The material of the wire of a potentiometer is

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:

$A$ potentiometer wire of length $10 \,m$ and resistance $20 \,\Omega$ is connected in series with a $25 \,V$ battery and an external resistance $30 \,\Omega$. $A$ cell of emf $E$ in the secondary circuit is balanced by a $250 \,cm$ long potentiometer wire. The value of $E$ (in volt) is $\frac{x}{10}$. The value of $x$ is.......

Two cells $A$ and $B$ are connected in the secondary circuit of a potentiometer one at a time,and the balancing lengths are $400 \ cm$ and $440 \ cm$ respectively. The emf of cell $A$ is $1.08 \ V$. The emf of the second cell $B$ in volts is:

In a potentiometer circuit, there is a cell of $e.m.f.$ $2\, V$, a resistance of $5\, \Omega$ and a wire of uniform thickness of length $1000\, cm$ and resistance $15\, \Omega$. The potential gradient in the wire is:

$A$ potentiometer wire has length $4\, m$ and resistance $8\, \Omega$. The resistance that must be connected in series with the wire and an accumulator of e.m.f. $2\, V$,so as to get a potential gradient of $1\, mV$ per $cm$ on the wire is ............. $\Omega$.