$A$ potentiometer is an ideal device for measuring potential difference because

$A$ potentiometer wire has a length of $4 \ m$ and a resistance of $10 \ \Omega$. It is connected to a cell of $2 \ V$ emf. The potential gradient (potential difference per unit length) of the wire is: (in $V/m$)

In a potentiometer experiment,it is found that no current passes through the galvanometer when the terminals of the cell are connected across $52 \ cm$ of the potentiometer wire. If the cell is shunted by a resistance of $5 \ \Omega$,a balance is found when the cell is connected across $40 \ cm$ of the wire. Find the internal resistance of the cell ........... $\Omega$.

The length of a potentiometer wire is $4 \,m$ and is connected in series with an accumulator. The e.m.f. of a cell balances against $1.5 \,m$ length of the wire. If the length of the potentiometer wire is doubled,then the new balancing length of the wire will be: (in $\,m$)

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

In the primary circuit of a potentiometer,the current is $0.2 \ A$. The resistivity and cross-sectional area of the potentiometer wire are $4 \times 10^{-7} \ \Omega \cdot m$ and $8 \times 10^{-7} \ m^2$ respectively. The potential gradient will be ......... $V/m$.