$A$ 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 the experiment of calibration of a voltmeter,a standard cell of $e.m.f. = 1.1 \text{ V}$ is balanced against $440 \text{ cm}$ of a potentiometer wire. The potential difference across a resistance is found to balance against $220 \text{ cm}$ of the wire. The corresponding reading of the voltmeter is $0.5 \text{ V}$. The error in the reading of the voltmeter will be ................. $V$.

In a potentiometer experiment, a cell of emf $1.25 \,V$ gives a balancing length of $30 \,cm$. If the cell is replaced by another cell, the balancing length is found to be $40 \,cm$. What is the emf of the second cell?

$A$ $10\,m$ long potentiometer wire has a potential gradient of $0.0025\,V/cm$. Calculate the distance of the null point when the wire is connected to a $1.025\,V$ standard cell.

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 (see figure) a balance is obtained at a length of $400 \ mm$ when using a known battery of emf $1.6 \ V$. After removing this battery,another battery of unknown emf is used and balance is obtained at a length of $650 \ mm.$ The emf of the unknown battery is ............. $V$. (in $V$)