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?

In the following circuit, a $10 \, m$ long potentiometer wire with resistance $1.2 \, \Omega/m$, a resistance $R_1$, and an accumulator of $emf$ $2 \, V$ are connected in series. When the $emf$ of the thermocouple is $2.4 \, mV$, the deflection in the galvanometer is zero at a balancing length of $5 \, m$. The current supplied by the accumulator 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 ...........

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 $cm$)

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:

$A$ potentiometer having a potential gradient of $2\, mV/cm$ is used to measure the potential difference across a resistance of $10\, \Omega$. If a length of $50\, cm$ of the potentiometer wire is required to obtain the null point,the current passing through the $10\, \Omega$ resistor is (in $mA$):