$A$ potentiometer wire is $4 \text{ m}$ long and a potential difference of $2 \text{ V}$ is maintained between its ends. The e.m.f. of the cell which balances against a length of $80 \text{ cm}$ of the potentiometer wire is: (in $\text{ V}$)

In a potentiometer wire experiment, the $emf$ of a battery in the primary circuit is $20\,V$ and its internal resistance is $5\,\Omega$. There is a resistance box in series with the battery and the potentiometer wire, whose resistance can be varied from $120\,\Omega$ to $170\,\Omega$. The resistance of the potentiometer wire is $75\,\Omega$. Which of the following potential differences can be measured using this potentiometer?

$A$ $10 \, m$ long wire of $20 \, \Omega$ resistance is connected with a battery of $3 \, V$ $e.m.f.$ (negligible internal resistance) and a $10 \, \Omega$ resistance is joined to it in series. The potential gradient along the wire in $V/m$ is:

In the experiment of a potentiometer,at balance,there is no current in the

$A$ potentiometer wire has length $10\, m$ and resistance $10\, \Omega$. It is connected to a battery of $EMF$ $11\, V$ and internal resistance $1\, \Omega$. The potential gradient in the wire is ............... $V/m$.

It is preferable to measure the $e.m.f.$ of a cell by a potentiometer rather than by a voltmeter because of the following possible reasons.
$(i)$ In the case of a potentiometer,no current flows through the cell.
$(ii)$ The length of the potentiometer wire allows for greater precision.
$(iii)$ Measurement by the potentiometer is quicker.
$(iv)$ The sensitivity of the galvanometer,when using a potentiometer,is not relevant.
Which of these reasons are correct?