In a potentiometer experiment,cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$ and the balancing length is $80 \ cm$. If the polarity of $E_2$ is reversed,the balancing length becomes $20 \ cm$. The ratio $E_1 / E_2$ is:

The length of a potentiometer wire is $L$. $A$ cell of e.m.f. $E$ is balanced at a length $\frac{L}{4}$ from the positive end of the wire. If the length of the original wire is increased by $\frac{L}{3}$,then using the same cell,the null point is obtained at:

$A$ potentiometer wire has a length of $5 \, m$ and a resistance of $5 \, \Omega$. If the null point is obtained at $300 \, cm$,what is the $emf$ $E$ of the cells (connected in parallel) in $V$?

In the potentiometer experiment shown in the figure,for the position $X$ of the jockey $J$,there occurs a null deflection in the galvanometer. Then the potential difference between points $A$ and $X$ is ................ $V$.

$A$ resistance of $R \; \Omega$ draws current from a potentiometer. The potentiometer has a total resistance $R_{0} \; \Omega$ (Figure). $A$ voltage $V$ is supplied to the potentiometer. Derive an expression for the voltage across $R$ when the sliding contact is in the middle of the potentiometer.

$A$ voltmeter reads the potential difference across the terminals of an old battery as $1.2 \, V$,while a potentiometer reads $1.4 \, V$. The internal resistance of the battery is $40 \, \Omega$. What is the resistance of the voltmeter in $\Omega$?