In a potentiometer experiment,the balancing length is $8 \ m$ when two cells $E_1$ and $E_2$ are joined in series. When two cells are connected in opposition,the balancing length is $4 \ m$. The ratio of the e.m.f. of the two cells $\left(\frac{E_1}{E_2}\right)$ is

The potential difference across the ends of a potentiometer wire is constant. Two cells are connected in such a way that they first assist each other and then oppose each other. They are balanced on the potentiometer wire at lengths of $120 \ cm$ and $60 \ cm$ respectively. The ratio of the electromotive forces $(EMF)$ of the cells is:

Two students $X$ and $Y$ perform a potentiometer experiment separately and the null point was obtained as shown in the diagram. During the experiment:
$(i)$ $X$ increases the value of $R$ (resistance).
(ii) $Y$ decreases the value of $S$ (resistance) without any other change.
The position of the null point obtained by students $X$ and $Y$ respectively:

$A$ potentiometer wire of length $1 \ m$ is connected in series with $495 \ \Omega$ resistance and $2 \ V$ battery. If $0.2 \ mV/cm$ is the potential gradient,then the resistance of the potentiometer wire is: (in $\Omega$)

In a potentiometer circuit, a cell of $EMF$ $1.5\, V$ gives a balance point at $36\, cm$ length of wire. If another cell of $EMF$ $2.5\, V$ replaces the first cell, then at what length of the wire will the balance point occur? (in $cm$)

In a potentiometer,a balance point is obtained when: