$A$ 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 $\text{ V}$)

$A$ wire of length $100\, cm$ is connected to a cell of emf $2\, V$ and negligible internal resistance. The resistance of the wire is $3\, \Omega$. The additional resistance required to produce a potential difference of $1\, mV/cm$ is ............. $\Omega$.

$A$ null point is found at $200\,cm$ in a potentiometer when the cell in the secondary circuit is shunted by $5\,\Omega$. When a resistance of $15\,\Omega$ is used for shunting,the null point moves to $300\,cm$. The internal resistance of the cell is $..............\,\Omega$.

In a potentiometer experiment, the null point is obtained at a particular point for a cell on a potentiometer wire of length $L$. If the length of the potentiometer wire is increased without changing the cell or the driving source, the balancing length will:

$A$ $10\,m$ 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.

$A$ potentiometer wire has a length of $5 \, m$ and a resistance of $5 \, \Omega$. If the null point is obtained at $300 \, cm$,what is the $emf$ $E$ of the cells (connected in parallel) in $V$?