Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

Two cells of emf $E_1$ and $E_2$ $(E_1 > E_2)$ are connected individually to a potentiometer and their corresponding balancing lengths are $625 \, cm$ and $500 \, cm$. Then the ratio $\frac{E_1}{E_2}$ is ...........

The null point of a potentiometer with a cell of emf $\varepsilon$ is obtained at a distance $l$ on the wire,then

$A$ cell,shunted by an $8 \; \Omega$ resistance,is balanced across a potentiometer wire of length $3 \; m$. The balancing length is $2 \; m$ when the cell is shunted by a $4 \; \Omega$ resistance. The value of internal resistance of the cell will be $\dots \; \Omega$.

When a potentiometer is connected between the points $A$ and $B$ as shown in the circuit,the balance point is obtained at $64 \ cm$. When it is connected between $A$ and $C$,the balance point is $8 \ cm$. If the potentiometer is connected between $B$ and $C$,the balance point will be: (in $cm$)

$A$ potentiometer wire $AB$ having length $L$ and resistance $12r$ is joined to a cell $D$ of $emf$ $\varepsilon$ and internal resistance $r$. $A$ cell $C$ having $emf$ $\varepsilon/2$ and internal resistance $3r$ is connected as shown in the figure. The length $AJ$ at which the galvanometer shows no deflection is