In a potentiometer experiment,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 $\frac{E_1}{E_2}$ is: (in $: 1$)

In a potentiometer (see figure) a balance is obtained at a length of $400 \ mm$ when using a known battery of emf $1.6 \ V$. After removing this battery,another battery of unknown emf is used and balance is obtained at a length of $650 \ mm.$ The emf of the unknown battery is ............. $V$. (in $V$)

The null point of a potentiometer with a cell of emf $\varepsilon$ is obtained at a distance $l$ on the wire,then

In a potentiometer experiment, the balancing length with a cell is $560 \, cm$. When an external resistance of $10 \, \Omega$ is connected in parallel to the cell, the balancing length changes by $60 \, cm$. Find the internal resistance of the cell.

$A$ wire of length $100\, cm$ is connected to a cell of $emf$ $2\, V$ and negligible internal resistance. The resistance of the wire is $3\, \Omega$. The additional resistance required to produce a potential drop of $1\, mV/cm$ is ............... $\Omega$.

To determine the internal resistance of a cell by using a potentiometer, the null point is at $1 \, m$ when the cell is shunted by $3 \, \Omega$ resistance and at a length $1.5 \, m$ when the cell is shunted by $6 \, \Omega$ resistance. The internal resistance of the cell is: