. If the length of the potentiometer wire is increased by a few centimeters without changing the cell or the driving source, the balancing length will: