In a potentiometer experiment,the balancing length is $8 \ m$ when two cells $E_1$ and $E_2$ are joined in series. When two cells are connected in opposition,the balancing length is $4 \ m$. The ratio of the e.m.f. of the two cells $\left(\frac{E_1}{E_2}\right)$ is

$A$ potentiometer $PQ$ is set up to compare two resistances as shown in the figure. The ammeter $A$ in the circuit reads $1.0\, A$ when the two-way key $K_3$ is open. The balance point is at a length $l_1\, cm$ from $P$ when the two-way key $K_3$ is plugged in between $2$ and $1$,while the balance point is at a length $l_2\, cm$ from $P$ when key $K_3$ is plugged in between $3$ and $1$. The ratio of two resistances $\frac{R_1}{R_2}$ is found to be

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 $V$):

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 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 .............. $\Omega$.

$A$ $10 \ m$ long wire of resistance $20 \ \Omega$ is connected in series with a battery of e.m.f. $3 \ V$ and a resistance of $10 \ \Omega$. The potential gradient along the wire in $V/m$ is