$A$ 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 $V$?

When a cell of $E.M.F.$ $E_1$ is connected to a potentiometer wire,the balancing length is $l_1$. Another cell of $E.M.F.$ $E_2$ $(E_1 > E_2)$ is connected such that the two cells oppose each other,then the balancing length is $l_2$. The ratio $E_1 : E_2$ is

$A$ potentiometer wire of length $L$ and a resistance $r$ are connected in series with a battery of e.m.f. $E_0$ and a resistance $r_1$. An unknown e.m.f. $E$ is balanced at a length $l$ of the potentiometer wire. The e.m.f. $E$ will be given by

$A$ $2 \, V$ 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 $V m^{-1}$)

To determine the internal resistance of a cell with a potentiometer,when the cell is shunted by a resistance of $5 \Omega$,the balancing length is $250 \ cm$. When the cell is shunted by $20 \Omega$,the balancing length of the potentiometer wire is $400 \ cm$. The internal resistance of the cell is: (in $\Omega$)

The material of the wire of a potentiometer is