Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

In a potentiometer experiment,the balancing length with a cell $E_{1}$ of unknown e.m.f. is $\ell_{1} \ cm$. By shunting the cell with a resistance $R \ \Omega$,the balancing length becomes $\frac{\ell_{1}}{2} \ cm$. The internal resistance $(r)$ of the cell is:

$A$ cell in a 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 $m$)

In a potentiometer experiment,two cells of $e.m.f.$ $E_1$ and $E_2$ are used in series and in conjunction,and the balancing length is found to be $58 \ cm$ of the wire. If the polarity of $E_2$ is reversed,the balancing length becomes $29 \ cm$. The ratio $\frac{E_1}{E_2}$ of the $e.m.f.$ of the two cells is: (in $:1$)

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 $10 \,m$ and resistance $20 \,\Omega$ is connected in series with a $25 \,V$ battery and an external resistance $30 \,\Omega$. $A$ cell of emf $E$ in the secondary circuit is balanced by a $250 \,cm$ long potentiometer wire. The value of $E$ (in volt) is $\frac{x}{10}$. The value of $x$ is.......