The potentiometer is best for measuring voltage,as

$A$ battery of internal resistance $1 \, \Omega$ and $emf$ $3 \, V$ sends a current through $1 \, m$ of uniform wire of resistance $5 \, \Omega$. The poles of a cell of $emf$ $1.4 \, V$ are connected to two points on the wire such that no current passes through this cell. The potential gradient of the wire 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 $\dots \; \Omega$.

$A$ galvanometer can be used as a voltmeter by connecting a

$A$ potentiometer wire is $4 \text{ m}$ long and a potential difference of $3 \text{ V}$ is maintained between the ends. The e.m.f. of the cell which balances against a length of $100 \text{ cm}$ of the potentiometer wire is: (in $\text{ V}$)

In the given potentiometer circuit,the length of the wire $AB$ is $3 \, m$ and its resistance is $R = 4.5 \, \Omega$. The length $AC$ for no deflection in the galvanometer is ............... $m$.