In potentiometer experiments,two cells of e.m.f. $E_1$ and $E_2$ are connected in series $(E_1 > E_2)$,the balancing length is $64 \ cm$ of the wire. If the polarity of $E_2$ is reversed,the balancing length becomes $32 \ cm$. The ratio $E_1 / E_2$ is

In the experiment to determine the internal resistance $(r)$ of a cell $(E_1)$ using a potentiometer,the resistance drawn from the resistance box is $R$. The potential difference across the balancing length of the wire is equal to the terminal potential difference $(V)$ of the cell. The value of the internal resistance $(r)$ of the cell is:

$A$ 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 $V$?

$A$ cell of internal resistance $3 \, \Omega$ and $emf$ $10 \, V$ is connected to a uniform wire of length $500 \, cm$ and resistance $3 \, \Omega$. The potential gradient in the wire is .............. $mV/cm$.

$A$ potentiometer consists of two wires $AC$ and $CB$ of the same material and of equal lengths but diameters in the ratio $3 : 1$. Then the potential gradients on the two wires will be in the ratio :-

The adjoining figure shows the connections of a potentiometer experiment to determine the internal resistance of a Leclanché cell. When the cell is on open circuit,the balancing length of the potentiometer wire is $3.4 \, m$,and on closing the key $K_2$,the balancing length becomes $1.7 \, m$. If the resistance $R$ through which current is drawn is $10 \, \Omega$,then the internal resistance of the cell is .............. $\Omega$.