(a) Working principle of potentiometer: The Potentiometer is an electric instrument that used to measure the potential difference of a given cell and the internal resistance of a cell. It is also used to compare EMFs of different cells.

The potentiometer consists of a long resistive wire $\mathrm{L}$ having uniform cross-section and a battery of known emf $\mathrm{V}$. This voltage is called as driver cell voltage. Two ends of the resistive wire $\mathrm{L}$ is connected to the battery terminals. This is a primary circuit arrangement. One terminal of another cell (whose emf $\mathrm{E}$ is to be measured) is at one end of the primary circuit and another end of the cell terminal is connected to any point on the resistive wire using a jockey $\mathrm{J}$ and a galvanometer $\mathrm{G}$. This arrangement is a secondary circuit.

The basic working principle of this circuit is based on the fact that the potential drop across any portion of the wire is directly proportional to the length of the wire.
If the current through the potentiometer is I and $\mathrm{R}$ is the total resistance of the potentiometer, then $\mathrm{V}=\mathrm{IR}$.
The resistivity $\mathit{\rho }$ and area of cross-section $\mathit{A}$ are constant. Also, the Current $\mathrm{I}$ is kept constant using the rheostat.
$\mathrm{R}\propto \frac{\mathrm{L}}{\mathrm{A}}$

Now, the jockey is so adjusted that for length $\mathit{X}$ of the wire there is no current flow through the galvanometer.
So, $\mathit{E}=\mathit{K}\mathit{X}$
$\mathrm{K}$ is known, $\mathrm{X}$ is measurable by a scale. So, $\mathrm{E}$ can be evaluated from the equation.
When EMF of two cells are to be considered, then in the above similar way
${\mathrm{E}}_1={\mathrm{KX}}_1$
${\mathrm{E}}_2={\mathrm{KX}}_2$
$\therefore \frac{{\mathrm{E}}_1}{{\mathrm{E}}_2}=\frac{{\mathrm{L}}_1}{{\mathrm{L}}_2}$
(b) Potential gradient $=\mathit{k}=\frac{5\mathrm{V}}{10\mathrm{m}}=0.5\mathrm{V}∕\mathrm{m}$
When ${\mathit{E}}_1$ and ${\mathit{E}}_2$ are joined with same polarity, then
${\mathrm{E}}_1-{\mathrm{E}}_2=\mathit{k}\times \frac{700}{100}=3.5\mathrm{V}$ .......(1)
When ${\mathit{E}}_1$ and ${\mathit{E}}_2$ are joined with opposite polarity, then
${\mathrm{E}}_1+{\mathrm{E}}_2=\mathit{k}\times \frac{100}{100}=7\mathrm{V}$ ......(2)
Adding eqns (1) and (2)
$2{\mathrm{E}}_1=10.5\mathrm{V}$
$\therefore {\mathrm{E}}_1=5.25\mathrm{V}$
Subtracting eqn (2) from eqn (1)
$2{\mathrm{E}}_2=3.5\mathrm{V}$
$\therefore {\mathrm{E}}_2=1.75\mathrm{V}$