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 $64 \ cm$. If the polarity of $E_{2}$ is reversed,the balancing length becomes $32 \ cm$. The ratio $\frac{E_{1}}{E_{2}}$ is:

$A$ $2 \, V$ battery, a $990 \, \Omega$ resistor, and a potentiometer of $2 \, m$ length are connected in series. If the resistance of the potentiometer wire is $10 \, \Omega$, then the potential gradient of the potentiometer wire is: (in $V m^{-1}$)

$A$ potentiometer wire $PQ$ of $1\,m$ length is connected to a standard cell $E_{1}$. Another cell $E_{2}$ of emf $1.02\,V$ is connected with a resistance $r$ and switch $S$ (as shown in figure). With switch $S$ open,the null position is obtained at a distance of $49\,cm$ from $Q$. The potential gradient in the potentiometer wire is.......$V/cm$.

In a potentiometer experiment,the balancing length with a cell is $240 \ cm$. On shunting the cell with a resistance of $2 \ \Omega$,the balancing length becomes $120 \ cm$. The internal resistance of the cell is ................. $\Omega$.

$A$ potentiometer consists of a wire of length $4\, m$ and resistance $10\,\Omega$. It is connected to a cell of $e.m.f.$ $2\, V$. The potential difference per unit length of the wire will be ............. $V/m$.

In the given figure,there is a circuit of a potentiometer of length $AB = 10 \, m$. The resistance per unit length is $0.1 \, \Omega/cm$. $A$ battery of $6 \, V$ and an internal resistance of $20 \, \Omega$ is connected across $AB$. The maximum value of emf that can be measured by this potentiometer is (in $V$):