$A$ potentiometer is used to measure the potential difference between $A$ and $B$. The null point is obtained at $0.9 \ m$. Now,the potential difference between $A$ and $C$ is measured,and the null point is obtained at $0.3 \ m$. The ratio $\frac{E_2}{E_1}$ is $\left(E_1 > E_2\right)$.

Two cells,when connected in series,are balanced on $8 \; m$ on a potentiometer. If the cells are connected such that the polarity of one of the cells is reversed,they balance on $2 \; m$. The ratio of the $e.m.f.$'s of the two cells is:

In a potentiometer experiment,the balancing length for a cell is $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes half the initial balancing length. The internal resistance of the cell is: (in $\Omega$)

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:

The length of a wire of a potentiometer is $100 \, cm$,and the $emf$ of its standard cell is $E \, volt$. It is employed to measure the $emf$ of a battery whose internal resistance is $0.5 \, \Omega$. If the balance point is obtained at $l = 30 \, cm$ from the positive end,the $emf$ of the battery is (where $i$ is the current in the potentiometer wire).

$A$ potentiometer wire of length $1 \, m$ is connected in series with a $490 \, \Omega$ resistance and a $2 \, V$ battery. If the potential gradient is $0.2 \, mV/cm$,then the resistance of the potentiometer wire is ................ $\Omega$. (in $.9$)