In a potentiometer circuit,a cell of $2\,V$ $e.m.f.$ and $5\,\Omega$ internal resistance is connected to a uniform wire of length $1000\,cm$ and resistance $15\,\Omega$. The potential gradient of the wire 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:

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 $\Omega$)

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:

$A$ cell,shunted by an $8 \; \Omega$ resistance,is balanced across a potentiometer wire of length $3 \; m$. The balancing length is $2 \; m$ when the cell is shunted by a $4 \; \Omega$ resistance. The value of internal resistance of the cell will be $\dots \; \Omega$.

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: