The resistivity of a potentiometer wire is $40 \times 10^{-8} \, \Omega \cdot m$ and its area of cross-section is $8 \times 10^{-6} \, m^2$. If $0.2 \, A$ current is flowing through the wire,the potential gradient will be:

$AB$ is a potentiometer wire of length $100\, cm$ and its resistance is $10\,\Omega$. It is connected in series with a resistance $R = 40\,\Omega$ and a battery of $e.m.f.$ $2\,V$ and negligible internal resistance. If a source of unknown $e.m.f.$ $E$ is balanced by $40\, cm$ length of the potentiometer wire,the value of $E$ is ................. $V$. (in $,V$)

$A$ battery of $6\,V$ is connected to the terminals of a $3\,m$ long wire of uniform thickness and resistance of the order of $100\,\Omega$. The difference of potential between two points separated by $50\,cm$ on the wire will be .......... $V$.

The potentiometer wire is $5 \ m$ long and a potential difference of $4 \ V$ is maintained between the ends. The e.m.f. of the cell which balances against a length of $200 \ cm$ of the potentiometer wire is: (in $V$)

The circuit shown here is used to compare the $e.m.f.$ of two cells ${E_1}$ and ${E_2}$ $(E_1 > E_2)$. The null point is at $C$ when the galvanometer is connected to ${E_1}$. When the galvanometer is connected to ${E_2}$,the null point will be

$A$ potentiometer balances at $44 \ cm$ when a cell of internal resistance $1 \ \Omega$ is in the secondary circuit. To obtain the balancing point at $40 \ cm$,the resistance to be connected in parallel to the cell is: (in $\Omega$)