When two cells are connected in series in a potentiometer circuit to assist each other,the balancing length is $6 \ m$. When they are connected in series to oppose each other,the balancing length is $2 \ m$. What is the ratio of the $EMF$ of the two cells?

The length of a potentiometer wire is $l$. $A$ cell of emf $E$ is balanced at a length $\left(\frac{l}{3}\right)$ from the positive end of the wire. If the length of the wire is increased by $\left(\frac{l}{2}\right)$,the distance at which the same cell gives the balancing point is (The cell in the primary circuit is ideal and no series resistance is present in the primary circuit.)

$A$ 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

For the arrangement of the potentiometer shown in the figure,the balance point is obtained at a distance $75\,cm$ from $A$ when the key $k$ is open. The second balance point is obtained at $60\,cm$ from $A$ when the key $k$ is closed. Find the internal resistance (in $\Omega$) of the battery $E_1$.

$A$ resistance of $4\,\Omega$ and a wire of length $5\,m$ and resistance $5\,\Omega$ are joined in series and connected to a cell of $e.m.f.$ $10\,V$ and internal resistance $1\,\Omega$. $A$ parallel combination of two identical cells is balanced across $300\,cm$ of the wire. The $e.m.f.$ $E$ of each cell is ........... $V$.

In a potentiometer experiment,a wire of length $10 \ m$ and resistance $5 \ \Omega$ is connected to a cell of emf $2.2 \ V$. If the potential difference between two points separated by a distance of $660 \ cm$ on the potentiometer wire is $1.1 \ V$,then the internal resistance of the cell is: (in $Omega$)