In a potentiometer arrangement,a cell of emf $1.5 \ V$ gives a balance point at $150 \ cm$ length of the wire. If the cell is replaced by another cell and the balance point shifts to $210 \ cm$,what is the emf of the second cell (in $V$)?

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 $\Omega$)

Two cells of e.m.f. $E_1$ and $E_2$ $(E_1 > E_2)$ are connected as shown in the figure. When a potentiometer is connected between points $A$ and $B$, the balancing length of the potentiometer wire is $412 \text{ cm}$. When the same potentiometer is connected between points $A$ and $C$, the balancing length is $103 \text{ cm}$. The ratio $E_1 : E_2$ is:

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

$A$ potentiometer wire is $100 \, cm$ long and a constant potential difference is maintained across it. Two cells are connected in series first to support one another and then in opposite direction. The balance points are obtained at $50 \, cm$ and $10 \, cm$ from the positive end of the wire in the two cases. The ratio of emfs is:

$A$ $10 \ m$ long wire of resistance $20 \ \Omega$ is connected in series with a battery of e.m.f. $3 \ V$ and a resistance of $10 \ \Omega$. The potential gradient along the wire in $V/m$ is