To determine the internal resistance of a cell by using a potentiometer,the null point is at $1 \ m$,when shunted by $3 \ \Omega$ resistance and at a length $1.5 \ m$,when cell is shunted by $6 \ \Omega$ resistance. The internal resistance of the cell is (in $\Omega$)

To measure the internal resistance of a battery,a potentiometer is used. For $R = 10 \ \Omega$,the balance point is observed at $\ell = 500 \ cm$ and for $R = 1 \ \Omega$,the balance point is observed at $\ell = 400 \ cm$. The internal resistance of the battery is approximately: (in $\Omega$)

In the circuit shown, a four-wire potentiometer is made of a $400\, cm$ long wire, which extends between $A$ and $B$. The resistance per unit length of the potentiometer wire is $r = 0.01\, \Omega /cm$. If an ideal voltmeter is connected as shown with jockey $J$ at $50\, cm$ from end $A$, the expected reading of the voltmeter will be: ............... $V$

In a potentiometer experiment,the balancing length with a cell is $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is ................. $\Omega$.

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?

$A$ potentiometer wire of length $300\,cm$ is connected in series with a resistance $780\,\Omega$ and a standard cell of emf $4\,V$. $A$ constant current flows through the potentiometer wire. The length of the null point for a cell of emf $20\,mV$ is found to be $60\,cm$. The resistance of the potentiometer wire is ... $\Omega$.