In an experiment to find the emf of a cell using a potentiometer, the length of the null point for a cell of emf $1.5 \text{ V}$ is found to be $60 \text{ cm}$. If this cell is replaced by another cell of emf $E$, the length of the null point increases by $40 \text{ cm}$. The value of $E$ is $x/10 \text{ V}$. 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:

$A$ potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery,used across the potentiometer wire,has an $EMF$ of $2.0\,V$ and a negligible internal resistance. The potentiometer wire itself is $4\,m$ long. When the resistance $R$,connected across the given cell,has values of $(i)$ infinity and $(ii)$ $9.5\,\Omega$,the balancing lengths on the potentiometer wire are found to be $3\,m$ and $2.85\,m$,respectively. The value of internal resistance of the cell is ............... $\Omega$.

In the primary circuit of a potentiometer,the current is $0.2 \ A$. The resistivity and cross-sectional area of the potentiometer wire are $4 \times 10^{-7} \ \Omega \cdot m$ and $8 \times 10^{-7} \ m^2$ respectively. The potential gradient will be ......... $V/m$.

Define potential gradient and write its $SI$ unit.

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