A battery of e.m.f. $2 \mathrm{V}$ and internal resistance $2 \mathrm{\Omega }$ is connected to an external resistance $8 \mathrm{\Omega }$. If the length of the conductor is $4 \mathrm{m}$, then potential gradient between the two ends of the wire is

A resistance of $990 \mathrm{\Omega }$ and a cell of emf $2 \mathrm{V}$ is connected in series with a potentiometer wire having a length $2 \mathrm{m}$ and resistance $10 \mathrm{\Omega }$. The potential gradient along the wire will be

An accumulator of $5 \mathrm{V}$ is connected through a resistance of $40 \mathrm{\Omega }$ to a potentiometer wire $10 \text{m}$ long and of resistance $10 \mathrm{\Omega }$. For a cell, the null point is found at a length of $8 \mathrm{m}$ from the common terminal. The current through the wire is

The resistance per unit length of a wire is $1 \mathrm{\Omega } {\mathrm{m}}^{-1}$. A Leclanche cell of e.m.f. $1.45 \mathrm{V}$ is balanced against $2.9 \mathrm{m}$ length of potentiometer wire. The current through the wire is

The e.m.f. $E$ of the battery is balanced by potential difference across $75 \mathrm{cm}$ of a potentiometer wire. For a standard cell of e.m.f. $1.02 \mathrm{V}$, the balancing length is $50 \mathrm{cm}.$ The value of $E$ is

A potentiometer wire of length $100 \mathrm{cm}$ has a resistance of $10 \mathrm{\Omega }$. It is connected in series with a resistance and an accumulator of e.m.f. $2$ $\mathrm{V}$ and negligible internal resistance. A source of e.m.f. $10 \mathrm{mV}$ is balanced against a $40 \mathrm{cm}$ length of the potentiometer wire. The value of the external resistance is

When two cells of e.m.f. $1.5 \mathrm{V}$ and $1.1 \mathrm{V}$ connected in series are balanced on a potentiometer, the balancing length is $260 \mathrm{cm}$. The balancing length, when they are connected in opposition is (in $\mathrm{cm}$ )