$A$ $2.5 \ V$ battery is connected to a potentiometer wire. $A$ cell of e.m.f. $1.08 \ V$ is balanced by the voltage drop across $2.16 \ m$ of wire. The length of the potentiometer wire is (in $m$)

Potentiometer measures the potential difference more accurately than a voltmeter because

Two cells having unknown e.m.f.s $E_{1}$ and $E_{2}$ $(E_{1} > E_{2})$ are connected in a potentiometer circuit so as to assist each other. The null point is obtained at $490 \ cm$ from the higher potential end. When cell $E_{2}$ is connected so as to oppose cell $E_{1}$,the null point is obtained at $90 \ cm$ from the same end. The ratio of the e.m.f.s of the two cells $(\frac{E_{1}}{E_{2}})$ is:

In potentiometer experiments,two cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$,the balancing length is $64 \ cm$ of the wire. If the polarity of $E_2$ is reversed,the balancing length becomes $32 \ cm$. The ratio $E_1 / E_2$ is

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 $\Omega$)

In a potentiometer experiment for the determination of the internal resistance of a cell,when an external resistance of $R$ is connected in parallel to the cell,the balancing length decreases by $10 \%$. The internal resistance of the cell is