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

$A$ potentiometer wire of length $4 \text{ m}$ and resistance $5 \Omega$ is connected in series with a resistance of $992 \Omega$ and a cell of e.m.f. $4 \text{ V}$ with internal resistance $3 \Omega$. The length of $0.75 \text{ m}$ on the potentiometer wire balances the e.m.f. of: (in $\text{ mV}$)

In a potentiometer arrangement,a cell of emf $1.5 \ V$ gives a balance point at $150 \ cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $210 \ cm$,what is the emf of the second cell (in $V$)?

In a potentiometer experiment, a 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 by a few centimeters without changing the cell or the driving source, the balancing length will:

$A$ 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 ............. $\Omega$.

$A$ cell is connected to a potentiometer, and the balance point is obtained at a length of $2 \, m$. When a resistance of $5 \, \Omega$ is connected in parallel with the cell, the balance point is obtained at a length of $3 \, m$. What is the internal resistance of the cell in $\Omega$?