$A$ 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 $V/m$)

If the length of the potentiometer wire is increased by keeping the potential difference across the wire constant,then:

$A$ wire of length $10 \ cm$ is connected to a cell of $emf$ $2 \ V$ and negligible internal resistance. The resistance of the wire is $3 \ \Omega$. The value of the resistance required to obtain a potential gradient of $1 \ mV/cm$ is ................ $\Omega$.

$A$ cell balances against a length of $150 \ cm$ on a potentiometer wire when it is shunted by a resistance of $5 \ \Omega$. But when it is shunted by a resistance of $10 \ \Omega$,then the balancing length increases by $25 \ cm$. The balancing length when the cell is in an open circuit is: (in $cm$)

$A$ $10\,m$ 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.

In a potentiometer experiment,the null point is obtained on the $7^{\text{th}}$ wire for a given cell. To shift the null point to the $9^{\text{th}}$ wire for the same cell,what should we do?