$A$ potentiometer is used for the comparison of $e.m.f.$ of two cells $E_1$ and $E_2$. For cell $E_1$,the null point is obtained at $20 \ cm$ and for $E_2$,the null point is obtained at $30 \ cm$. The ratio of their $e.m.f.$s will be:

In a potentiometer, the area of cross-section of the wire is $4 \, cm^2$, the current flowing in the circuit is $1 \, A$ and the potential gradient is $7.5 \, V/m$, then the resistivity of the potentiometer wire is

In a potentiometer experiment,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 $\frac{E_1}{E_2}$ is: (in $: 1$)

$A$ $2.5 \ V$ 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 $m$)

The length of a potentiometer wire is $\ell$. $A$ cell of emf $E$ is balanced at a length $\ell/3$ from the positive end of the wire. If the length of the wire is increased by $\ell/2$,at what distance will the same cell give a balanced point?

As shown in the figure,a potentiometer wire of resistance $20\,\Omega$ and length $300\,cm$ is connected with a resistance box ($R$.$B$.) and a standard cell of emf $4\,V$. For a resistance '$R$' of the resistance box introduced into the circuit,the null point for a cell of $20\,mV$ is found to be $60\,cm$. The value of '$R$' is $.....\Omega$