In a potentiometer experiment,when two cells of emfs $E_1$ and $E_2$ $(E_2 > E_1)$ are connected in series,the balancing length is $160 \ cm$. If one of the cells is reversed,the balancing length decreases by $75 \%$. If $E_1 = 1.2 \ V$,then $E_2 =$ (in $V$)

In the given potentiometer circuit,the length of the wire $AB$ is $3 \, m$ and its resistance is $R = 4.5 \, \Omega$. The length $AC$ for no deflection in the galvanometer is ............... $m$.

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

The length of a potentiometer wire is $4 \,m$ and is connected in series with an accumulator. The e.m.f. of a cell balances against $1.5 \,m$ length of the wire. If the length of the potentiometer wire is doubled,then the new balancing length of the wire will be: (in $\,m$)

$A$ potentiometer wire of length $L$ and a resistance $r$ are connected in series with a battery of e.m.f. $E_0$ and a resistance $r_1$. An unknown e.m.f. $E$ is balanced at a length $l$ of the potentiometer wire. The e.m.f. $E$ will be given by

In an experiment to find the $emf$ of a cell using a potentiometer,the length of the null point for a cell of $emf$ $1.5 \ V$ is found to be $60 \ cm$. If this cell is replaced by another cell of $emf$ $E$,the length of the null point increases by $40 \ cm$. The value of $E$ is $\frac{x}{10} \ V$. The value of $x$ is $............$