The plot of the variation of potential difference across a combination of three identical cells in series, versus current is shown alongside. What is the emf and internal resistance of each cell?

Hint: EMF or the electromotive force is the energy which gets transferred. It is similar to the potential difference and is often used interchangeably. Also, resistance is the property of the material to resist the flow of current through it.
Formula used: $E_{s}=E_{1}+E_{2}+E_{3}$, $R_{s}=R_{1}+R_{2}+R_{3}$and $E=IR$

Complete step-by-step solution:
Given the combination of three identical cells in series. In a series connection, the potential of the combination is the sum of the potentials due to the individual resistances. Or we can say that potentially gets distributed in the series connection of resistances. Similarly, the combination of resistance is the sum of the individual resistances.
Let the emf of the series combination be $E_{s}$ and the individual EMFs be $E_{1}=E_{2}=E_{3}=E$
Then we can say that, $E_{s}=E_{1}+E_{2}+E_{3}=3E$
From the graph, it is clear that, $E_{s}=6V$
Then reducing, we get $6=3E$ or $E=2V$
We know from ohms law that, $E=IR$ where $E$ is the emf, $I$ is the current in the circuit and $R$ is the resistance of combination.
From the graph, we get $I=1A$ for the combination.
Then, we get $R_{s}=\dfrac{E_{s}}{I}=\dfrac{6}{1}=6\Omega$
We know that in series connection, if the resistance of the series combination be $R_{s}$ and the individual resistances be $R_{1}=R_{2}=R_{3}=R$, then we get
$R_{s}=R_{1}+R_{2}+R_{3}=3R$
Then reducing, we get $3R=6$ or $R=2\Omega$
Thus for the given combination, we get,$R=2\Omega$ and $E=2V$.

Note: To read the graph, we must understand a few basics. Like here, the x-axis denotes the current flow. The maximum current flowing through the circuit is $I=1A$. Similarly, the y-axis denotes the potential drop in the circuit. The maximum potential drop possible is given as $E_{s}=6V$. We can then use this to solve the question.