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 $80 \ cm$. If the polarity of $E_2$ is reversed,the balancing length becomes $20 \ cm$. The ratio $E_1 / E_2$ is:

To compare the $EMF$ of two cells using a potentiometer, the balancing lengths obtained are $200 \ cm$ and $150 \ cm$. The least count of the scale is $1 \ cm$. The percentage error in the ratio of the EMFs is . . . . . . .

$A$ wire,$10 \ m$ long,has a resistance of $40 \ \Omega$. It is connected in series with a resistance box of resistance $R$ and a $2 \ V$ storage cell. If the potential gradient along the wire is $0.1 \ mV/cm$,then the value of $R$ is (in $\Omega$)

The length of a potentiometer wire is $l$. $A$ cell of emf $E$ is balanced at a length $\left(\frac{l}{3}\right)$ from the positive end of the wire. If the length of the wire is increased by $\left(\frac{l}{2}\right)$,the distance at which the same cell gives the balancing point is (The cell in the primary circuit is ideal and no series resistance is present in the primary circuit.)

In a potentiometer experiment,cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$. The balancing length is $64 \ cm$ of the wire. If the polarity of $E_2$ is reversed,the balancing length becomes $32 \ cm$. The ratio $\frac{E_1}{E_2}$ is: (in $: 1$)

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